According to the Centers for Disease Control and Prevention (February 18, 2016),

Question

According to the Centers for Disease Control and Prevention (February 18, 2016), 1 in 3 American adults don’t get enough sleep. A researcher wants to determine if Americans are sleeping less than the recommended 7 hours of sleep on weekdays. He takes a random sample of 150 Americans and computes the average sleep time of 6.7 hours on weekdays. Assume that the population is normally distributed with a known standard deviation of 2.1 hours. Test the researcher’s claim at = 0.01. (You may find it useful to reference the appropriate table:z table or t table)

a. Select the relevant null and the alternative hypotheses.
  

H0: = 7; HA: 7

H0: 7; HA: 7

H0: 7; HA: < 7

b. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)

Test statistic = ?

c. Find the p-value.
  

p-value < 0.01

0.01 p-value < 0.025

0.025 p-value < 0.05

0.05 p-value < 0.10

p-value 0.10

d. What is the conclusion at = 0.01?

Reject H0 since the p-value is greater than .

Reject H0 since the p-value is smaller than .

Do not reject H0 since the p-value is greater than .

Do not reject H0 since the p-value is smaller than .

e. Make an inference.

There is insufficient evidence to suggest that Americans sleep less than the recommended 7 hours of sleep.

There is sufficient evidence to suggest that Americans sleep less than the recommended 7 hours of sleep.

Explanation / Answer

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 7.0
Alternative hypothesis: u < 7.0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.17146
DF = n - 1

D.F = 149

b)

t = (x - u) / SE

t = - 1.75

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of - 1.75.

c) 0.025 p-value < 0.05

Thus the P-value in this analysis is 0.041.

Interpret results. Since the P-value (0.041) is greater than the significance level (0.01), we cannot reject the null hypothesis.

d) Reject H0 since the p-value is greater than .

e) There is insufficient evidence to suggest that Americans sleep less than the recommended 7 hours of sleep.

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