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Chapter 9: Q 41. (page 581)

Significance tests A test of Ho:p=0.5versus Ha:p>0.5based on

a sample of size 200yields the standardized test statistic z=2.19. Assume that the conditions for performing inference are met.

a. Find and interpret the P-value.

b. What conclusion would you make at the 伪=0.01 significance level? Would

your conclusion change if you used 伪=0.05 instead? Explain your reasoning.

c. Determine the value of p^= the sample proportion of successes.

Short Answer

Expert verified

a. Hypothesis is rejected.

b. 伪=0.01:fail to reject the null hypothesis.and 伪=0.05:reject the null hypothesis

c.p^=0.57774

Step by step solution

01

Part (a) Step 1: Given Information

It is given that z=2.19

n=200

伪=0.5

H0:p=0.5

Ha:p>0.5

02

Part (a) Step 2: Explanation

As z=2.19

P(x<z)=0.98574

P(x>z)=1-0.98574=0.014262

When Pvalue鈮�伪, null hypothesis is rejected.

From above P=0.014262

P=0.014262<伪, null hypothesis is rejected. H0:p=0.5

03

Part (b) Step 1: Given Information

It is given that伪=0.01,伪=0.05

04

Part (b) Step 2: Explanation

Higher the significance level, higher is probability to reject null hypothesis.

As P=0.014262<0.5, so null hypothesis is rejected.

If 伪=0.01P=0.014262>0.01, null hypothesis is not rejected.

If 伪=0.05, P=0.014262>0.05, null hypothesis is not rejected.

Therefore,Pvalue changes with change in significance levels.

05

Part (c) Step 1: Given Information

It is given that n=200

p=0.5

z=2.19

06

Part (c) Step 2: Explanation

We know that z=p^-pp(1-p)n

Hence, 2.19=p^-0.50.5(1-0.5)200

2.19=p^-0.50.5(0.5)200

p^=(2.19脳0.03535)+0.5

Therefore,p^=0.57774

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Most popular questions from this chapter

Walking to school Refer to Exercise 36.

a. Explain why the sample result gives some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and P-value.

c. What conclusion would you make?

An opinion poll asks a random sample of adults whether they favor banning ownership of handguns by private citizens. A commentator believes that more than half of all adults favor such a ban. The null and alternative hypotheses you would use to test this claim are

邪.H0:p鈭�=0.5;Ha:p鈭�>0.5H0:p^=0.5;Ha:p^>0.5.

b. H0:p=0.5;Ha:p>0.5H0:p=0.5;Ha:p>0.5.

c. H0:p=0.5;Ha:p<0.5H0:p=0.5;Ha:p<0.5.

d. H0:p=0.5;Ha:p鈮�0.H0:p=0.5;Ha:p鈮�0.5.

e. H0:p>0.5;Ha:p=0.5H0:p>0.5;Ha:p=0.5.

Candy! A machine is supposed to fill bags with an average of 19.2 ounces of candy. The manager of the candy factory wants to be sure that the machine does not consistently underfill or overfill the bags. So the manager plans to conduct a significance test at the 伪=0.10significance level of

H0:渭=19.2Ha:渭notequalto19.2

where 渭=the true mean amount of candy (in ounces) that the machine put in all bags filled that day. The manager takes a random sample of 75 bags of candy produced that day and weighs each bag. Check if the conditions for performing the test are met.

Side effects A drug manufacturer claims that less than 10%of patients who take its new drug for treating Alzheimer鈥檚 disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300out of 5000Alzheimer鈥檚 patients whose families have given informed consent for the patients to participate in the study. In all, 25of the subjects experience nausea.

a. Describe a Type I error and a Type II error in this setting, and give a possible

consequence of each.

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Which of choices (a) through (d) is not a condition for performing a significance test about a population proportion p?

a. The data should come from a random sample from the population of interest.

b. Both np0and n(1-p0)should be at least 10.

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d. The population distribution should be approximately Normal unless the sample size is large.

e. All of the above are conditions for performing a significance test about a population proportion.
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