Answer to Question #350533 in Statistics and Probability for Jane

Question

The label on a can of pineapple slices that the mean carbohydrate content per serving of canned pineapple is less than 50 grams. It may be assumed that the standard deviation of the carbohydrate content is 4 grams. A random sample of forty servings has a mean carbohydrate content of 52.3 grams. Is the company correct in its claim? Use level of significance 0.05.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge50$

$H_1:\mu<50$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

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

The rejection region for this left-tailed test is $R = \{z:z<-1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{52.3-50}{4/\sqrt{40}}=3.6366$$

Since it is observed that $z=3.6366>-1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(z<3.6366)=0.999862,$ and since $p=0.999862>0.05=\alpha,$ 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 50, at the $\alpha = 0.01$ significance level.