Question

A "sleep habits" survey answered by 48 randomly selected New Yorkers contained the question "How much sleep do you get per night?" The sample average was 7.85 hours, with a corresponding sample standard deviation of 0.82 hours. We want to test against the null hypothesis that New Yorkers get, on average, 8 hours of sleep per night. α=0.05.
b. The t test statistic is: _______ (Round your answer to 3 decimal places)

c. The approximate 95% CI is: ______ to ______ (Round your answer to 3 decimal places)

d. The statistical decision and its justification are:

(You have two attempts at this question. You should check that your previous answers are correct before answering this question.)

|Test statistic| < 2 and 95% CI does not contain null value, so reject H₀

|Test statistic| < 2 and 95% CI contains null value, so reject H₀   

|Test statistic| > 2 and 95% CI does not contain null value, so reject H₀

|Test statistic| > 2 and 95% CI contains null value, so reject H₀

|Test statistic| < 2 and 95% CI does not contain null value, so fail to reject H₀|

Test statistic| < 2 and 95% CI contains null value, so fail to reject H₀

|Test statistic| > 2 and 95% CI does not contain null value, so fail to reject H₀

|Test statistic| > 2 and 95% CI contains null value, so fail to reject H₀

Answer 1

Answer: A "sleep habits" survey answered by 48 randomly selected New Yorkers contained the question "How much sleep do you get per night?" The sample average was 7.85 hours, with a corresponding sample standard deviation of 0.82 hours. We want to test against the null hypothesis that New Yorkers get, on average, 8 hours of sleep per night. α=0.05.

Solution:

n = 48

x̄ = 7.85

s = 0.82

α = 0.05

The hypothesis test:

Ho: μ = 8

Ha: μ ≠ 8

Test statistic Z:

Z = x̄ - μ / s / √n

Z = 7.85 - 8 / 0.82 / √48

Z = -1.2674

Test statistic Z = - 1.267

At 95% confidence interval, α = 0.05

Z critical = Z(α/2)

Z critical = 1.96

The 95% confidence interval :

CI = x̄ ± Z critical * s/√n

CI = 7.85 ± 1.96 * 0.82 / √48

CI = 7.85 ± 0.232

CI = (7.618, 8.082)

Therefore, the 95% confidence interval is between 7.618 and 8.082.

Decision:

|Test statistic| > 2 and 95% CI contains null value, so fail to reject Ho.