Answer to Question #342464 in Statistics and Probability for Ari

Question #342464

A shop owner claims that on average, his shop has a daily turnover of R75 000. Test this claim at the 5% significance level, if the average daily turnover for a sample of 88 days was R71 500. Assume a population standard deviation of R4 450.


1
Expert's answer
2022-05-24T11:58:12-0400

The following null and alternative hypotheses need to be tested:

"H_0:\\mu=75000"

"H_1:\\mu\\not=75000"

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{71500-75000}{4450\/\\sqrt{88}}\\approx-7.3782"

Since it is observed that "|z|=7.3782>1.96=z_c," it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is "p=2P(z<-7.3782)=0," and since "p= 0<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 75000, at the "\\alpha = 0.05" significance level.


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