Answer to Question #206629 in Statistics and Probability for Baneng

Question

According to last year's report, a Filipino household spends an average of Php400 per day for food. Suppose you recently took random samples of 25 households. You determined how much each household spent for food each day and the results revealed a mean of Php390 and a standard deviation of Php21.50. Using a 0.01 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:

$H_0: \mu\geq 400$

$H_1:\mu<400$

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 $\alpha=0.01$ and $df=n-1=25-1=24$ degrees of freedom the critical value for a left-tailed test is $t_c=-2.492159.$

The rejection region for this left-tailed test is $R=\{t:t<-2.492159\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{390-400}{21.5/\sqrt{25}}\approx-2.325581$$

Since it is observed that $t=-2.325581>-2.492159=t_c,$ 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 400, at the $\alpha=0.01$  significance level.

Using the P-value approach: The p-value for left-tailed $t=-2.325581, df=24, \alpha=0.01$ is $p=0.014395,$ and since $p=0.014395>0.01=\alpha,$ 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 400, at the $\alpha=0.01$  significance level.