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]).

first option

a function is a relation in which an input is paired with exactly one output.

so, let's find the right answer.

in our first option, our inputs are january, february, and march. the outputs are 30, 28, and 25. each input has it's own output. this is correct!

our second option we know is incorrect, because ms. valverade has two outputs, dogs and cats. this is not a function.

our third option is incorrect as well. the yellow square has two ouputs, the square and the triangle.

lastly, in our fourth option, we see the number 7 two times with a different output. therefore, this is not a function.

remember: each input must be paired with only one output!

d≈7.07in

step-by-step explanation: