University of Massachusetts Amherst



Stat 501 HW#2 & #3

Probelm 4.

(a) One-Sample T: Prob4-tb4.2

Variable N Mean StDev SE Mean 99% CI

Prob4-tb4.2 25 48.00 10.30 2.06 (42.24, 53.76)

The 99% confidence interval for the average time is (42.24, 53.76).

(b) One-Sample T: Prob4-tb4.2

Test of mu = 50 vs not = 50

Variable N Mean StDev SE Mean 99% CI T P

Prob4-tb4.2 25 48.00 10.30 2.06 (42.24, 53.76) -0.97 0.341

Since P=0.341, at significance level of 0.05, the mean is not significantly different from 50 hours.

(c) Power and Sample Size

1-Sample t Test

Testing mean = null (versus not = null)

Calculating power for mean = null + difference

Alpha = 0.05 Assumed standard deviation = 10.3

Sample

Difference Size Power

2 25 0.153994

The power is only 0.153 for a difference of 2 hours.

Power Curve

[pic]

Problem 4.7

(a) One-Sample Z

Test of mu = 40 vs > 40

The assumed standard deviation = 11

95% Lower

N Mean SE Mean Bound Z P

30 44.00 2.01 40.70 1.99 0.023

Conclusions: As P=0.023 < 0.05, the mean is significantly greater than 40 pounds.

So, I would support the program's claim.

(b) Power and Sample Size

1-Sample t Test

Testing mean = null (versus > null)

Calculating power for mean = null + difference

Alpha = 0.05 Assumed standard deviation = 11

Sample Target

Difference Size Power Actual Power

3 85 0.8 0.802124

The power is 0.802 for a mean difference of 3 pounds.

Power Curve for 1-Sample t Test

[pic]

(c) Power Curve for 1-Sample t Test

Power and Sample Size

1-Sample t Test

Testing mean = null (versus > null)

Calculating power for mean = null + difference

Alpha = 0.05 Assumed standard deviation = 11

Sample

Size Power Difference

30 0.8 5.11440

Power Curve for 1-Sample t Test [pic]

Problem 4.8

One-Sample T: Prob8-tb4.5

Test of mu = 65 vs > 65

99% Lower

Variable N Mean StDev SE Mean Bound T P

Prob8-tb4.5 12 69.33 7.58 2.19 63.38 1.98 0.037

Conclusions: Significantly greater than 65 mils/hr at 5% level, but not a 1% level.

Problem 4.9

(a) One-Sample T

Test of mu = 70 vs not = 70

N Mean StDev SE Mean 99% CI T P

19 69.700 2.168 0.497 (68.268, 71.132) -0.60 0.554

Conclusion: Since P=.554 >>0.05, the mean is not significantly different from 70 even at 0.5 level of significant.

(b) One-Sample T

N Mean StDev SE Mean 99% CI

19 69.700 2.168 0.497 (68.268, 71.132)

A 99% CI for the true population mean is (68.268, 71.132).

Problem 4.10

(a) Inverse Cumulative Distribution Function

Student's t distribution with 27 DF

P( X  1.70) = 0.05.

(b) Inverse Cumulative Distribution Function

Student's t distribution with 27 DF

P( X  1.313) = 0.10

(c) Inverse Cumulative Distribution Function

Student's t distribution with 27 DF

P( X  0.9

95% Lower Exact

Sample X N Sample p Bound P-Value

1 484 528 0.916667 0.894146 0.112

Conclusions: P=.112, the approval proportion is not significantly greater than 90% at 0.05 level.

Problem 4.14

(a) Test and CI for One Proportion

Test of p = 0.8 vs p > 0.8

95% Lower Exact

Sample X N Sample p Bound P-Value

1 1274 1286 0.990669 0.984925 0.000

Conclusion: Since P=0.000 which is extremely small, the evidence is very strong to suggest that more than 80% of high school math teachers use calculators in their classroom.

(b) Power and Sample Size

Test for One Proportion

Testing proportion = 0.8 (versus > 0.8)

Alpha = 0.05

Alternative Sample Target

Proportion Size Power Actual Power

0.85 286 0.7 0.700377

The sample size required is 286.

Power Curve for Test for One Proportion

[pic]

Problem 4.15

Test and CI for Two Proportions

Sample X N Sample p

1. (hi-sch) 4988 9275 0.537790

2 (coll) 5245 10286 0.509916

a) Proportion of Hi-school graduates earning more than 40K

= 0.537790

(b) Proportion of college graduates earning more than 40K

=0.509916

(c)Test and CI for Two Proportions

Difference = p (1) - p (2)

Estimate for difference: 0.0278734

95% lower bound for difference: 0.0161158

Test for difference = 0 (vs < 0): Z = 3.90 P-Value = 0.999

Fisher's exact test: P-Value = 0.999

Conclusions: Since P=0.99, proportion for college graduates who earn more than 40K can not be significantly higher than that for high school graduates.

