Answer to Question #338198 in Statistics and Probability for JohnMark

Question

A consumer advocacy group suspects that a local supermarket’s 500 grams of sugar actually weigh less than 50 grams. The group look a random sample of 20 such packages, weigh each one, and found the mean weight for the sample to be 496 grams with standard deviation of 8 grams. Using 1% significance level, would you conclude that the mean weight is less than 500 grams? Also, find the 99% confidence interval of the true mean.

Answer 1

We need to construct the $99\%$ confidence interval for the population mean $\mu.$

The critical value for $\alpha = 0.01$ and $df = n-1 = 19$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.860935.$

The corresponding confidence interval is computed as shown below:

$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$

$=(496-2.860935\times\dfrac{8}{\sqrt{20}},$

$496+2.860935\times\dfrac{8}{\sqrt{20}})$

$=(490.8822,501.1178)$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $490.8822 < \mu < 501.1178,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(490.8822,501.1178).$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge500$

$H_a:\mu<500$

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 $\alpha = 0.01,$ $df=n-1=19$ 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:

$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{496-500}{8/\sqrt{20}}\approx-2.236068$

Since it is observed that $t = -2.236068 \ge -2.539483=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for left-tailed test, $df=19$ degrees of freedom, $t=-2.236068$ is $p = 0.01877,$ and since $p = 0.01877 \ge 0.01=\alpha,$ 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 500, at the $\alpha = 0.01$ significance level.