Answer to Question #207501 in Statistics and Probability for zaimsquah

Question

A cat food company estimates that a salmon based product that they produce has a mean shelf life of 25 months. Some customers claim that the product went bad earlier than the estimated expiration date and the mean shelf life is actually shorter. To test this claim, the company monitors 70 packages of this product and finds that the average shelf life was 23 with a standard deviation of 7 months.

a. Test the customers’ claim for α=0.05

b. Calculate the power of the test if the true mean is 23.5 months

Answer 1

a. The following null and alternative hypotheses need to be tested:

$H_0:\mu=25$

$H_1:\mu<25$

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 $\alpha=0.05,$

$df=n-1=70-1=69$ degrees of freedom, and the critical value for a left-tailed test is $t_c=-1.667239.$

The rejection region for this left-tailed test is $R=\{t:t<-1.667239\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{23-25}{7/\sqrt{70}}\approx-2.390457$$

Since it is observed that $t=-2.390457<-1.667239=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 25, at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value for left-tailed, $df=69, \alpha=0.05,$ $t=-2.390457$ is $p=0.009792,$ and since $p=0.009792<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 25, at the $\alpha=0.05$ significance level.

b.

The significance level is $\alpha=0.05,$ and the critical value for a left-tailed test is $z_c=-1.6449.$

$$-1.6449=\dfrac{\bar{X}-25}{7/\sqrt{70}}$$

$$\bar{X}=23.6238$$

$$\beta=P(\bar{X}>23.6238|\mu_1=23.5)$$

$$=P(Z>\dfrac{23.6238-23.5}{7/\sqrt{70}})\approx P(Z>0.1480)$$

$$\approx0.44118$$

$$Power=1-\beta=1-0.44118\approx0.5588$$