Answer to Question #207959 in Statistics and Probability for Lovely

Question #207959

An association of city mayons 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 a standard deviation of 2 use 5% significance level to test that the population mean is not equal to 5 assume that the population is normally distributed.


1
Expert's answer
2021-06-18T04:47:02-0400

The following null and alternative hypotheses need to be tested:

"\\mu=5"

"\\mu\\not=5"

This corresponds to a two-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=0.05" and "df=n-1=40-1=39" degrees of freedom, and the critical value for a two-tailed test is "t_c=2.022691."

The rejection region for this two-tailed test is "R=\\{t: |t|>2.022691\\}."

The t-statistic is computed as follows:


"t=\\dfrac{\\bar{x}-\\mu}{s\/\\sqrt{n}}=\\dfrac{4-5}{2\/\\sqrt{40}}=-3.162278"

Since it is observed that "|t|=3.162275>2.022691=t_c," it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean "\\mu" is different than "5," at the "\\alpha=0.05" significance level.

Using the P-value approach: The p-value for two-tailed, "\\alpha=0.05, df=39,"

"t=-3.162278," is "p=0.003027," and since "p=0.003027<0.05=\\alpha," it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean "\\mu" is different than "5," at the "\\alpha=0.05" significance level.



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