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Chapter 10: Q.10 (page 622)

Explain why the conditions for using two-sample z procedures to perform inference about p1-p2are not met in the settings

Broken crackers We don鈥檛 like to 铿乶d broken crackers when we open the package. How can makers reduce breaking? One idea is to microwave the crackers for 30seconds right after baking them. Breaks start as hairline cracks called 鈥渃hecking.鈥� Assign 65newly baked crackers to the microwave and another 65to a control group that is not microwaved. After one day, none of the microwave group and 16 of the control group show checking.8

Short Answer

Expert verified

From the given information,

The success rate in the first sample, i.e. the microwave group, is 0(less than 10). It means that the normalcy assumption isn't met in this case.

As a result, the two-proportion z-test is inappropriate in this situation.

Step by step solution

01

Given Information

It is given in the question that, the values arex1=0,n1=65,x2=16,n2=65

02

Step 2: Explanation

The number of successes in both samples must be greater than 10to execute the two-proportion z-test.

The success rate in the first sample, i.e. the microwave group, is 0(less than 10). It means that the normalcy assumption isn't met in this case.

As a result, the two-proportion z-test is inappropriate in this situation.

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Most popular questions from this chapter

A better drug? In a pilot study, a company's new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing

evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood (mg/dl) greater than with the current drug, the company will begin the expensive process of mass-producing the new drug. For the 14 subjects who were assigned at random to the current drug, the mean cholesterol reduction was 54.1mg/dlwith a standard deviation of 11.93mg/dl.For the 15 subjects who were randomly assigned to the new drug, the mean cholesterol reduction was 68.7mg/dlwith a standard deviation of13.3mg/dl.Graphs of the data reveal no outliers or strong skewness.

(a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is notH0:渭1-渭2=0-

(b) Based on your conclusion in part (a), could you have made a Type I error or a Type Il error? Justify your answer.

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(a) Both populations are Normally distributed.

(b) The data come from two independent samples.

(c) Both samples were chosen at random.

(d) The counts of successes and failures are large enough to use Normal calculations.

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(a) A Fathom dotplot of the data is shown below. Use the graph to answer these questions:

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(b) Construct and interpret a 90% confidence interval for the difference in mean percent change in polyphenol levels for the red wine and white wine treatments.
(c) Does the interval in part (b) suggest that red wine is more effective than white wine? Explain.

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