Question

Assume that, on average, healthy young adult’s dream 90 minutes each night, as inferred from a number of measures, including rapid eye movement (REM) sleep. An investigator wishes to determine whether drinking coffee just before going to sleep affects the amount of dream time. After drinking a standard amount of coffee, dream time is monitored for each of 28 healthy young adults in a random sample. Results show a sample mean, X, of 88 minutes and a sample standard deviation, s, of 9 minutes. (a) Use t to test the null hypothesis at the .05 level of significance (b) If appropriate (because the null hypothesis has been rejected), construct a 95 per-cent confidence interval and interpret this interval

Answer 1

Explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 90

For the alternative hypothesis,

µ < 90

If drinking coffee just before going to sleep affects the amount of dream time, then the amount of dream time would be less than 90 minutes. It means that it is left tailed.

Since the number of samples is 28 and no population standard deviation is given, the distribution is a student's t.

Since n = 28,

Degrees of freedom, df = n - 1 = 28 - 1 = 27

t = (x - µ)/(s/√n)

Where

x = sample mean = 88

µ = population mean = 90

s = samples standard deviation = 9

t = (88 - 90)/(9/√28) = - 1.176

We would determine the p value using the t test calculator. It becomes

p = 0.124

Since alpha, 0.05 < than the p value, 0.124, then we would not reject the null hypothesis. Therefore, At a 5% level of significance, the sample data did

not show significant evidence that drinking coffee just before going to sleep affects the amount of dream time.