Question

a.

In general, high school and college students are the most pathologically sleep-deprived segment of the population. Their alertness during the day is on par with that of untreated narcoleptics and those with untreated sleep apnea. Not surprisingly, teens are also 71 percent more likely to drive drowsy and/or fall asleep at the wheel compared to other age groups. (Males under the age of twenty-six are particularly at risk.)

The accompanying data set represents the number of hours 25 college students at a small college in the northeastern United States slept and is from a random sample. Enter this data into C1 of Minitab Express.

6 9 7 7 6 7 7 5 8 6 6 6 8 8 8 5 4 6 7 8 5 8 7 6 7

For the analyses that follow, we shall use

·         90%, 95%, and 99% as the confidence levels for the confidence interval.

·      5% as the level of significance ( ) for the hypothesis test.

·         7 hours sleep as the null hypothesis (according to The Sleep Foundation).

l.     Using a 5% level of significance, α = 0.05, make a statistical DECISION regarding the plausibility of the hypotheses; that is, would you reject or fail to reject the null hypothesis? Justify your answer.

a.    Describe what the p-value measures in the context of this study. This is also referred to as “interpreting the p-value.” S

Answer 1

We have Run the problem in MINITAB and get the output as

the confidence interval as ·         90%, 95%, and 99% as the confidence levels for the confidence interval.

95% CI
(6.178, 7.182)

90% CI

(6.264, 7.096)

99% CI

(6.000, 7.360)

Now we will test

H0: mean hour of sleep = 7

vs

H1: not H0

the test output from MINITAB is

Variable	N	Mean	StDev	SE Mean	95%	CI	T	P
data	25	6.680	1.215	0.243	(6.178,	7.182)	-1.320	0.200

Using a 5% level of significance, α = 0.05, make a statistical DECISION regarding the plausibility of the hypotheses; that is, would you reject or fail to reject the null hypothesis? Justify your answer.

so we will reject H0 is |T|> t_n-1,0.05/2

The 95% confidence the critical point is t_n-1,0.05/2= 2.06390

here |T|=1.32 <2.0639

then will accept H0 and claim that the mean sleep hour is 7hour

Describe what the p-value measures in the context of this study.

From the table of test output we can note that p-value = 0.20 >0.05 , hence by the p-value also we can claim the mean is 7 at 95% confidence.