Stat501 Hw#5 Solution



Stat501 Hw#5 Solution Hints

9.10 a If the airline is to determine whether or not the flight is unprofitable, they are interested in finding out whether or not [pic](since a flight is profitable if [pic]is at least 60). Hence, the alternative hypothesis is [pic]and the null hypothesis is [pic]. Formally,

[pic] vs. [pic]

b Since only small values of [pic](and hence, negative values of z) would tend to disprove H0 in favor of Ha, this is a one-tailed test.

c For this exercise, [pic].

One-Sample T

Test of mu = 0.6 vs < 0.6

95% Upper

N Mean StDev SE Mean Bound T P

120 0.5800 0.1100 0.0100 0.5966 -1.99 0.024

Since the p-value of 0.024 is small ( .01). Using the guidelines for significance given in Section 9.3 of the text, we declare the results statistically significant, but not highly significant.

9.34 a The hypothesis of interest is two-tailed:

[pic]

b (Get a Minitab output and make conclusions in stead of using following)

With [pic]and [pic], so that [pic], the test statistic is

[pic]

with[pic]or [pic]. Since this p-value is less than .01, H0 is rejected at the 1% level of significance and the results are declared highly significant. There is evidence that the proportion of red flowered plants is not .75.

9.35 a-b Since the survival rate without screening is [pic], the survival rate with an effective program may be greater than 2/3. Hence, the hypothesis to be tested is

[pic]

c With [pic], the test statistic is

[pic]

The rejection region is one-tailed, with [pic] or [pic]and H0 is rejected. The screening program seems to increase the survival rate.

d For the one-tailed test,

[pic]

That is, H0 can be rejected for any value of [pic]. The results are highly significant.

9.38 a The hypothesis of interest is

[pic]

With [pic], the test statistic is

[pic]

The rejection region is two-tailed [pic], or [pic] and H0 is not rejected. There is insufficient evidence to indicate that the claim is incorrect.

b The hypothesis of interest is

[pic]

With [pic], the test statistic is

[pic]

The rejection region is two-tailed [pic], or [pic] and H0 is not rejected. There is insufficient evidence to indicate that the claim is incorrect.

c Unless the experimenter had some preconceived idea that the proportion might be greater or less than claimed, there would be no reason to run a one-tailed test.

9.40 The hypothesis of interest is

[pic]

with [pic], the test statistic is

[pic]

The rejection region with [pic]=.01 is [pic] and the null hypothesis is not rejected. (Alternatively, we could calculate [pic]. Since this p-value is greater than .01, the null hypothesis is not rejected.) There is insufficient evidence to indicate that the percentage of adults who say that they always vote is different from the percentage reported in Time.

9.45 a The hypothesis of interest is:

[pic]

Calculate [pic],[pic]and [pic]. The test statistic is then

[pic]

The rejection region, with [pic], is [pic]and H0 is rejected. There is evidence of a difference in the proportion of survivors for the two groups.

b From Section 8.7, the approximate 95% confidence interval is

[pic]

9.46 a The hypothesis of interest is

[pic]

Calculate [pic],[pic], and [pic]. The test statistic is then

[pic]

The rejection region, with [pic], is [pic]and H0 is not rejected. There is no evidence of a difference in the proportion of frequent moviegoers in the two demographic groups.

b A difference in the proportions might mean that the advertisers would choose different products to advertise before this movie.

9.50 a Since the two treatments were randomly assigned, the randomization procedure can be implemented as each patient becomes available for treatment. Choose a random number between 0 and 9 for each patient. If the patient receives a number between 0 and 4, the assigned drug is aspirin. If the patient receives a number between 5 and 9, the assigned drug is clopidogrel.

b Assume that [pic]and [pic]. It is given that [pic], [pic], so that

[pic].

The test statistic is then

[pic]

with[pic]. Since the p-value is less than .01, the results are statistically significant. There is sufficient evidence to indicate a difference in the proportions for the two treatment groups.

c Clopidogrel would be the preferred treatment, as long as there are no dangerous side effects.

9.51 The hypothesis of interest is

[pic]

Calculate [pic], [pic], and [pic]. The test statistic is then

[pic]

with[pic]. Since the p-value is less than .01, the results are reported as highly significant at the 1% level of significance. There is evidence to confirm the researcher’s conclusion.

9.60 a The hypothesis of interest is

[pic]

with [pic], and the test statistic is

[pic]

Since no value of [pic]is specified in advance, we calculate [pic]. Since this p-value is less than .10, you can reject H0 at the 1% level (highly significant) and reject the DA’s claim of 50% or greater.

b If you take [pic]and [pic]with [pic], the 95% confidence interval for p is approximately

[pic]

or [pic]

c Even with the conservative value of x in part b, you can see that all the possible values for p are greater than [pic]. You cannot conclude that the DA’s claim of 50% or greater is wrong – in fact, it appears that the DA is correct!

