Question

A local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle. In order to analyze the accuracy of the bottling process, he takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce. Use a confidence level of 99.
Use the critical value approach to test the bottler's concern at α = 0.01.
(A) Select the null and the alternative hypotheses for the test.

a) H0: μ = 22; HA: μ ≠ 22
b) H0: μ ≤ 22; HA: μ > 22
c) H0: μ ≥ 22; HA: μ < 22

(B) Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

Answer 1

Answer:

(A)  $H_0$ : μ = 22 ; $H_A$ : μ ≠ 22

(B) Value of the test statistic = -2.69

Step-by-step explanation:

We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle. He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.

We have to test the accuracy of the bottling process.

(A) Let, NULL HYPOTHESIS, $H_0$ : $\mu$ = 22 ounces  {means that an average of 22 ounces of passion fruit juice is used to fill each bottle}

ALTERNATE HYPOTHESIS, $H_A$ : $\mu \neq$ 22 ounces  {means that an average of 22 ounces of passion fruit juice is used to fill each bottle}

The test statistics that will be used here is One-sample z-test;

             T.S. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weight of the passion fruit juice = 21.54 ounces

            $\sigma$ = population standard deviation = 1.38 ounces

            n = sample of bottles = 65

(B) So, test statistics =  $\frac{21.54-22}{\frac{1.38}{\sqrt{65} } }$

                                  = -2.69

Hence, the value of test statistics is -2.69.