A facilities manager at a university reads in a research report that the mean amount of time spent in the shower by an adult is

Question

3 years ago

11

A facilities manager at a university reads in a research report that the mean amount of time spent in the shower by an adult is

5 minutes. He decides to collect data to see if the mean amount of time that college students spend in the shower is significantly different from 5 minutes. In a sample of 11 students, he found the average time was 4.52 minutes and the standard deviation was 0.75 minutes. Using this sample information, conduct the appropriate hypothesis test at the 0.1 level of significance. Assume normality. (can you please show how to do this without a calculator or excel i just dont want answer but want to know how to do it).
a) What are the appropriate null and alternative hypotheses?
A) H0: μ = 5 versus Ha: μ < 5
B) H0: μ = 5 versus Ha: μ ≠ 5
C) H0: x = 5 versus Ha: x ≠ 5
D) H0: μ = 5 versus Ha: μ > 5
b) What is the test statistic? Give your answer to four decimal places.
c) What is the P-value for the test? Give your answer to four decimal places.
d) What is the appropriate conclusion?
A) Fail to reject the claim that the mean time is 5 minutes because the P-value is larger than 0.01.
B) Reject the claim that the mean time is 5 minutes because the P-value is larger than 0.01.
C) Reject the claim that the mean time is 5 minutes because the P-value is smaller than 0.01.
D) Fail to reject the claim that the mean time is 5 minutes because the P-value is smaller than 0.01.

Mathematics

1 answer:

Zolol [24] · Answer 1 · 2022-02-06T10:08:24+03:00

Answer:

A) Null Hypothesis;H0: μ = 5

Alternative Hypothesis;Ha: μ ≠ 5

B) test statistic = -2.1226

C) p-value = 0.0598

D) Option A is correct

Step-by-step explanation:

We are given;

x = 4.52 minutes

s = 0.75 minutes

μ = 5 minutes

n = 11

degree of freedom = n - 1 = 11 - 1 = 10

A) The hypotheses are;

Null Hypothesis;H0: μ = 5

Alternative Hypothesis;Ha: μ ≠ 5

B) t-statistic = (x - μ)/(s/√n)

(4.52 - 5)/(0.75/√11) = -2.1226

C) From the t-score calculator results attached, the p-value is approximately 0.0598.

D) The P-value of 0.0598 is is greater than the significance level of 0.01, thus we fail to reject the null hypothesis, and we say that the result is statistically nonsignificant. So option A is correct.