4: Probability and Probability Distributions



10: Inference from Small Samples

10.1 Refer to Table 4, Appendix I, indexing df along the left or right margin and [pic]across the top.

a [pic]with 5 df b [pic]with 8 df

c [pic]with 18 df c [pic]with 30 df

10.2 The value [pic]is the tabled entry for a particular number of degrees of freedom.

a For a two-tailed test with [pic], the critical value for the rejection region cuts off [pic]in the two tails of the t distribution shown below, so that [pic]. The null hypothesis H0 will be rejected if [pic]or [pic](which you can also write as [pic]).

[pic]

b For a right-tailed test, the critical value that separates the rejection and nonrejection regions for a right tailed test based on a t-statistic will be a value of t (called [pic]) such that [pic]and [pic]. That is, [pic]. The null hypothesis H0 will be rejected if [pic].

c For a two-tailed test with [pic]and [pic], H0 will be rejected if [pic].

d For a left-tailed test with[pic]and[pic], H0 will be rejected if [pic].

10.3 a The p-value for a two-tailed test is defined as

[pic]

so that

[pic]

Refer to Table 4, Appendix I, with [pic]. The exact probability, [pic]is unavailable; however, it is evident that [pic]falls between [pic]and[pic]. Therefore, the area to the right of [pic]must be between .01 and .025. Since

[pic]

the p-value can be approximated as

[pic]

b For a right-tailed test, [pic]with [pic]. Since the value [pic]is larger than [pic], the area to its right must be less than .005 and you can bound the p-value as

[pic]

c For a two-tailed test, [pic], so that [pic]. From Table 4 with [pic], [pic]is smaller than [pic]so that

[pic]

d For a left-tailed test, [pic]with [pic]. Since the value [pic]is larger than [pic], the area to its right must be less than .005 and you can bound the p-value as

[pic]

10.4 a The stem and leaf plot is shown below. Notice the mounded shape of the data, which justifies the normality assumption.

Stem-and-Leaf Display: Scores

Stem-and-leaf of Scores N = 20

Leaf Unit = 1.0

1 5 7

2 6 2

5 6 578

8 7 123

(4) 7 5567

8 8 244

5 8 669

2 9 13

b Using the formulas given in Chapter 2 or your scientific calculator, calculate

[pic]

[pic]

c Small sample confidence intervals are quite similar to their large sample counterparts; however, these intervals must be based on the t-distribution. Thus, the confidence interval for the single population mean described in this exercise will be

[pic]

where [pic]is a value of t (Table 4) based on [pic]degrees of freedom that has area [pic]to its right. For this exercise, [pic]and [pic]with [pic]degrees of freedom is [pic]. Hence the 95% confidence interval is

[pic]

or [pic]. Intervals constructed using this procedure will enclose [pic]95% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses[pic].

10.5 a Using the formulas given in Chapter 2, calculate [pic]and [pic]. Then

[pic]

[pic]

b With [pic], the appropriate value of t is [pic](from Table 4) and the 99% upper one-sided confidence bound is

[pic]

or [pic]. Intervals constructed using this procedure will enclose [pic]99% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses[pic].

c The hypothesis to be tested is

[pic]

and the test statistic is

[pic]

The rejection region with [pic]and [pic]degrees of freedom is located in the lower tail of the t-distribution and is found from Table 4 as [pic]. Since the observed value of the test statistic falls in the rejection region, H0 is rejected and we conclude that [pic]is less than 7.5.

d Notice that the 99% upper one-sided confidence bound for [pic]does not include the value [pic]. This would confirm the results of the hypothesis test in part c, in which we concluded that [pic]is less than 7.5.

10.6 a Using the formulas given in Chapter 2, calculate [pic]and [pic]. Then

[pic]

[pic]

With [pic], the appropriate value of t is [pic](from Table 4) and the 95% confidence interval is

[pic]

or [pic]. Intervals constructed using this procedure will enclose [pic]95% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses[pic].

b Calculate [pic]and [pic]

and the 95% confidence interval is

[pic] or [pic].

The interval is narrower than the interval in part a, even though the sample size is smaller, because the data is so much less variable.

c For white tuna in water,

[pic]and [pic]

and the 95% confidence interval is

[pic] or [pic].

For light tuna in oil,

[pic]and [pic]

and the 95% confidence interval is

[pic] or [pic].

[pic]

The plot of the four treatment means shows substantial differences in variability. The cost of light tuna in water appears to be the lowest, and quite difference from either of the white tuna varieties.

10.7 Similar to previous exercises. The hypothesis to be tested is

[pic]

Calculate [pic]

[pic]

The test statistic is

[pic]

The critical value of t with [pic]and [pic]degrees of freedom is [pic]and the rejection region is [pic]. Since the observed value does not fall in the rejection region, H0 is not rejected. There is no evidence to indicate that the dissolved oxygen content is less than 5 parts per million.

10.8 Calculate [pic]

[pic]

The 95% confidence interval based on [pic]is

[pic]

or [pic].

10.9 a Similar to previous exercises. The hypothesis to be tested is

[pic]

Calculate [pic]

[pic]

The test statistic is

[pic]

The critical value of t with [pic]and [pic]degrees of freedom is [pic]and the rejection region is [pic]. The null hypothesis is rejected and we conclude that [pic]is less than 100 DL.

b The 95% upper one-sided confidence bound, based on [pic]degrees of freedom, is

[pic]

This confirms the results of part a in which we concluded that the mean is less than 100 DL.

10.10 a Notice that the stem and leaf plot is fairly mound-shaped, with the peak very close to the center and with no gaps. For a small sample of size [pic], the data appear to be approximately mound-shaped,so that we have no reason to conclude that there is a problem with the normality assumption.

b Calculate [pic]

[pic]

c The 95% confidence interval with [pic]is

[pic]

or [pic].

10.11 Calculate [pic]

[pic]

The 95% confidence interval based on [pic]is

[pic]

or [pic].

10.12 The problem of selecting a proper sample size to achieve a given bound is now complicated by the fact that the t value, which is used in calculating the correct sample size, changes as the value of n changes. In Chapter 8, the procedure was to choose n so that the half-width of the [pic]confidence interval was less than some given bound, B. That is,

[pic]

Now, the inequality to be solved is

[pic]

and the [pic]value must be based on [pic]degrees of freedom. Since n is unknown, the procedure is as follows:

1. Calculate n using [pic]instead of [pic]. If the value for n is large (that is, [pic]), this sample size will be valid.

2. If the value for n is small, we are not justified in using the value [pic]. This value must be replaced by the appropriate t-value with [pic]degrees of freedom. If the inequality holds, the sample size is valid; if not, it is necessary to pick larger values of n until the inequality will hold. This repetitive procedure is usually not necessary, because the [pic]value will usually yield a satisfactory approximation to the required sample size.

In this exercise, we want to estimate [pic]to within .06 with probability .95. Hence, the following inequality must hold:

[pic]

Consider the sample size obtained by replacing [pic]with [pic].

