Statistical Analysis: Confidence Intervals and Hypothesis Testing, Exams of Data Analysis & Statistical Methods

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1. An interview with 221 randomly chosen investors revealed that 73 believed in “momentum investing,” investing only when the market has been going up regularly. a) Describe the parameter of interest in this situation. b) Set a 95% confidence interval on this parameter. Use the plus-4 method.

ANS: .333 .

c) How large a sample would be necessary in order that the margin of error of your 95% confidence interval be only .03? You may assume you have the results above to aid you. ANS: n = 944 d) In part c), suppose no preliminary estimate was available. Determine a sample size that would be adequate regardless of the true value of p.

ANS: n= 1068

2. To estimate the average amount spent per week by students on lattes/coffee

at CWU, a random sample of 64 students was taken; the summary data are

recorded below.:

X =$14.86, S= $3.

a) What is the parameter of interest in this problem? Be specific!

b) Find the best point estimate of this parameter.

ANS: X = 14.

c) Find a 95% confidence interval on this parameter.

ANS: 14.86  .78 or the interval is 14.08 <^ ^ < 15.

d) Test, as easily as possible, the hypothesis that average amount spent is $

against the alternative that it is not $12. Use a 5% level of significance.

ANS: reject H 0 : because $12 is not in the 95% C.I. Conclude^ ^ not

only greater than 12, but also^ ^ > 14.08.

e) The 95% margin of error in the above interval was .78. How large a sample

would have been needed in order to have the margin of error of a 90%

confidence interval be .75?

Math 311 Fall 2008 Some Review Problems for Final

ANS: Use n = 50

3. To estimate the average beginning salary for actuarial graduates in CY 2008 (who have passed two

exams), a random sample of size 6 produced the following summary statistics: S^ 4.8^  x ^344

(data are in $000). a) Assuming that the beginning salary of actuarial graduates follows a normal

distribution, use these data to set a 90% confidence interval on the true unknown mean salary of all

beginning actuarial graduates who have two exams by the time they graduate.

ANS: 57.333  2.015* 4.8/ 6 =57.333  3.95, or $53383 <^ ^ < $

b) Test H 0 :^ ^ = 62000 against Ha:^ ^ < 62000. Use^ ^ = .05.

ANS: Use appropriate side of your two-sided 90% C.I. as a one-sided 95% C.I. Here, it is^ ^ < $61283. Because

’62,000’ is not in this interval, we reject H 0 : and conclude^ ^ < 62000, but we can “sharpen” this to^ ^ < 61283.

4. 80 randomly chosen individuals who are severely overweight were put on two weight-loss diets: A and B---40 for each diet. At the conclusion of a 12-week period, the results below were obtained: Diet A Diet B

X = 24.4 lbs^ Y = 20.8 lbs

S = 2.44 S = 2.15 lbs

a) Find the best point estimate for ^ A – ^ B.

Ans: 3.6 lbs.

b) Set a 95% confidence interval on ^ A – ^ B.

ANS: 3.6  1.

5.954 4.

 = 3.6^ ^ 1.01 or^ 2.59 <^ ^ A –^ ^ B < 4.

c) Test H 0 : ^ A – ^ B =0 versus Ha: ^ A – ^ B 0. Use^ ^ =.05.

ANS: Use your 95% C.I. Check for ‘0’ in the interval. Not there! Conclude ^ A – ^ B > 0, but better yet,

conclude ^ A – ^ B > 2.59.

d) Test H 0 : ^ A – ^ B ≤ 2.2 against Ha: ^ A – ^ B > 2.2 Use^ ^ =.025.

ANS: Use your 97.5% C.I. obtained by choosing the bottom half (because it reads in the same direction as

Ha: ) of the interval in part b: 2.59 < ^ A – ^ B. Because 2.2 is not in this interval, reject H 0 : and conclude that

 A –  B > 2, and you can sharpen this to  A –  B > 2.

