Answer to Question #346601 in Statistics and Probability for CriselJoy

Question #346601

2. 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 20 families were asked and found a mean of 5 times in a week and 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
2022-06-01T03:34:27-0400

The following null and alternative hypotheses need to be tested:

"H_0:\\mu=5"

"H_1:\\mu\\not=5"

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is "\\alpha = 0.05," and the critical value for a two-tailed test is "z_c = 1.96."

The rejection region for this two-tailed test is "R = \\{z:|z|>1.96\\}."

The z-statistic is computed as follows:


"z=\\dfrac{\\bar{x}-\\mu}{\\sigma\/\\sqrt{n}}=\\dfrac{4-5}{2\/\\sqrt{20}}\\approx-2.236"

Since it is observed that "|z|=2.236>1.6449=z_c," it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is "p=2P(z<-2.236)= 0.0253529," and since "p= 0.0253529<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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