The average number of days required by a college’s undergraduate students to complete a project was 22 days in the previous year. The college administration wishes to determine whether the average differs this year. A sample of 20 students produced a sample mean of 20 days this year, with a sample standard deviation of 6 days. Assume that the number of days to complete the project is normally distributed. Is there enough evidence at the 5% significance level to determine whether the mean has increased? a. No, because the test statistic falls outside the rejection region [1.645, [infinity]).
b. Yes, because the test statistic falls in the rejection region (-[infinity],-1.96].
c. No, because the test statistic falls outside the rejection region [1.960, [infinity]).
d. No, because the test statistic falls outside the rejection region [1.729, [infinity]).
e. None of the above

Step-by-step explanation:

We would set up the hypothesis test.

For the null hypothesis,

H0: µ = 22

For the alternative hypothesis,

H1: µ ≠ 22

This is a two tailed test

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

Since n = 20,

Degrees of freedom, df = n - 1 = 20 - 1 = 19

t = (x - µ)/(s/√n)

Where

x = sample mean = 22

µ = population mean = 20

s = samples standard deviation = 6

t = (22 - 20)/(6/√20) = 1.49

For a 5% level of significance, the critical value is determined from the t test distribution table.

α/2 = 0.05/2 = 0.025

From the t distribution table, the rejection region is - 1.729 and 1.729

hypothesis.

In order to reject the null hypothesis, the test statistic must be smaller than - - 1.729 or greater than 1.729

Since - 1.49 > - 1.729 and 1.49 < 1.729, we would fail to reject the null hypothesis. Therefore, the correct option is

d. No, because the test statistic falls outside the rejection region [1.729, [infinity]).

ok

step-by-step explanation:

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step-by-step explanation: