Question #348572

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
2022-06-07T10:56:41-0400

The following null and alternative hypotheses need to be tested:

H0:μ=22.2H_0:\mu=22.2

H1:μ<22.2H_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 α=0.05,\alpha = 0.05, df=n1=19df=n-1=19 and the critical value for a left-tailed test is tc=1.729133.t_c =-1.729133.

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

The t-statistic is computed as follows:



t=xˉμs/n=21.922.21.42/20=0.9448t=\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=tc,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=19df=19 degrees of freedom, t=0.9448t=-0.9448 is p=0.178311,p=0.178311, and since p=0.178311>0.05=α,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 μ\mu

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


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