[pic]

Since this sample size is greater than 30, the sample size [pic]is valid.

10.13 a The hypothesis to be tested is

[pic]

The test statistic is

[pic]

The critical value of t with [pic]and [pic]degrees of freedom is [pic]and the rejection region is [pic]. Since the observed value does falls in the rejection region, H0 is rejected, and we conclude that pre-treatment mean is less than 25.

b The 95% confidence interval based on [pic]is

[pic]

or [pic].

c The pre-treatment mean looks considerably smaller than the other two means.

10.14 Calculate [pic]

[pic]

The 95% confidence interval based on [pic]is

[pic]

or [pic].

10.15 a The t test of the hypothesis

[pic]

is not significant, since the [pic] associated with the test statistic

[pic]

is greater than .10. There is insufficient evidence to indicate that the mean weight per package is different from one-pound.

b In fact, the 95% confidence limits for the average weight per package are

[pic]

or [pic]. These values agree (except in the last decimal place) with those given in the printout. Remember that you used the rounded values of [pic]and s from the printout, causing a small rounding error in the results.

10.16 a Answers will vary. A typical histogram generated by Minitab shows that the data are approximately mound-shaped.

[pic]

b Calculate [pic]

[pic]

Table 4 does not give a value of t with area .025 to its right. If we are conservative, and use the value of t with df = 29, the value of t will be [pic], and the approximate 95% confidence interval is

[pic]

or [pic].

10.17 Refer to Exercise 10.16. If we use the large sample method of Chapter 8, the large sample confidence interval is

[pic]

or [pic]. The intervals are fairly similar, which is why we choose to approximate the sampling distribution of [pic] with a z distribution when [pic].

10.18 The degrees of freedom for s2, the pooled estimator of [pic]are [pic].

a [pic]

b [pic]

c [pic]

10.19 a [pic]

b [pic]

10.20 When the actual data are given and s2 must be calculated, the calculation is done by

1 Using your scientific calculator to first obtain s1 and s2 and substituting into the formula for s2, OR

2 Using the computing formula for s2, noting that

[pic]

Hence, for the pooled estimator,

[pic]

The preliminary calculations are shown below:

Sample 1 Sample 2

[pic] [pic]

[pic] [pic]

[pic] [pic]

Then

[pic]

b A 90% confidence interval for [pic]is given as

[pic]

c As in Chapter 9, the hypothesis to be tested is

[pic]

The test statistic, under the assumption that [pic]is calculated using the pooled value of s2 in the t-statistic shown below:

[pic]

The rejection region is one-tailed, based on [pic]degrees of freedom. With [pic]from Table 4, the rejection region is [pic]. Since the observed value, [pic]does not fall in the rejection region, H0 is not rejected. We do not have sufficient evidence to indicate that [pic].

10.21 a The hypothesis to be tested is: [pic]

b The rejection region is two-tailed, based on [pic]degrees of freedom. With [pic], from Table 4, the rejection region is [pic].

c The pooled estimator of [pic]is calculated as

[pic]

and the test statistic is

[pic]

d The p-value is

[pic], so that [pic].

From Table 4 with [pic], [pic]is greater than the largest tabulated value ([pic]). Therefore, the area to the right of [pic]must be less than .005 so that

[pic]

e Comparing the observed [pic]to the critical value [pic]or comparing the p-value ( < .01) to [pic], H0 is rejected and we conclude that [pic].

10.22 Refer to Exercise 10.21. A 99% confidence interval for [pic]is given as

[pic]

10.23 a If you check the ratio of the two variances using the rule of thumb given in this section you will find:

[pic]

which is less than three. Therefore, it is reasonable to assume that the two population variances are equal.

b From the Minitab printout, the test statistic is [pic]with [pic].

c The value of [pic]is labeled “Pooled StDev” in the printout, so that [pic].

d Since the [pic]is greater than .10, the results are not significant. There is insufficient evidence to indicate a difference in the two population means.

e A 95% confidence interval for [pic]is given as

[pic]

Since the value [pic]falls in the confidence interval, it is possible that the two population means are the same. There insufficient evidence to indicate a difference in the two population means.

10.24 a If the antiplaque rinse is effective, the plaque buildup should be less for the group using the antiplaque rinse. Hence, the hypothesis to be tested is

[pic]

b The pooled estimator of [pic]is calculated as

[pic]

and the test statistic is

[pic]

The rejection region is one-tailed, based on [pic]degrees of freedom. With [pic], from Table 4, the rejection region is [pic]and H0 is rejected. There is evidence to indicate that the rinse is effective.

c The p-value is

[pic]

From Table 4 with [pic], [pic]is between two tabled entries [pic]and [pic], we can conclude that

[pic]

10.25 a The hypothesis to be tested is

[pic]

From the Minitab printout, the following information is available:

[pic]

and the test statistic is

[pic]

The rejection region is two-tailed, based on [pic]degrees of freedom. With [pic], from Table 4, the rejection region is [pic]and H0 is not rejected. There is not enough evidence to indicate a difference in the population means.

b It is not necessary to bound the p-value using Table 4, since the exact p-value is given on the printout as P-Value = .260.

c If you check the ratio of the two variances using the rule of thumb given in this section you will find:

[pic]

which is less than three. Therefore, it is reasonable to assume that the two population variances are equal.

10.26 a The hypothesis to be tested is

[pic]

where [pic]is the average compartment pressure for runners, and [pic]is the average compartment pressure for cyclists. The pooled estimator of [pic]is calculated as

[pic]

and the test statistic is

[pic]

The rejection region is two-tailed, based on [pic]degrees of freedom. With [pic], from Table 4, the rejection region is [pic]. We do not reject H0; there is insufficient evidence to indicate a difference in the means.

b Calculate

[pic]

A 95% confidence interval for [pic]is given as

[pic]

c Check the ratio of the two variances using the rule of thumb given in this section:

[pic]

which is greater than three. Therefore, it is not reasonable to assume that the two population variances are equal. You should use the unpooled t test with Satterthwaite’s approximation to the degrees of freedom.

10.27 a Check the ratio of the two variances using the rule of thumb given in this section:

[pic]

which is greater than three. Therefore, it is not reasonable to assume that the two population variances are equal.

b You should use the unpooled t test with Satterthwaite’s approximation to the degrees of freedom for testing

[pic]

The test statistic is

[pic]

with

[pic]

With [pic], the p-value for this test is bounded between .02 and .05 so that H0 can be rejected at the 5% level of significance. There is evidence of a difference in the mean number of uncontaminated eggplants for the two disinfectants.

