_____ 18. The jobless rate for Nevada in July 2010 was 13%. In a random sample o
ID: 3369196 • Letter: #
Question
_____ 18. The jobless rate for Nevada in July 2010 was 13%. In a random sample of 50 employable
adults in one city, 8 were unemployed. Which is true?
a. p = 0.16 and ˆp = 0.13
b. The 95% confidence interval for this city’s jobless rate is 0.044 to 0.235.
c. The sample size is not large enough to use a Normal model for the sampling
distribution.
d. This situation does not meet the 10% Condition for using a Normal model to model
the sampling distribution.
_____ 20. Nationally, 29% of high school students report riding in a car driven by someone who had been drinking during the 30 days preceding their survey. A state youth council believes this proportion is lower in their state where there had been an intense media campaign against riding with drivers who have been drinking. A survey of randomly chose students from across the state shows 304 of 1214 students answering yes to this question. What are the hypotheses in this survey?
a. H0: p = 0.29; HA: p < 0.29
b. H0: p = 0.29; HA: p > 0.29
c. H0: p = 0.71; HA: p > 0.71
d. H0: p = 0.71; HA: p < 0.71
_____ 21. A large company randomly samples 5% of its customers and compares their rate of satisfaction to last year’s satisfaction rate. The z-score for the observed satisfaction rate in this sample is z = 2.12. What should the company conclude?
a. The satisfaction rate has only increased.
b. The satisfaction rate has only decreased.
c. The satisfaction rate has changed.
d. There has been no change in the satisfaction rate.
Explanation / Answer
Solution 19:
Option d is correct.
We have p = 0.13 and p^ = X/n = 8/50 = 0.16
Since np = 50*0.13 = 6.5 and n (1 - p) = 50*(1-0.13) = 43.5 both are not greater 10.
This situation does not meet the 10% Condition for using a Normal model to model
the sampling distribution.
Solution 20:
Option a is correct.
Null Hypothesis (Ho): p = 0.29
Alternative Hypothesis (Ha): p < 0.29
Solution 21:
Option a is correct.
Using Z-tables, the z-score at 0.05 is 1.645 and the observed z-score is now 2.12. Therefore, satisfaction rate has increased.
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