The manager of a fund that provides loans for college students has estimated that the average monthly loan repayment for students borrowing from the fund is $75.00. You are to test this estimate. You take a sample of 20 students and find that the mean monthly payment is $69.46 with a standard deviation of $9.78. Which of the following statements is true about this test?

a. The value of the test statistic is -.57; therefore, the null hypothesis is rejected for mean = 0.02.

b. The value of the test statistic is - 2.53; therefore, the null hypothesis is rejected for mean = 0.02.

c. The value of the test statistic is - 2.53; therefore, the null hypothesis is rejected for mean = 0.05 but not for mean = 0.02.

d. The value of the test statistic is -.57; therefore, the null hypothesis is rejected for mean = 0.05 but not for mean = 0.02.

Question

Mathematics

Macie Werner

12 April, 02:06

The manager of a fund that provides loans for college students has estimated that the average monthly loan repayment for students borrowing from the fund is $75.00. You are to test this estimate. You take a sample of 20 students and find that the mean monthly payment is $69.46 with a standard deviation of $9.78. Which of the following statements is true about this test?

a. The value of the test statistic is -.57; therefore, the null hypothesis is rejected for mean = 0.02.

b. The value of the test statistic is - 2.53; therefore, the null hypothesis is rejected for mean = 0.02.

c. The value of the test statistic is - 2.53; therefore, the null hypothesis is rejected for mean = 0.05 but not for mean = 0.02.

d. The value of the test statistic is -.57; therefore, the null hypothesis is rejected for mean = 0.05 but not for mean = 0.02.

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Answers (1)

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Giuliana Dixon · Answer 1

Step-by-step explanation:

We would set up the hypothesis test.

For the null hypothesis,

µ = 75

For the alternative hypothesis,

µ ≠ 75

Since the number of samples is 20 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 = $69.46

µ = population mean = $75

s = samples standard deviation = $9.78

t = (69.46 - 75) / (9.78/√20) = - 2.53

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

p = 0.01

Since alpha, 0.05 > than the p value, 0.01, then the null hypothesis is rejected.

Therefore,

The value of the test statistic is - 2.53; therefore, the null hypothesis is rejected for level of significance = 0.05