Answer to Question #347947 in Statistics and Probability for Criceljoice

Question #347947

An association of City Mayors conducted a study to determine the average number of times a family went to buy necessities in a week. They found that the mean is 4 times in a week. A random sample of 25 families were asked and found a mean of 5 times in a week and a standard deviation of 2. Use 1% significance level to test that the population mean is greater than 4 hours. Assume that the population is normally distributed.

1
Expert's answer
2022-06-07T09:35:42-0400

The following null and alternative hypotheses need to be tested:

"H_0:\\mu\\le4"

"H_1:\\mu>4"

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is "\\alpha = 0.01," "df=n-1=24" and the critical value for a right-tailed test is "t_c =2.492159."

The rejection region for this right-tailed test is "R = \\{t:t>2.492159\\}."

The t-statistic is computed as follows:


"t=\\dfrac{\\bar{x}-\\mu}{s\/\\sqrt{n}}=\\dfrac{5-4}{2\/\\sqrt{25}}=2.5"


Since it is observed that "t=2.5>2.492159=t_c," it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, "df=24" degrees of freedom, "t=2.5" is "p=0.009827," and since "p= 0.009827<0.01=\\alpha," it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean "\\mu"

is greater than 4, at the "\\alpha = 0.01" significance level.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS