Answer to Question #343551 in Statistics and Probability for Amelie

Question

A government agency is investigating a complaint from some concerned citizens who said that there is short-weight selling of rice in a certain town. An agent manufacturer took a random sample of 20 sacks of "50 kilo" sacks of rice from a large shipment and finda that the mean weight is 49.7 kilogram with the standard deviation of 0.35 kilogram. The level of significance is 0.01

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge50$

$H_a:\mu<50$

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 $\alpha = 0.01,$ $df=n-1=19$ degrees of freedom, and the critical value for a left-tailed test is $t_c =-2.539483.$The rejection region for this left-tailed test is $R = \{t:t<-2.539483\}.$

The t-statistic is computed as follows:

$$t=\dfrac{49.7-50}{0.35/\sqrt{20}}=-3.83326$$

Since it is observed that $t =-3.83326<-2.539483=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed $df=19$ degrees of freedom, $t=-3.83326$ is $p= 0.000561,$ and since $p=0.000561<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 50, at the $\alpha = 0.01$ significance level.