Answer to Question #343125 in Statistics and Probability for Nene

Question

Consider the problem: According to last year's report, a Filipino household spends an average of P500 per day for food. Suppose you recently took random samples of 30 households. You determined how much each households spent for food each day and the results revealed a mean of P450 and the standard deviation of P25.50. Using a 0.05 level of significance, can it be concluded that the average amount spent per day for food of a Filipino household has decreased? Assume normality over the population

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq 500$

$H_1:\mu<500$

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 $\alpha=0.05$ and $df=n-1=30-1=29$ degrees of freedom the critical value for a left-tailed test is $t_c=-1.699$

The rejection region for this left-tailed test is $R=\{t:t<-1.699\}.$

The t-statistic is computed as follows:

$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{450-500}{25.5/\sqrt{30}}\approx-0.358$

Since it is observed that $t=-0.358>-1.699=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 500, at the $\alpha=0.05$  significance level.

Using the P-value approach: The p-value for left-tailed $t=-0.358, df=29, \alpha=0.05$ is $p=0.36147$ (we can find the p-value using the t-Distribution table or using this site https://www.statology.org/t-score-p-value-calculator/ ) and since $p=0.36147>0.05=\alpha,$ 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 500, at the $\alpha=0.05$  significance level.