Answer to Question #333869 in Statistics and Probability for sheshe

Question

In particularly community,it is claimed that the mean household water usage for a particular month is 48 cubic meters. The following year,a country wide water conservation campaign was conducted.Forty five homes were randomly selected and found that the mean comsumption is 52 cubic meters with a standard deviation of 4 cubic meters.Is there enough evidence to say that the mean household water usage per month is higher than 48 cubic meters at a=0.01?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le48$

$H_1:\mu>48$

This corresponds to a right-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 $\alpha=0.01,$ $df=n-1=45-1=44$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 2.414134.$

The rejection region for this right-tailed test is $R = \{t: t > 2.414134\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{52-48}{4/\sqrt{45}}\approx6.708204$$

Since it is observed that $t = 6.708204 > 2.414134=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, $df=44$ degrees of freedom, $t=6.708204$ is $p=0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 48, at the $\alpha = 0.01$ significance level.