Problem 4.16

(a) Descriptive Statistics: prob16-tb4.8

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3

prob16-tb4.8 24 0 15.392 0.136 0.667 13.800 14.900 15.350 15.800

Variable Maximum

prob16-tb4.8 17.100

Answer:: Mean=15.392, SD=0.667

(b) One-Sample T: prob16-tb4.8

Test of mu = 15 vs not = 15

Variable N Mean StDev SE Mean 99% CI T P

prob16-tb4.8 24 15.392 0.667 0.136 (15.009, 15.774) 2.88 0.009

Conclusions: As P-value=0.009 0.05 implies that the average distance of accidents from home is not significantly less than 25 miles.

Power and Sample Size

(c)

1-Sample t Test

Testing mean = null (versus < null)

Calculating power for mean = null + difference

Alpha = 0.05 Assumed standard deviation = 13.37

Sample Target

Difference Size Power Actual Power

-5 46 0.8 0.803201

Answers: It needs a sample of size 46 to detect a difference of 5 miles with probability 0.80.

Power Curve for 1-Sample t Test

[pic]

Problem 4.21

(a) Descriptive Statistics: Weight Before, Weight After

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3

Weight Before 24 0 182.21 5.97 29.24 129.00 164.75 176.50 203.75

Weight After 24 0 178.25 5.79 28.38 125.00 159.25 174.50 203.50

Variable Maximum

Weight Before 254.00

Weight After 238.00

Answers:

Weight Before : mean=182.21 pounds, SD=29.24 pounds;

Weight After: mean=178.25 pounds, SD=28.38 pounds.

(b) Before - After

0

4

-2

2

8

6

16

8

2

3

3

2

9

4

4

6

-1

4

7

1

0

2

6

1

(c)

Paired T-Test and CI: Weight Before, Weight After

Paired T for Weight Before - Weight After

N Mean StDev SE Mean

Weight Before 24 182.21 29.24 5.97

Weight After 24 178.25 28.38 5.79

Difference 24 3.958 3.906 0.797

95% lower bound for mean difference: 2.592

T-Test of mean difference = 0 (vs > 0): T-Value = 4.96 P-Value = 0.000

Conclusion: The average loss is 3.958 pounds, with P-value of 0.000 (almost 0), we can conclude

that participants of the weight loss program lost significant amount of weight in average.

(d)

One-Sample T: Before-After

Test of mu = 0 vs > 0

99% Lower

Variable N Mean StDev SE Mean Bound T P

Before-After 24 3.958 3.906 0.797 1.965 4.96 0.000

Conclusions are the same as that in (c).

Problem 4.22

(a) Use Calc to find difference first No. in 2002 minus No. in 1997 = column (Difference)

Do 95% confidence interval of the differences.

One-Sample T: Difference

Variable N Mean StDev SE Mean 95% CI

Difference 12 17704 25392 7330 (1570, 33837)

(b)

One-Sample T: Difference

Variable N Mean StDev SE Mean 98% CI

Difference 12 17704 25392 7330 (-2220, 37627)

(c) The 95% confidence interval is narrower than the 98% interval, and

the former does not include 0, while the latter includes 0.

(d) One-Sample T: Difference

Test of mu = 0 vs not = 0

Variable N Mean StDev SE Mean 98% CI T P

Difference 12 17704 25392 7330 (-2220, 37627) 2.42 0.034

Conclusions: Since P=0.034 0.01, the mean differences are not significantly different at 0.01 level, though

it is significantly different at 0.05 level.

(f)

Paired T-Test and CI: Number of Businesses 2002, Number of Businesses 1997

Paired T for Number of Businesses 2002 - Number of Businesses 1997

N Mean StDev SE Mean

Number of Businesses 200 12 402087 320011 92379

Number of Businesses 199 12 384384 314465 90778

Difference 12 17704 25392 7330

99% lower bound for mean difference: -2220

T-Test of mean difference = 30000 (vs > 30000): T-Value = -1.68 P-Value =

0.939

Conclusion: No, since P=.939, too large a P-value. The average of 2002 is not significantly greater

than the average number in 1997 by more than 30,000.

(g) Comments on (d) and (e): whether significantly different or not depends on the level of significance.

Larger level of significance will lead more likely to rejecting the null hypothesis.

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