9.65 a-b The hypothesis to be tested is

[pic]

Calculate [pic], [pic], and [pic]. The test statistic is

then

[pic]

and the p-value is [pic].

c Since the observed p-value, .4364, is greater than [pic], H0 cannot be rejected. There is insufficient evidence to support the researcher’s belief.

9.66 Refer to the figure below, which represents the two probability distributions, one assuming that [pic]and one assuming that [pic].

[pic]

The right curve is the true distribution of the random variable [pic]and consequently any probabilities that we wish to calculate concerning the random variable must be calculated as areas under the curve to the right. The objective of this exercise is to find a common sample size so that [pic]and [pic]. For [pic]consider the critical value of [pic]that separates the rejection and acceptance regions. This value will be denoted by [pic]. Recall that the random variable [pic]measures the distance from a particular value x to the mean (in units of standard deviation). Since the z-value corresponding to [pic]is [pic], we have

[pic]

or

[pic]

Now [pic]which is the area under the right hand curve to the left of [pic]. Since it is required that [pic]we must find the z-value corresponding to [pic], which is [pic](see Table 3). Then

[pic]

where [pic]. Substituting for [pic],

[pic]

The following two assumptions will allow us to calculate the appropriate sample size:

1 [pic].

2 The maximum value of [pic]will occur when [pic]. Since values of [pic]and [pic]are unknown, the use of [pic]will provide a valid sample size, although it may be slightly larger than necessary.

Then, solving for n, we obtain

[pic]

[pic]or [pic]. Hence, a common sample size for the researcher’s test will be [pic].

9.71 a The hypothesis to be tested is

[pic]

and the test statistic is

[pic]

The rejection region, with [pic] is one-tailed or [pic]and the null hypothesis is rejected. There is a difference in mean yield for the two types of spray.

b An approximate 95% confidence interval for[pic]is

[pic]

9.72 a Let [pic]be the proportion of cells in which RNA developed normally when treated with a .6 micrograms per millimeter concentration of Actinomysin-D, and [pic]be the proportion of normal cells treated with the higher concentration of Actinomysin-D. The hypothesis to be tested is

[pic]

Calculate [pic], [pic], and [pic]. The test statistic is

then

[pic]

and the p-value is [pic].

b Since the observed p-value < .0004, is must be less than [pic], and H0 is rejected. We can conclude that there is a difference in the rate of normal RNA synthesis for cells exposed to the two different concentrations of Actinomysin-D.

9.73 a The hypothesis to be tested is

[pic].

The test statistic is

[pic]

and the p-value is

[pic]

Since the p-value, .3576, is greater than [pic], and H0 cannot be rejected and we cannot conclude that the average verbal score for California students in 2005 is different from the national average.

b The hypothesis to be tested is

[pic].

The test statistic is

[pic]

and the p-value is

[pic]

Since the p-value, .6744, is greater than [pic], and H0 cannot be rejected and we cannot conclude that the average math score for California students in 2005 is different from the national average.

c Since the same students are used to measured verbal and math scores, there would not be two independent samples, and the two sample z-test would not be appropriate.

9.75 The hypothesis to be tested is

[pic]

and the test statistic is

[pic]

The rejection region with [pic]is [pic]. Since the observed value, [pic], does not fall in the rejection region and H0 is not rejected. The data do not provide sufficient evidence to indicate that the mean ppm of PCBs in the population of game birds exceeds the FDA’s recommended limit of 5 ppm.

9.76 Refer to Exercise 9.75, in which the rejection region was given as [pic]where

[pic]

Solving for [pic]we obtain the critical value of [pic]necessary for rejection of H0.

[pic]

The probability of a Type II error is defined as

[pic]

Since the acceptance region is [pic]from part a, [pic]can be rewritten as

[pic]

Several alternative values of [pic]are given in this exercise.

(Get a Minitab output using power calculation and make conclusions to do the followings)

Insert your output

a For [pic],(difference = 6-5=1 in Minitab dialog)

[pic]

and [pic].

b For [pic],

[pic]

and [pic].

c For [pic],

[pic]

For [pic],

[pic]

For [pic],

[pic]

For [pic],

[pic]

d The power curve is shown on the next page.

[pic]

You can see that the power becomes greater than or equal to .90 for a value of [pic]a little smaller than [pic]. To find the exact value, we need to solve for [pic] in the equation:

[pic]

From Table 3, the value of z that cuts off .10 in the lower tail of the z-distribution is [pic], so that

[pic]

9.78 a The hypothesis to be tested is

[pic]

and the test statistic is

[pic]

The rejection region, with [pic] is one-tailed or [pic]and the null hypothesis is rejected. There is sufficient evidence to indicate that the average height for males is greater than females.

b An approximate 99% one-sided confidence bound for[pic]is approximately

[pic]

Males are at least 4.03 inches taller than females on average.

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.

10.25 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.

10.27 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.

10.40 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].

10.43 a A paired-difference test is used, since the two samples are not random and independent (at any location, the ground and air temperatures are related). The hypothesis of interest is

[pic]

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

|Location |1 |2 |3 |4 |5 |Total |

|[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.

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

[pic]

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

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

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.

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