Consider a class survey of 259 students at Queens Academy in
Consider a class survey of 259 students at Queen’s Academy in which students reported studying an average of 350 minutes on a typical weeknight. The sample standard deviation is found to be 67 minutes. Regard these students as an SRS from the population of all first-year students at this school.
Does the study give good evidence that students report studying less than six hours per weeknight on average?
A) Perform a hypothesis test, report the P-value, and state the conclusion (based on the = .01 significance level).
B) Find a 99% confidence interval for the mean amount of time that first-year Queen’s Academy students would report studying on a typical weeknight.
Solution
A.
Formulating the null and alternative hypotheses,
Ho: u = 360
Ha: u < 360
As we can see, this is a 1 tailed test.
Thus, getting the critical z,
zcrit = -2.326347874
Getting the test statistic, as
X = sample mean = 350
uo = hypothesized mean = 360
n = sample size = 259
s = standard deviation = 67
Thus, z = (X - uo) * sqrt(n) / s = -2.402011483
Also, the p value is
p = 0.008152598
Comparing z and zcrit (or, p and significance level), we REJECT THE NULL HYPOTHESIS.
There is a significant evidence that the students sleep less than six hours per weeknight on average.
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B.
Note that
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
X = sample mean = 350
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 67
n = sample size = 259
Thus,
Lower bound = 339.2763656
Upper bound = 360.7236344
Thus, the confidence interval is
(339.2763656 , 360.7236344) [ANSWER]