10.28 a Use your scientific calculator or the computing formulas to find:

[pic]

Since the ratio of the variances is less than 3, you can use the pooled t test, calculating

[pic]

and the test statistic is

[pic]

For a two-tailed test with [pic], the p-value can be bounded using Table 4 so that

[pic]

Since the p-value is greater than .10, [pic]is not rejected. There is insufficient evidence to indicate that there is a difference in the mean titanium contents for the two methods.

b A 95% confidence interval for [pic]is given as

[pic]

Since [pic]falls in the confidence interval, the conclusion of part a is confirmed. This particular data set is very susceptible to rounding error. You need to carry as much accuracy as possible to obtain accurate results.

10.29 a The Minitab stem and leaf plots are shown below. Notice the mounded shapes which justify the assumption of normality.

Stem-and-Leaf Display: Generic, Sunmaid

Stem-and-leaf of Generic N = 14 Stem-and-leaf of Sunmaid N = 14

Leaf Unit = 0.10 Leaf Unit = 0.10

1 24 0 1 22 0

4 25 000 1 23

(5) 26 00000 5 24 0000

5 27 00 7 25 00

3 28 000 7 26

7 27 0

6 28 0000

2 29 0

1 30 0

b Use your scientific calculator or the computing formulas to find:

[pic]

Since the ratio of the variances is greater than 3, you must use the unpooled t test with Satterthwaite’s approximate df.

[pic]

c For testing [pic], the test statistic is

[pic]

For a two-tailed test with [pic], the p-value can be bounded using Table 4 so that

[pic]

Since the p-value is greater than .10, [pic]is not rejected. There is insufficient evidence to indicate that there is a difference in the mean number of raisins per box.

10.30 a The hypothesis of interest is

[pic]

and the preliminary calculations are as follows:

Sample 1 (Above) Sample 2 (Below)

[pic] [pic]

[pic] [pic]

[pic] [pic]

Then

[pic]

Also, [pic] and [pic]

The test statistic is

[pic]

For a one-tailed test with [pic]and [pic]the rejection region is [pic], and H0 is not rejected. There is insufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above.

b A 95% confidence interval for [pic]is given as

[pic]

10.31 a If swimmer 2 is faster, his(her) average time should be less than the average time for swimmer 1. Therefore, the hypothesis of interest is

[pic]

and the preliminary calculations are as follows:

Swimmer 1 Swimmer 2

[pic] [pic]

[pic] [pic]

[pic] [pic]

__________________________________

Then

[pic]

Also, [pic] and [pic]

The test statistic is

[pic]

For a one-tailed test with [pic], the p-value can be bounded using Table 4 so that

[pic], and H0 is not rejected. There is insufficient evidence to indicate that swimmer 2’s average time is still faster than the average time for swimmer 1.

10.32 Refer to Exercise 10.31. A 95% lower one-sided confidence bound for [pic]is given as

[pic]

Since the value [pic] is in the interval, it is possible that the two means might be equal. We do not have enough evidence to indicate that there is a difference in the means.

10.33 a The hypothesis of interest is

[pic]

and the preliminary calculations are as follows:

Favre Manning_______

[pic] [pic]

[pic] [pic]

[pic] [pic]

[pic] [pic]

[pic] [pic]

__________________________________

Then

[pic]

The test statistic is

[pic]

The degrees of freedom for this test are [pic], we estimate the rejection region using a value of t with df = 29 and the rejection region is [pic]. The null hypothesis is not rejected; there is insufficient evidence to indicate that the average number of completed passes is different for Brett Favre and Peyton Manning.

b Again, we estimate the value of t with df = 29 and the 95% confidence interval for [pic]is given as

[pic]

Since the value [pic] is in the interval, it is possible that the two means might be equal. We do not have enough evidence to indicate that there is a difference in the means.

10.34 The hypothesis of interest is

[pic]

and the preliminary calculations are as follows:

Island Thorns Ashley Rails_______

[pic] [pic]

[pic] [pic]

[pic] [pic]

[pic] [pic]

[pic] [pic]

__________________________________

Then

[pic]

The test statistic is

[pic]

The rejection region with [pic]and [pic] is [pic]and the null hypothesis is not rejected. There is insufficient evidence to indicate a difference in the average percentage of aluminum oxide at the two sites.

10.35 a The test statistic is

[pic]

with [pic]degrees of freedom. The p-value is then

[pic]so that [pic]

Since the value [pic]falls between two tabled entries for [pic]([pic]and [pic]), you can conclude that

[pic]

Since the p-value is less than [pic], the null hypothesis is rejected and we conclude that there is a difference in the two population means.

b A 95% confidence interval for [pic]is

[pic]

or [pic].

c Using [pic] and B = .1, the inequality to be solved is approximately

[pic]

Since this value of n is greater than 30, the sample size, [pic]pairs, will be valid.

10.36 a The hypothesis of interest is

[pic]

b The test statistic is

[pic]

with [pic]degrees of freedom. The rejection region is with [pic] is [pic], and H0 is not rejected. We cannot conclude that [pic].

10.37 a It is necessary to use a paired-difference test, since the two samples are not random and independent. The hypothesis of interest is

[pic]

The table of differences, along with the calculation of [pic]and [pic], is presented below.

|[pic] |.1 |.1 |0 |.2 |–.1 |[pic] |

|[pic] |.01 |.01 |.00 |.04 |.01 |[pic] |

[pic] and [pic]

The test statistic is

[pic]

with [pic]degrees of freedom. The rejection region with [pic] is [pic], and H0 is not rejected. We cannot conclude that the means are different.

b The p-value is

[pic]

c A 95% confidence interval for [pic]is

[pic]

or [pic].

d In order to use the paired-difference test, it is necessary that the n paired observations be randomly selected from normally distributed populations.

10.38 a A paired-difference test is used, since the two samples are not independent (for any given city, Allstate and 21st Century premiums will be related).

b The hypothesis of interest is

[pic]

where [pic]is the average for Allstate insurance and [pic]is the average cost for 21st Century insurance. The table of differences, along with the calculation of [pic]and [pic], is presented below.

|City |1 |2 |3 |4 |Totals |

|[pic] |389 |207 |222 |357 |1175 |

|[pic] |151,321 |42,849 |49,284 |127,449 |370,903 |

[pic] and

[pic]

The test statistic is

[pic]

with [pic]degrees of freedom. The rejection region with [pic] is [pic], and H0 is rejected. There is sufficient evidence to indicate a difference in the average premiums for Allstate and 21st Century.

c [pic]. Since [pic]is greater than [pic],

[pic].

d A 99% confidence interval for [pic]is

[pic]

or [pic].

e The four cities in the study were not necessarily a random sample of cities from throughout the United States. Therefore, you cannot make valid comparisons between Allstate and 21st Century for the United States in general.

