State the null and alternative hypotheses to determine if the average number of hours until spoilage begins differs for the preservatives A and B. Assume population variances are equal. Calculate the pooled variance and the value of the test statistic. t: Determine the rejection region at a = .05 and write the proper conclusion. Rejection region: Decision: Conclusion:

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A FOOD PROCESSOR WANTS TO COMPARE TWO PRESERVATIVES FOR THEIR EFFECTS ON RETARDING SPOILAGE. SUPPOSE
16 CUTS OF FRESH MEAT ARE TREATED WITH PRESERVATIVE A AND 16 ARE TREATED WITH PRESERVATIVE B, AND THE
NUMBER OF HOURS UNTIL SPOILAGE BEGINS IS RECORDED FOR EACH OF THE 32 CUTS OF MEAT. THE RESULTS ARE
SUMMARIZED IN THE TABLE BELOW
4.1
4.2
4.3
Sample Mean
Sample Standard Deviation
Preservative A
108.7 hours
t:
10.5 hours
Preservative B
98.7 hours
13.6 hours
State the null and alternative hypotheses to determine if the average number of hours until
spoilage begins differs for the preservatives A and B.
Assume population variances are equal. Calculate the pooled variance and the value of the
test statistic.
Determine the rejection region at a= .05 and write the proper conclusion.
Rejection region:
Decision:
Conclusion:
Transcribed Image Text:A FOOD PROCESSOR WANTS TO COMPARE TWO PRESERVATIVES FOR THEIR EFFECTS ON RETARDING SPOILAGE. SUPPOSE 16 CUTS OF FRESH MEAT ARE TREATED WITH PRESERVATIVE A AND 16 ARE TREATED WITH PRESERVATIVE B, AND THE NUMBER OF HOURS UNTIL SPOILAGE BEGINS IS RECORDED FOR EACH OF THE 32 CUTS OF MEAT. THE RESULTS ARE SUMMARIZED IN THE TABLE BELOW 4.1 4.2 4.3 Sample Mean Sample Standard Deviation Preservative A 108.7 hours t: 10.5 hours Preservative B 98.7 hours 13.6 hours State the null and alternative hypotheses to determine if the average number of hours until spoilage begins differs for the preservatives A and B. Assume population variances are equal. Calculate the pooled variance and the value of the test statistic. Determine the rejection region at a= .05 and write the proper conclusion. Rejection region: Decision: Conclusion:
Expert Solution
Step 1

It is given that

For preservative A, sample size n1 = 16, sample mean = 108.7 and sample SD s1 = 10.5

For preservative B, sample size n2 = 16, sample mean = 98.7 ans sample SD s2 = 13.6

Level of significance = 0.05

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