Oct. 29, 2003 ANSWER KEY
ANSWERS ON ATTENDANCE QUIZ: all = B except #2, 4, and 8 = A
Activity 1: During the summer of 2002, about 70% of the general public has said they approve of
President Bush’s job performance. Historically, college students have been less approving of the president’s job performance than the general public.
In a Stat 200 survey done in late August of 2002, 130 of 209 students (62.2%) said they approved of
President Bush’s job performance. a. Consider p = proportion of college students who approved of President Bush’s job performance Using appropriate statistical notation, write null and alternative hypotheses for testing whether p was the same as it was for the general public or whether it was less than the value for the general public. b. Use Minitab to conduct the test of hypothesis in this situation. The steps are:
(1) Use Stat>Basic Stats>1-proportion, (2) Click on “Summarized Data”, enter 209 as “Number of trials” and “130” as “Number of Successes”, (3) click the Options button, enter the null value of p where it says “Test proportion,” AND select the appropriate type of alternative hypothesis AND click on “Use test and interval based on normal distribution”. Finally, click OK enough times to get the output.
What is the value given in the output for the z-statistic? _________
What value is given for the P-value? _____________
Based on this p -value, which hypothesis is favored? ________________
State a conclusion in the context of the “real world” situation. That is, what do the data indicate about how students compare to the general population? d. The z-statistic is computed as: z

Sample
Null p -
Std
Null p
Error
 pˆ
 p
0 p
0
( 1
 p
0
) n
Give values for pˆ , p
0
, and n. e. Refer to the previous part. Draw a sketch of a bell curve that shows the connection between the value of z (

2.46) and the p -value (0.007) in this problem.
--------------------------------------------------------------------------
ANSWERS: a. H
0
: p = .70 versus H a
: p < .70 b. z =

2.46
p-value = 0.007
reject the null (decide alternative)
College students are less likely to approve of the President’s job performance.
No part c d. p-hat = .622 p
0
= .70 n = 209 e. Draw a bell curve. Show the p-value as a shaded area under the curve to the left of

2.46 (less than region)

Activity 2 : On Oct. 1, 2002 the USA Today newspaper had a graph showing that 34% of U.S. high school females think they are overweight. In a Stat 200 survey we asked a question about how students felt about their weight. Among 113 women who responded, 35 women said they felt overweight. a. Let p = proportion of Penn State women who feel overweight. Use the Stat 200 survey data to test whether p = .34 (as for high school females) or whether p is not .34.
Step 1: Write the null and alternative hypotheses.
Step 2: Check necessary conditions and compute the test statistic. To compute z , use Minitab (almost) as you did for Activity 1
What are the values of np
0 and n (1
 p
0
)? Are these values sufficiently large?
What are we assuming about the sample?
Value of z -statistic = ___________ (from output).
Step 3: p -value = _______ (from output)
Step 4: Make a conclusion abut the null and alternative hypotheses
Step 5:
State a conclusion about the “real world” situation. b. Draw a sketch of a bell curve that illustrates the connection between the value of z and the p -value in this situation. c. The Minitab output will include a 95% confidence interval for p, the population proportion.
What is the interval given?
Explain why this interval is evidence that the null hypothesis cannot be rejected n this situation. c. Suppose the alternative hypothesis had been Ha: p < .34. What would be the p-value for this onesided test?
ANSWERS a. Step 1 H
0
: p = .34 versus H a
: p

.34
Step 2: np
0
= 113(.34), n(1
 p
0
) = 113(1-.34) both bigger than 10.
We assume the sample is a random sample.
Z =

0.68
Step 3: p-value = 0.497
Step 4: Cannot reject the null
Step 5: We cannot reject the possibility that 0.34 is the proportion of college women who think they are overweight. b. Draw a bell curve. P-value is the combined area to the left of -0.68 and to the right of +0.68.
1 st
part c: interval is .225 to .395
Inteval includes .34, the null value, so it cannot be rejected as a possible value of p.
2 nd
part c: .497/2 = .2485
It would be the area to the left of -0.68.