10.39 a The hypothesis of interest is [pic]for the two independent samples of runners and cyclists before exercise. Since the ratio of the sample variances is greater than 3, the population variances cannot be assumed to be equal and you must use the unpooled t test with Satterthwaite’s approximate df. Calculate

[pic]

which has a t distribution with [pic]

A two-tailed rejection region is then [pic] and H0 is not rejected. A 95% confidence interval for [pic]is given as

[pic]

b Similar to part a. The hypothesis of interest is [pic]for the two independent samples of runners and cyclists after exercise. Since the ratio of the sample variances is greater than 3, you must again use the unpooled t test with Satterthwaite’s approximate df. Calculate

[pic]

which has a t distribution with [pic]

A two-tailed rejection region is then [pic] and H0 is rejected. A 95% confidence interval for [pic]is given as

[pic]

c To test the difference between runners before and after exercise, you use a paired difference test, and the hypothesis of interest is

[pic]

It is given that [pic]and[pic], so that the test statistic is

[pic]

The rejection region with[pic]and [pic] is [pic], and H0 is rejected. We can conclude that the means are different.

d The difference in mean CPK values for the 10 cyclists before and after exercise uses [pic]with[pic]. The 95% confidence interval for [pic]is

[pic]

or [pic]. Since the interval contains the value [pic], we cannot conclude that there is a significant difference between the means.

10.40 a-b The table of differences, along with the calculation of [pic]and[pic], is presented below.

|Week |1 |2 |3 |4 |Totals |

|di |–1.77 |–15.03 |–23.22 |127.05 |–67.07 |

[pic]

[pic] and [pic]

The hypothesis of interest is

[pic]

and the test statistic is

[pic]

Since [pic]with [pic]falls between the two tabled values, [pic]and [pic],

[pic]

for this two tailed test and H0 is not rejected. We cannot conclude that the means are different.

c The 99% confidence interval for [pic]is

[pic]

or [pic].

10.41 a Each subject was presented with both signs in random order. If his reaction time in general is high, both responses will be high; if his reaction time in general is low, both responses will be low. The large variability from subject to subject will mask the variability due to the difference in sign types. The paired-difference design will eliminate the subject to subject variability.

b The hypothesis of interest is

[pic]

The table of differences, along with the calculation of [pic]and [pic], is presented below.

|Driver |1 |2 |3 |4 |5 |6 |

|[pic] |–.4 |–2.7 |–1.6 |–1.7 |–1.5 |–7.9 |

[pic]

[pic] and [pic]

and the test statistic is

[pic]

A rejection region with [pic] and [pic]is [pic], and H0 is rejected at the 5% level of significance. We conclude that the air-based temperature readings are biased.

b The 95% confidence interval for [pic]is

[pic]

or [pic].

c The inequality to be solved is

[pic]

We need to estimate the difference in mean temperatures between ground-based and air-based sensors to within .2 degrees centigrade with 95% confidence. Since this is a paired experiment, the inequality becomes

[pic]

With [pic] and n represents the number of pairs of observations, consider the sample size obtained by replacing [pic]by [pic].

[pic]

Since the value of n is greater than 30, the use of [pic]for [pic]is justified.

10.44 A paired-difference test is used, since the two samples are not random and independent (within any sample, the dye 1 and dye 2 measurements are related). The hypothesis of interest is

[pic]

The table of differences, along with the calculation of [pic]and [pic], is presented below.

|Sampl|1 |2 |3 |4 |5 |6 |7 |

|e | | | | | | | |

Calculate [pic], [pic], and the test statistic is

[pic]

Since [pic]with [pic]is smaller than the smallest tabled value [pic],

[pic]

for this one-tailed test and H0 is not rejected. We cannot conclude that the average time outside the office is less when music is piped in.

10.48 It is necessary to test

[pic]

This will be done using s2, the sample variance, which is a good estimator for [pic]. Refer to Section 10.6 of the text and notice that the quantity

[pic]

possesses a chi-square distribution in repeated sampling. This distribution is shown below.

[pic]

Notice that the distribution is nonsymmetrical and that the random variable

[pic]

takes on values commencing at zero (since s2, [pic]and [pic]are never negative). The test statistic is

[pic]

A one-tailed test of an hypothesis is required. Hence, a critical value of [pic](denoted by [pic]) must be found such that

[pic]

Indexing [pic] with [pic]degrees of freedom (see Table 5), the critical value is found to be [pic](see the above figure). The value of the test statistic does not fall in the rejection region. Hence, H0 is not rejected . We cannot conclude that the variance exceeds 15.

10.49 For this exercise, [pic]and [pic].

[pic]

A 90% confidence interval for [pic]will be

[pic]

where [pic]represents the value of [pic]such that 5% of the area under the curve (shown in the figure above) lies to its right. Similarly, [pic]will be the [pic]value such that an area .95 lies to its right.

Hence, we have located one-half of [pic]in each tail of the distribution. Indexing [pic]and [pic] with [pic]degrees of freedom in Table 5 yields

[pic] and [pic]

and the confidence interval is

[pic] or [pic]

10.50 a Calculate [pic], [pic]and [pic]. Then

[pic]

b Indexing [pic]and [pic] with [pic]degrees of freedom in Table 5 yields

[pic] and [pic]

and the 95% confidence interval is

[pic] or [pic]

c It is necessary to test

[pic]

and the test statistic is

[pic]

The two-tailed rejection region with [pic] and [pic]degrees of freedom is

[pic]

and H0 is not rejected. There is insufficient evidence to indicate that [pic]is different from .8.

d The p-value is found by approximating [pic]and then doubling that value to account for an equally small value of s2 which might have produced a value of the test statistic in the lower tail of the chi-square distribution. The observed value, [pic], is smaller than [pic]in Table 5. Hence,

[pic]

10.51 The hypothesis of interest is

[pic]

or equivalently [pic]

Calculate

[pic]

The test statistic is

[pic]

The one-tailed rejection region with [pic]and [pic]degrees of freedom is[pic]

and H0 is rejected. There is sufficient evidence to indicate that [pic]is greater than .49.

10.52 Similar to previous exercises. Indexing [pic]and [pic] with [pic]degrees of freedom in Table 5 yields

[pic] and [pic]

and the 90% confidence interval is

[pic]

[pic] or [pic]

10.53 a The hypothesis to be tested is

[pic]

Calculate [pic] [pic]

and the test statistic is

[pic]

The rejection region with [pic]and [pic]degrees of freedom is found from Table 4 as [pic]. Since the observed value of the test statistic does not fall in the rejection region, H0 is not rejected. There is insufficient evidence to show that the mean differs from 5 mg/cc.

b The manufacturer claims that the range of the potency measurements will equal .2. Since this range is given to equal [pic], we know that [pic]. Then

[pic]

The test statistic is

[pic]

and the one-tailed rejection region with [pic]and [pic]degrees of freedom is

[pic]

H0 is rejected; there is sufficient evidence to indicate that the range of the potency will exceed the manufacturer’s claim.

10.54 In order to construct a 95% confidence interval, values of [pic] and [pic]with [pic]degrees of freedom are needed. Using the values for 60 degrees of freedom as approximate values, the confidence interval is approximately

[pic]

[pic] or [pic]

10.55 a The force transmitted to a wearer, x, is known to be normally distributed with [pic]and [pic]. Hence,

[pic]

It is highly improbable that any particular helmet will transmit a force in excess of 1000 pounds.

b Since [pic], a large sample test will be used to test

[pic]

The test statistic is

[pic]

and the rejection region with [pic]is [pic]. H0 is rejected and we conclude that [pic].

