A local bottler in Hawaii wishes to ensure that an average of 26 ounces of passi

Question

A local bottler in Hawaii wishes to ensure that an average of 26 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 43 bottles. The mean weight of the passion fruit juice in the sample is 25.70 ounces. Assume that the population standard deviation is 1.19 ounce. Use Standard Normal Curve Area.

Use the critical value approach to test the bottler's concern at = 0.05.

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

a. H0: = 26; HA: 26

b. H0: 26; HA: > 26

c. H0: 26; HA: < 26

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

Test statistic ????

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

Critical value(s) ± ????

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2-3. What is the conclusion?

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

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

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

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

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3. Make a recommendation to the bottler. The accuracy of the bottling process is (not compromised/compromised).

Explanation / Answer

1) a. H0: = 26; HA: 26

2-1) std error of mean =std deviation/(n)1/2 =1.19/(43)1/2 =0.1815

test statistic =z=(X-mean)/std error =(25.70-26)/0.1815=-1.65

2.2) critical values +/- 1.96

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

3)

The accuracy of the bottling process is (not compromised

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