Answer to Question #348570 in Statistics and Probability for max

Question #348570

A random sample of 20 drinks from a soft-drink machine has an average content of 21.9 deciliters, with a standard deviation of 1.42 deciliters. At 0.05 level of significance, test the hypothesis that μ = 22. 2 against the alternative hypothesis μ < 22.2. Assume that the distribution is normal.


1
Expert's answer
2022-06-07T10:57:34-0400

The following null and alternative hypotheses need to be tested:

"H_0:\\mu=22.2"

"H_1:\\mu<22.2"

This corresponds to a left-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," "df=n-1=19" and the critical value for a left-tailed test is "t_c =-1.729133."

The rejection region for this left-tailed test is "R = \\{t:t<-1.729133\\}."

The t-statistic is computed as follows:




"t=\\dfrac{\\bar{x}-\\mu}{s\/\\sqrt{n}}=\\dfrac{21.9-22.2}{1.42\/\\sqrt{20}}=-0.9448"


Since it is observed that "t=-0.9448>-1.729133=t_c," it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, "df=19" degrees of freedom, "t=-0.9448" is "p=0.178311," and since "p= 0.178311>0.05=\\alpha," it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean "\\mu"

is less than 22.2, at the "\\alpha = 0.05" significance level.


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