10.56 Refer to Exercise 10.55. The hypothesis of interest is

[pic]

and the test statistic is

[pic]

The one-tailed rejection region with [pic]and [pic]degrees of freedom (approximated with 40 degrees of freedom) is[pic], and H0 is rejected. There is sufficient evidence to indicate that [pic]is greater than 40.

10.57 The hypothesis of interest is [pic]

Calculate

[pic]

and the test statistic is [pic]. The one-tailed rejection region with [pic]and [pic]degrees of freedom is[pic], and H0 is not rejected. There is insufficient evidence to indicate that he is meeting his goal.

10.58 When the assumptions for the F distribution are met, then [pic]possesses an F distribution with [pic]and [pic]degrees of freedom. Note that df1 and df2 are the degrees of freedom associated with [pic]and [pic], respectively. The F distribution is non-symmetrical with the degree of skewness dependent on the above-mentioned degrees of freedom. Table 6 presents the critical values of F (depending on the degrees of freedom) such that [pic], respectively. Because right-hand tail areas correspond to an upper-tailed test of an hypothesis, we will always identify the larger sample variance as [pic](that is, we will always place the larger sample variance in the numerator of [pic]). Hence, an upper-tailed test is implied and the critical values of F will determine the rejection region. If we wish to test the hypothesis

[pic]

There will be another portion of the rejection region in the lower tail of the distribution. The area to the right of the critical value will represent only [pic], and the probability of a Type I error is [pic].

a In this exercise, the hypothesis of interest is

[pic]

and the test statistic is [pic].

The rejection region (two-tailed) will be determined by a critical value of F based on [pic]and [pic]degrees of freedom with area .025 to its right. That is, from Table 6, [pic]. The observed value of F does not fall in the rejection region, and we cannot conclude that the variances are different.

b The student will need to find critical values of F for various levels of [pic]in order to find the approximate p-value. The critical values with [pic]and [pic]are shown below from Table 6.

|[pic] |.10 |.05 |.025 |.01 |.005 |

|[pic] |1.86 |2.23 |2.62 |3.15 |3.59 |

Hence,

[pic]

10.59 Refer to Exercise 10.58. From Table 6, [pic]and [pic]. The 95% confidence interval for [pic]is

[pic]

10.60 a The hypothesis of interest is [pic]

and the test statistic is

[pic].

The rejection region (one-tailed) with [pic]and[pic]degrees of freedom is [pic], and H0 is not rejected. We cannot conclude that the variances are different.

b The critical values of F for various values of [pic]are given below.

|[pic] |.10 |.05 |.025 |.01 |.005 |

|[pic] |2.15 |2.69 |3.28 |4.16 |4.91 |

Hence, the [pic] lies between .05 and .10, and the results are not significant.

10.61 The hypothesis of interest is [pic]

and the test statistic is

[pic].

The critical values of F for various values of [pic]are given below using [pic]and [pic].

|[pic] |.10 |.05 |.025 |.01 |.005 |

|[pic] |2.01 |2.46 |2.95 |3.66 |4.25 |

Hence,

[pic]

Since the p-value is so large, H0 is not rejected. There is no evidence to indicate that the variances are different.

10.62 a The hypothesis of interest is [pic]

and the test statistic is

[pic].

The rejection region (one-tailed) with [pic]and [pic]degrees of freedom by interpolation in Table 6. The value [pic] is roughly halfway between [pic]and [pic]; therefore, we reject H0 if [pic]. The observed value of the test statistic falls in the rejection region and we conclude that the “suspect line” possesses a larger variance.

b The student must obtain various critical levels of F from Table 6. We “roughly” interpolate [pic]as halfway between [pic]and [pic].

|[pic] |.05 |.025 |.01 |.005 |

|[pic] |1.61 |1.775 |1.975 |2.13 |

In any event,

[pic]

10.63 Refer to Exercise 10.62. Noting that [pic], the 90% confidence interval for [pic]is

[pic]

10.64 a The assumption of equal variances [pic]was made.

b The hypothesis of interest is [pic]

and the test statistic is

[pic].

The upper portion of the rejection region with , [pic], [pic]and [pic]is [pic]and H0 is not rejected. There is no reason to believe that the assumption has been violated.

10.65 For each of the three tests, the hypothesis of interest is

[pic]

and the test statistics are

[pic] [pic] and [pic]

The critical values of F for various values of [pic]are given below using [pic]and [pic].

|[pic] |.10 |.05 |.025 |.01 |.005 |

|[pic] |2.44 |3.18 |4.03 |5.35 |6.54 |

Hence, for the first two tests,

[pic]

while for the last test,

[pic]

There is no evidence to indicate that the variances are different for the first two tests, but H0 is rejected for the third variable. The two-sample t-test with a pooled estimate of [pic]cannot be used for the third variable.

10.66 a The hypothesis of interest is

[pic]

and the test statistic is

[pic].

The upper portion of the rejection region with[pic]is [pic](from Table 6) and H0 is not rejected. There is insufficient evidence to indicate that the supplier’s shipments differ in variability.

b The 99% confidence interval for [pic]is

[pic]

[pic] or [pic]

Intervals constructed in this manner enclose [pic]99% of the time. Hence, we are fairly certain that[pic] is between .036 and .488.

10.67 A Student’s t test can be employed to test an hypothesis about a single population mean when the sample has been randomly selected from a normal population. It will work quite satisfactorily for populations which possess mound-shaped frequency distributions resembling the normal distribution.

10.68 As in the case of the single population mean, random samples must be independently drawn from two populations which possess normal distributions with a common variance, [pic]. Consequently, it is local that information in the two sample variances, [pic]and [pic], should be pooled in order to give the best estimate of the common variance, [pic]. In this way, all of the sample information is being utilized to its best advantage.

10.69 Paired observations are used to estimate the difference between two population means in preference to an estimation based on independent random samples selected from the two populations because of the increased information caused by blocking the observations. We expect blocking to create a large reduction in the standard deviation, if differences do exist among the blocks.

Paired observations are not always preferable. The degrees of freedom that are available for estimating [pic]are less for paired than for unpaired observations. If there were no difference between the blocks, the paired experiment would then be less beneficial.

10.70 a The hypothesis to be tested is

[pic]

and the test statistic is

[pic].

Critical value approach: The rejection region with[pic]and [pic]degrees of freedom is located in the upper tail of the t-distribution and is found from Table 4 as [pic]. Since the observed value falls in the rejection region, H0 is rejected and we conclude that [pic]is greater than .05.

p-value approach: The[pic]. Since the value [pic]falls between [pic]and [pic], the p-value can be bounded as

[pic]

In any event, the p-value is less than .05 and H0 can be rejected at the 5% level of significance.