5. A 90% confidence interval on a population mean yields 21.8  0.88. What is the name for ‘0.88’?

6. You wish to choose the brightest color white paint for your house, and you have narrowed your search to two competing brands, A and B. You find someone to measure the brightness of both paints several times and the results are shown below, with higher values indicating more brightness. Paint A Paint B n 1 =16 n 2 = 21

X 1 = 104.4^ X 2 =97.

S 1 = 2.5 S 2 =1.

a) What assumptions are necessary in order to set a confidence interval on^ ^1 -^ ^2?

ANS: Both populations must be normally distributed

b) Set an ordinary two-sided 90% confidence interval on^ ^1 -^ ^2.

ANS: 6.8  1.753 (.738)= 6.8  1.29 or 5.51 <^ ^ 1 -^ ^ 2 < 8.

c) Test H 0 :^ ^1 –^ ^2 ≤ 3 against Ha:^ ^1 –^ ^2 > 3. Use^ ^ = .05. Use a confidence interval!

ANS: The one-sided 95% C.I> is^ ^1 -^ ^2 > 5.51. Because 3 is not in this interval, we reject H 0 : and

conclude^ ^1 -^ ^2 > 3, and in fact conclude^ ^1 -^ ^2 > 5.

d) Test H 0 :^ ^1 -^ ^2 ≤ 5 against Ha:^ ^1 -^ ^2 >5 by using a test statistic. Calculate or bracket the P-

value as appropriate, and state your conclusion for^ ^ = .10.

ANS: The test stat is (^2 2 )

* 2.

2.5 1.

T



 



. The degrees of freedom to use in deciding how significant this result is are provided by the smaller of the two d.f., namely 15. Checking your t- tables at 15 d.f., we see that 2.438 falls between 2.249 and 2.602, which means the best we can say is

that .01 < P-value <.02. But we can reject for any predetermined^ ^ greater than the P-value, so

we can easily reject at^ ^ = .10. Note that if someone wanted to make a decision for^ ^ = .014, we

could not be sure of our conclusion without using a computer to get more precise values from the t- distribution.

7. A random sample of size 81 is taken from the Snorbel distribution, a highly skewed distribution used in studying the fertility rites of Amazon chickens. The population mean of this distribution is 14.1 and the population standard deviation is 2.2.

a) Tell very specifically what distribution you would use in modeling the sampling distribution of X

b) What is the probability that an X for a random sample of size 81 would exceed 14.4?

Ans:

14.4 14.

[ 14.4] [ ] [ 1.23].

.

P X P Z P Z



     

8. A high fidelity dealer is comparing two defective rates for two brands of Stereo Receivers. He samples 20 of brand 1 and 16 of brand 2, and finds that 6 of the 20 from brand 1 are defective out of the box, while

only 2 of the 16 from brand 2 are defective. Letting p 1 represent the true unknown defective rate for all

receivers of this model from brand 1, and p 2 be the same thing for brand 2, answer the following:

a) Set a 90% confidence interval on p 1^ - p 2^.

ANS: p ^1 = .318 p ^2 = .167. The 90% confidence interval is

(.364)(.636) (.167)(.833)

(.318 .167) 1.645 .151.

     or -.071 < p 1^ - p 2^ < .373.

b) Test H 0 : p 1 - p 2 =0 against Ha: p 1 - p 2  0. Use^ ^ = .10.

ANS: Use your 90% two-sided C.I. Because ‘0’ is in this interval, cannot reject H 0 :.

c) Test H 0 : p 1 - p 2 ≤ -.1 against Ha: p 1 - p 2 > -.1. Use^ ^ =.

Ans: Use the 95% one-sided interval -.071 < p 1^ - p 2^ and see if the hypothesized value of -.1 lies

within the interval. It does not, so reject H 0 : and conclude that p 1^ - p 2^ > -.1, and in fact can

conclude p 1^ - p 2^ > -.071.