Answer to Question #350771 in Statistics and Probability for john

Question #350771

A canteen owner claims that the average meal cost of his usual clients is

Php180. In order to test his own claim, he took a random sample of 30

receipts and computed the mean cost of Php210 with a standard deviation of

Php25. Test the hypothesis at 0.01 level of significance. Show all steps


1
Expert's answer
2022-06-15T17:02:09-0400

The following null and alternative hypotheses need to be tested:

H0:μ=180H_0:\mu=180

H1:μ180H_1:\mu\not=180

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 α=0.01,\alpha = 0.01, df=n1=29df=n-1=29 and the critical value for a two-tailed test is tc=2.756386.t_c = 2.756386.

The rejection region for this two-tailed test is R={t:t>2.756386}.R = \{t:|t|> 2.756386\}.

The t-statistic is computed as follows:


t=xˉμs/n=21018025/30=6.57267t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{210-180}{25/\sqrt{30}}=6.57267


Since it is observed that t=6.57267>2.756386=tc,|t|=6.57267> 2.756386=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, df=29df=29 degrees of freedom, t=6.57267t=6.57267 is p=0,p=0, and since p=0<0.01=α,p=0<0.01=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu is different than 180, at the α=0.01\alpha = 0.01 significance level.


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