10.71 The 90% confidence interval is

[pic]

or [pic].

10.72 Using the formulas given in Chapter 2 or your scientific calculator, calculate

[pic]

[pic]

The 99% confidence interval is

[pic]

or [pic].

10.73 Since it is necessary to determine whether the injected rats drink more water than noninjected rates, the hypothesis to be tested is

[pic]

and the test statistic is

[pic].

Using the critical value approach, the rejection region with[pic]and [pic]degrees of freedom is located in the upper tail of the t-distribution and is found from Table 4 as [pic]. Since the observed value of the test statistic falls in the rejection region, H0 is rejected and we conclude that the injected rats do drink more water than the noninjected rats. The 90% confidence interval is

[pic]

or [pic].

10.74 Using the formulas given in Chapter 2 or your scientific calculator, calculate

[pic]

[pic]

The 95% confidence interval is then

[pic]

or [pic].

10.75 The hypothesis of interest is

[pic]

and the test statistic is

[pic].

The critical values of F for various values of [pic]are given below using [pic](since there are no tabled values for [pic]) and [pic].

|[pic] |.10 |.05 |.025 |.01 |.005 |

|[pic] |1.62 |1.85 |2.09 |2.41 |2.66 |

Hence,

[pic]

Since the p-value is so small, H0 is rejected. There is evidence to indicate that increased maintenance of the older system is needed.

10.76 The student may use the rounded values for [pic]and s given in the display, or he may wish to calculate [pic]and s and use the more exact calculations for the confidence intervals. The calculations are shown below.

a [pic] [pic]

[pic]and the 95% confidence interval is

[pic]

or [pic].

b [pic] [pic]

[pic]and the 95% confidence interval is

[pic]

or [pic].

c [pic] [pic]

[pic]and the 95% confidence interval is

[pic]

or [pic].

d No. The relationship between the confidence intervals is not the same as the relationship between the original measurements.

10.77 From Exercise 10.76, the best estimate for [pic]is [pic]. Then, with [pic], solve for n in the following inequality:

[pic]

which is approximately

[pic]

Since n is greater than 30, the sample size, [pic], is valid.

10.78 A paired-difference analysis must be used. The hypothesis of interest is

[pic]

The table of differences is presented below. Use your calculator to find [pic]and [pic],

|[pic]|(3 |(3 |2 |(1 |(1 |1 |(3 |

Calculate [pic], [pic], and the test statistic is

[pic]

The one-tailed rejection region with [pic]is [pic] and H0 is not rejected. There is insufficient evidence to indeicat that the mean reaction time is greater after consuming alcohol.

10.79 Use the computing formulas or your calculator to calculate

[pic] [pic]

[pic]and the 95% confidence interval is

[pic]

or [pic].

10.80 a The hypothesis to be tested is

[pic]

The following information is available:

[pic] [pic] and [pic]

[pic] [pic] and [pic]

Since the ratio of the variances is less than 3, you can use the pooled t test. The pooled estimator of [pic]is calculated as

[pic]

and the test statistic is

[pic]

Critical value approach: The rejection region is two-tailed, based on[pic]degrees of freedom. With[pic], from Table 4, the rejection region is [pic]and H0 is rejected. There is sufficient evidence to indicate a difference in the mean absorption rates between the two drugs.

b p-value approach: The p-value is[pic]for a two-tailed test with 18 degrees of freedom. Since [pic]exceeds the largest tabled value, [pic], we have

[pic]

Since the p-value is less than [pic], H0 can be rejected at the 5% level of significance, confirming the results of part a.

c A 95% confidence interval for [pic]is given as

[pic]

Since the confidence interval does not contain the value [pic], you can conclude that there is a difference in two population means. This confirms the results of Exercise 10.76.

10.81 The inequality to be solved is

[pic]

In this exercise, it is necessary to estimate the difference in means to within 1 minute with 95% confidence. The inequality is then

[pic]

We assume that [pic]and [pic]from Exercise 10.76. Consider the sample size obtained by replacing [pic]by [pic].

[pic]

Since the value of n is greater than 30, the sample size is valid.

b Consider the inequality from part a,

[pic].

When [pic], this becomes

[pic]

To reduce the sample size necessary to achieve this inequality, we must reduce the quantity

[pic]

If we are willing to lower the level of confidence (or equivalently increase the value of [pic]), the value of [pic]will be smaller. This will decrease the size of n, but at the price of decreased confidence. The only other quantity in the expression which is not fixed is s2. If this variable can be made smaller, the required sample size will also be reduced. From the definition of s2, it can be seen that s2 includes the variability associated with the difference in drugs A and B as well as the variability in absorption rates among the people in the experiment. The experiment could be run a bit differently by using the same people for both drugs. Each person would receive a dose of drug A and drug B. The difference in absorption rates for the two drugs would be observed within each individual. Such a design would eliminate the variability in absorption rates from person to person. The value of s2 would be reduced and a smaller sample size would be required. The resulting experiment would be analyzed as a paired-difference experiment, as discussed in Section 10.5.

10.82 a To check for equality of variances, we can use the rule of thumb or the formal F-test. First, calculate

[pic] and [pic]

The ratio

[pic]

is greater than 3, which is an indication that the variances are unequal. Using a two-sided F-test with [pic], the p-value for the test is bounded as

[pic]

Since the p-value is close to .05, we choose to assume that the variances are not equal, and use Satterthwaite’s approximation.

b The hypothesis of interest is [pic]for the two independent samples of male and female pheasants. Calculate

[pic]

which has a t distribution with [pic]

A one-tailed rejection region is then [pic] and H0 is rejected. There is sufficient evidence to indicate that the average weight of male pheasants exceeds that of females by more than 300 grams.

10.83 a The range of the first sample is 47 while the range of the second sample is only 16. There is probably a difference in the variances.

b The hypothesis of interest is

[pic]

Calculate [pic] [pic]

and the test statistic is

[pic].

The critical values with [pic] and [pic]are shown below from Table 6.

|[pic] |.10 |.05 |.025 |.01 |.005 |

|[pic] |3.62 |5.41 |7.76 |12.06 |16.53 |

Hence,

[pic]

Since the p-value is smaller than .01, H0 is rejected at the 1% level of significance. There is a difference in variability.

c Since the Student’s t test requires the assumption of equal variance, it would be inappropriate in this instance. You should use the unpooled t test with Satterthwaite’s approximation to the degrees of freedom.

10.84 a Use the computing formulas or your scientific calculator to calculate

[pic] [pic]

[pic]and the 99% confidence interval is

[pic]

or [pic].

b Intervals constructed using this procedure will enclose [pic]99% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses[pic].

c The sample must be randomly selected, or at least behave as a random sample from the population of interest. For the chemist performing the analysis, this means that he or she must be certain that there is no unknown factor which is affecting the measurements, thus causing a biased sample rather than a representative sample.

