question archive A recent report about sleep habits of Americans noted that sleep deprivation causes a number of problems, including highway deaths
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A recent report about sleep habits of Americans noted that sleep deprivation causes a number of problems, including highway deaths. The report claims that 53% of adult drivers admit to driving while drowsy. Suppose you want to test the hypothesis that more than 53% of the population of night shift workers admit to driving while drowsy. You select a sample of 800 night shift workers, of whom 452 admit to driving while drowsy. First state your null and alternative hypotheses and the proceed with the following questions.
34 What is the value of the test statistic?
a 1.98 b 1.85 c 1.76 d 1.69
35 Determine the p-value. The p-value is,
a 0.0643 At a 5% level of significance you do not conclude that that more than 53% of the population of night shift workers admit to driving while drowsy
b 0.0643 At a 10% level of significance conclude that that more than 53% of the population of night shift workers admit to driving while drowsy
c 0.0239 At a 1% level of significance do not conclude that that more than 53% of the population of night shift workers admit to driving while drowsy
d 0.0239 At a 5% level of significance conclude that that more than 53% of the population of night shift workers admit to driving while drowsy
e Both a and b are correct.
f Both c and d are correct.
?H0?:π≤0.53Ha?:π>0.53?
34) a 1.98
35) fBoth c and d are correct.
Step-by-step explanation
You want to test the hypothesis that more than 53% of the population of night shift workers admit to driving while drowsy
This can be written as ?π>0.53?
Since we cannot have strict inequality in the null hypothesis, this becomes the alternate hypothesis and the opposite becomes the null hypothesis Therefore, we get
?H0?:π≤0.53Ha?:π>0.53?
Question 34
Sample proportion = 452/800 = 0.565 = p
Standard deviation = ?σ=nπ(1−π)??=8000.53(1−0.53)??=0.0176?
Test statistic = ?σp−π?=0.01760.565−0.53?? = 1.98
Question 35
Since this is a right tailed test,
p-value = P(z > test statistic) = P(z > 1.98) = 1 - P(?z≤1.98? ) = 1 - 0.97615 = 0.0239
At ?α? = 0.01, p-value > 0.01. Therefore, we fail to reject the null hypothesis.
That is,
At a 1% level of significance do not conclude that that more than 53% of the population of night shift workers admit to driving while drowsy
At ?α? = 0.05, p-value < 0.05. Therefore, we reject the null hypothesis.
At a 5% level of significance conclude that that more than 53% of the population of night shift workers admit to driving while drowsy
These are options c and d. Both are correct.