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A local bottler in Hawaii wishes to ensure that an average of 15 ounces of passi

ID: 3269272 • Letter: A

Question

A local bottler in Hawaii wishes to ensure that an average of 15 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 54 bottles. The mean weight of the passion fruit juice in the sample is 14.67 ounces. Assume that the population standard deviation is 1.19 ounce. Use the critical value approach to test the bottler's concern at = 0.10.

a. Select the null and the alternative hypotheses for the test.

H0: 15; HA: > 15

H0: 15; HA: < 15

H0: = 15; HA: 15

b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.)

Test statistic

b-2. Find the critical value(s). (Round your answer to 2 decimal places.)

Critical value(s) ±

b-3. What is the conclusion?

Reject H0 since the value of the test statistic is not less than the negative critical value.

Reject H0 since the value of the test statistic is less than the negative critical value.

Do not reject H0 since the value of the test statistic is not less than the negative critical value.

Do not reject H0 since the value of the test statistic is less than the negative critical value.

c.Make a recommendation to the bottler.

The accuracy of the bottling process is compromised or not compromised

c.Make a recommendation to the bottler.

The accuracy of the bottling process is compromised or not compromised

Explanation / Answer

(a)

The hypotheses are:

H0: = 15; HA: 15

(b)

Given data:

Sample mean, m = 14.67

Standard deviation, S = 1.19

Sample size, n = 54

Calculating standard error, SE = S/n0.5 = 1.19/540.5 = 0.162

Calculating test statistic:

z = (m-15)/SE = (14.67-15)/0.162 = -2.037

(c)

The critical value for this two tailed test at significance level of 0.1 is:

zc = -1.64

(d)

Since |z| < |zc|, the decision is to reject the null hypothesis.

Hope this helps !

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