10.85 a The leaf measurements probably come from mound-shaped, or approximately normal populations, since their length, width, thickness, and so on can be thought of as being a composite sum of many factors which affect their growth (see the Central Limit Theorem). The values of the sample variances are not very different, and we would not question the assumption of equal variances. Finally, since the plants were all given the same experimental treatment, they can be considered random and independent samples within a treatment group.

b The hypothesis to be tested is

[pic]

Since the ratio of the variances is less than 3, you can use the pooled t test. The pooled estimator of [pic]is calculated as

[pic]

and the test statistic is

[pic]

The two-tailed p-value with [pic]can be bounded as

[pic]

and H0 is rejected. There is a difference in the means.

c The hypothesis to be tested is

[pic]

Since the ratio of the variances is slightly greater than 3, you should use the pooled t test with the test statistic

[pic]

The p-value is two-tailed, based on

[pic]

degrees of freedom and is bounded as

[pic]

The results are highly significant and H0 is rejected. There is a difference in the means.

10.86 a The hypothesis of interest is

[pic]

The test statistic is

[pic].

The two-sided rejection region with [pic] and [pic]and [pic] is [pic]and the null hypothesis is not rejected. There is insufficient evidence to indicate a difference in the precision of the two machines.

b The 95% confidence interval for the ratio of the two variances is

[pic]

Since the possible values for [pic]includes the value 1, it is possible that there is no difference in the precision of the two machines. This confirms the results of part a.

10.87 A paired-difference test is used, since the two samples are not random and independent. The hypothesis of interest is

[pic]

and the table of differences, along with the calculation of [pic]and [pic], is presented below.

|Pair |1 |2 |3 |4 |Totals |

|[pic] |–1 |5 |11 |7 |22 |

[pic] [pic] and [pic]

and the test statistic is

[pic]

The one-tailed p-value with [pic]can be bounded between .05 and .10. Since this value is greater than .10, H0 is not rejected. The results are not significant; there is insufficient evidence to indicate that lack of school experience has a depressing effect on IQ scores.

10.88 A paired-difference analysis is used. The differences are shown below.

47, 44, 38, 46, 37, 56, 35, 34

Calculate

[pic] [pic] and [pic]

and the 95% confidence interval for [pic]is

[pic]

or [pic].

10.89 A paired-difference analysis is used. The hypothesis of interest is

[pic]

and the differences are shown below.

–206, 47, 155, –61, 12

Calculate

[pic] [pic]

and the test statistic is

[pic]

The two-tailed p-value with[pic]is greater than [pic]. Since this value is greater than .05, H0 is not rejected. The results are not significant; there is insufficient evidence to indicate a difference in the average cost for front versus rear bumper repairs.

10.90 a Because of the mounded shape of the data shown in the stem and leaf plot, you can be assured that the assumption of normality is valid.

b Using the summary statistics given in the printout, the 99% confidence interval is

[pic]

or [pic].

10.91 The object is to determine whether or not there is a difference between the mean responses for the two different stimuli to which the people have been subjected. The samples are independently and randomly selected, and the assumptions necessary for the t test of Section 10.4 are met. The hypothesis to be tested is

[pic]

and the preliminary calculations are as follows:

[pic] and [pic]

[pic] and [pic]

Since the ratio of the variances is less than 3, you can use the pooled t test. The pooled estimator of [pic]is calculated as

[pic]

and the test statistic is

[pic]

The two-tailed rejection region with [pic]and [pic]is [pic], and H0 is not rejected. There is insufficient evidence to indicate that there is a difference in means.

10.92 Note that this experiment has been designed differently from the experiment performed in Exercise 10.91. That is, each of the eight people was subjected to both stimuli (in random order). A paired-difference analysis is used to test the hypothesis

[pic]

and the table of differences, along with the calculation of [pic]and [pic], is presented below.

|Person|1 |2 |3 |4 |5 |

|[pic] |2.01 |2.46 |2.95 |3.66 |4.25 |

Hence,

[pic]

This is too large to reject H0.

b From the table in part a, [pic]. The 99% confidence interval for [pic]is

[pic]

Intervals constructed using this procedure will enclose the ratio [pic] 99% of the time in repeated sampling. Hence, we are fairly confident that the interval, .452 to 8.1685, encloses [pic].

10.105 The underlying populations are ratings and can only take on the finite number of values, 1, 2, …, 9, 10. Neither population has a normal distribution, but both are discrete. Further, the samples are not independent, since the same person is asked to rank each car design. Hence, two of the assumptions required for the Student’s t test have been violated.

10.106 A paired-difference test is used. To test [pic], where [pic]is the mean before the safety program and [pic]is the mean after the program, calculate the differences:

7, 6, –1, 5, 6, 1

Then [pic] [pic] and [pic]

and the test statistic is

[pic]

For a one-tailed test with[pic], the rejection region with [pic]is [pic], and H0 is not rejected. We cannot conclude that the drug significantly increases reaction time.

10.107 A paired-difference test is used. To test [pic], calculate the differences:

29, 31, –35, –17, 99, 73, 54

The test statistic is given on the Minitab printout as

[pic]

with [pic]. Since this value is greater than .10, the results are not significant, and H0 is not rejected. There is insufficient evidence to indicate a greater mean demand for one of the entrees.

10.108 Indexing [pic]and [pic] with [pic]degrees of freedom in Table 5 yields

[pic] and [pic]

and the 90% confidence interval is

[pic] or [pic]

10.109 The Minitab printout below shows the summary statistics for the two samples:

Descriptive Statistics: Method 1, Method 2

Variable N Mean SE Mean StDev

Method 1 5 137.00 4.55 10.17

Method 2 5 147.20 3.29 7.36

Since the ratio of the two sample variances is less than 3, you can use the pooled t test to compare the two methods of measurement, using the remainder of the Minitab printout below:

Two-Sample T-Test and CI: Method 1, Method 2

Difference = mu (Method 1) - mu (Method 2)

Estimate for difference: -10.20

95% CI for difference: (-23.15, 2.75)

T-Test of difference = 0 (vs not =): T-Value = -1.82 P-Value = 0.107 DF = 8

Both use Pooled StDev = 8.8798

The test statistic is [pic]with[pic]and the results are not significant. There is insufficient evidence to declare a difference in the two population means.

10.110 .a The hypothesis of interest is

[pic]

and the test statistic is

[pic]

The one-tailed rejection region with [pic]and [pic]degrees of freedom is[pic], and H0 is rejected. There is sufficient evidence to indicate that [pic]is greater than .03.

b Indexing [pic]and [pic] with [pic]degrees of freedom in Table 5 yields

[pic] and [pic]

and the confidence interval is

[pic] or [pic]

10.111 a Calculate [pic]

[pic]

The 99% confidence interval based on [pic]is

[pic]

or [pic].

b The sample must have been randomly selected from a normal population.

10.112 Calculate [pic]

[pic]

The 95% confidence interval based on [pic]is

[pic]

or [pic]. Intervals constructed in this manner enclose the true value of μ 95% of the time. Hence, we are fairly certain that this interval contains the true value of μ.

10.113 The hypothesis to be tested is

[pic]

The test statistic is

[pic]

The critical value of t with [pic]and [pic]degrees of freedom is [pic]and the rejection region is [pic]. Since the observed value falls in the rejection region, H0 is rejected. There is sufficient evidence to indicate that the average number of calories is greater than advertised.

10.114 a The hypothesis of interest is

[pic]

and the test statistic is [pic].

The upper portion of the two-tailed rejection region with [pic] is [pic] and H0 is not rejected. There is insufficient evidence to indicate that the population variances are different.

b The hypothesis to be tested is

[pic]

Based on the results of part a, you can use the pooled t test. The pooled estimator of [pic]is calculated as

[pic]

and the test statistic is

[pic]

The rejection region with [pic] and [pic](approximated with df = 29) is [pic] and H0 is not rejected. There is insufficient evidence to indicate a difference in the two populaton means.

10.115 a A paired difference test is required, since the costs are paired according to the type of drug. To test [pic], the 9 differences are

111, 201, 45, 14, 175, 105, 288, 94, 79

Calculate

[pic] [pic] and [pic]

The test statistic is

[pic]

For a two-tailed test with[pic], the rejection region with [pic]is [pic], and H0 is rejected. There is sufficient evidence to indicate that the average cost of prescription drugs in the United States is different from the average cost in Canada.

b The observed value of the test statistic, t = 4.38 is larger than [pic], so that the [pic] Since [pic], the p-value would cause us to reject H0, as in part a.

10.116 Use the Student’s t Probabilities applet. Select the proper df using the slider on the right and use either the one- or two-tailed applet, depending on the probability to be calculated. Type the positive value of t into the box marked “Student t” and press enter. The answers are shown below.

a .1419 b .0054 c .0734 d .2798

10.117 Use the Student’s t Probabilities applet. Select the proper df using the slider on the right and use either the one- or two-tailed applet, depending on the type of rejection region. Type the positive value of [pic]into the box marked “prob:” and press enter. The answers are shown below.

a [pic] b [pic] c [pic]

10.118 Use the Interpreting Confidence Intervals applet.

a The 95% confidence interval with [pic]or [pic]is calculated as

[pic]

b Answers will vary. Students should use the formula in part a, substituting the appropriate values for [pic]and s. Remember that the applet rounds to the nearest integer!

10.119 Use the Interpreting Confidence Intervals applet. Answers will vary from student to student. The widths of the ten intervals will not be the same, since the value of s changes with each new sample. The student should find that approximately 95% of the intervals in the first applet contain [pic], while roughly 99% of the intervals in the second applet contain [pic].

10.120 Use the Interpreting Confidence Intervals applet. Answers will vary from student to student. The student should find that approximately 95% of the intervals in the first applet contain [pic], while roughly 99% of the intervals in the second applet contain [pic].

10.121 Use the Small Sample Test of a Population Mean applet. A screen capture is shown below. The test statistic

[pic]

has a two-tailed p-value of .1782 and H0 is not rejected. There is insufficient evidence to indicate that [pic]differs from 48.

[pic]

10.122 Use the Small Sample Test of a Population Mean applet.

a The test statistic for testing [pic]is

[pic]

has a two-tailed p-value of .7274 and H0 is not rejected. There is insufficient evidence to indicate that [pic]differs from 508.

b The test statistic for testing [pic]is

[pic]

has a two-tailed p-value of .8741 and H0 is not rejected. There is insufficient evidence to indicate that [pic]differs from 520.

10.123 Use the Two Sample t Test: Independent Samples applet. The hypothesis to be tested concerns the differences between mean recovery rates for the two surgical procedures. Let [pic]be the population mean for Procedure I and [pic]be the population mean for Procedure II. The hypothesis to be tested is

[pic]

Since the ratio of the variances is less than 3, you can use the pooled t test. Enter the appropriate statistics into the applet and you will find that test statistic is

[pic]

with a two-tailed p-value of .0030. Since the p-value is very small, H0 can be rejected for any value of [pic]greater than .003 and the results are judged highly significant. There is sufficient evidence to indicate a difference in the mean recovery rates for the two procedures.

[pic]

10.124 Use the Two Sample t Test: Independent Samples applet. The hypothesis to be tested is

[pic]

Since the ratio of the variances is less than 3, you can use the pooled t test. Enter the appropriate statistics into the applet and you will find that test statistic is

[pic]

with a two-tailed p-value of .0004. Since the p-value is very small, H0 can be rejected for any value of [pic]greater than .0004 and the results are judged highly significant. There is sufficient evidence to indicate a difference in the average prices of the two common stocks.

Case Study: How Would You Like a Four-Day Work Week?

1 A paired-difference test is used to test [pic], where [pic]is the mean number of personal leave-days taken on the conventional workweek and [pic]is the mean on the four-day workweek. Using Minitab to perform the paired difference test, you obtain the following printout.

Paired T-Test and CI: PL-Yr2, PL-Yr1

Paired T for PL-Yr2 - PL-Yr1

N Mean StDev SE Mean

PL-Yr2 11 22.3636 7.8648 2.3713

PL-Yr1 11 18.8182 11.2855 3.4027

Difference 11 3.54545 15.59079 4.70080

95% lower bound for mean difference: -4.97456

T-Test of mean difference = 0 (vs > 0): T-Value = 0.75 P-Value = 0.234

The test statistic of [pic]with [pic]indicates that the results are not significant. There is not enough evidence to indicate that there is a reduction in the average number of personal leave-days.

2 For the paired difference test of [pic], where [pic]is the mean number of sick-days taken on the conventional workweek and [pic]is the mean on the four-day workweek, a directional alternative might not be desirable. It could be that working long hours (10 hours per day) might increase rather than decrease the average number of sick-days. A two-tailed test of this hypothesis is provided by Minitab.

Paired T-Test and CI: SL-Yr2, SL-Yr1

Paired T for SL-Yr2 - SL-Yr1

N Mean StDev SE Mean

SL-Yr2 11 58.4545 25.8587 7.7967

SL-Yr1 11 46.0909 25.8126 7.7828

Difference 11 12.3636 23.0316 6.9443

95% CI for mean difference: (-3.1092, 27.8365)

T-Test of mean difference = 0 (vs not = 0): T-Value = 1.78 P-Value = 0.105

The test statistic of [pic]with [pic]indicates that the results are not significant. There is not enough evidence to indicate that there is a reduction in the average number of sick-days.

3 From the printout in part 2, the 95% confidence interval for the difference in the average number of sick-days in these two years is

[pic].

Based on this interval, which contains the value [pic], there is no evidence to indicate a difference between the conventional and four-day work week.

4 Based on the analysis of these two variables, there is little or no difference in the two types of work schedules.

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