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Testing Claims About Proportions. In Exercises 9鈥32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Births A random sample of 860 births in New York State included 426 boys. Use a 0.05 significance level to test the claim that 51.2% of newborn babies are boys. Do the results support the belief that 51.2% of newborn babies are boys?

Short Answer

Expert verified

Null hypothesis: The proportion of babies that are boys is equal to 51.2%.

Alternative hypothesis:The proportion of babies that are boys is not equal to 51.2%.

Test statistic: -0.977

Critical value: 1.96

P-value: 0.3286

The null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the proportion of babies that are boys is equal to 51.2%.

Yes, the results support the belief that 51.2% of the babies born are boys.

Step by step solution

01

Given information

A sample of 860 births is selected, out of which 426 are boys. It is claimed that 51.2% of newborn babies are boys.

02

Hypotheses

The null hypothesis is written as follows.

The proportion of newborn babies that are boys is equal to 51.2%.

H0:p=0.512

The alternative hypothesis is written as follows.

The proportion of newborn babies that are boys is not equal to 51.2%.

H1:p0.512

The test is two-tailed.

03

Sample size, sample proportion,and population proportion

The sample size is n=860.

The sample proportion of babies that are boys is computed below.

p^=NumberofboysTotalnumberofbabies=426860=0.495

The population proportion of babies that are boys is equal to 0.512.

04

Test statistic

The value of the test statistic is computed below.

z=p^-ppqn=0.495-0.5120.5121-0.512860=-0.977

Thus, z=-0.977.

05

Critical value and p-value

Referring to the standard normal table, the critical value of z at =0.05 for a two-tailed test is equal to 1.96.

Referring to the standard normal table, the p-value for the test statistic value of -0.977 is equal to 0.3286.

As the p-value is greater than 0.05, the null hypothesis is failed to reject.

06

Conclusion of the test

There is not enough evidence to reject the claim that the proportion of babies that are boys is equal to 51.2%.

The results support the belief that 51.2% of the babies born are boys.

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Most popular questions from this chapter

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then the conclusion about the null hypothesis, as well as the final conclusion that address the original claim. Assume that a simple random sample is selected from a normally distributed population.

Birth Weights A simple random sample of birth weights of 30 girls has a standard deviation of 829.5 hg. Use a 0.01 significance level to test the claim that birth weights of girls have the same standard deviation as birth weights of boys, which is 660.2 hg (based on Data Set 4 鈥淏irths鈥 in Appendix B).

Confidence interval Assume that we will use the sample data from Exercise 1 鈥淰ideo Games鈥 with a 0.05 significance level in a test of the claim that the population mean is greater than 90 sec. If we want to construct a confidence interval to be used for testing the claim, what confidence level should be used for the confidence interval? If the confidence interval is found to be 21.1 sec < < 191.4 sec, what should we conclude about the claim?

Testing Claims About Proportions. In Exercises 9鈥32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Is Nessie Real? This question was posted on the America Online website: Do you believe the Loch Ness monster exists? Among 21,346 responses, 64% were 鈥測es.鈥 Use a 0.01 significance level to test the claim that most people believe that the Loch Ness monster exists. How is the conclusion affected by the fact that Internet users who saw the question could decide whether to respond?

TV Viewing. According to Communications Industry Fore cast & Report, published by Veronis Suhler Stevenson, the average person watched 4.55hours of television per day in 2005. A random sample of 20people gave the following number of hours of television watched per day for last year.

At the 10%significance level, do the data provide sufficient evidence to conclude that the amount of television watched per day last year by the average person differed from that in 2005? (Note: x=4.760hours,s=2.297hours)

P-Values. In Exercises 17鈥20, do the following:

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value. (See Figure 8-3 on page 364.)

c. Using a significance level of = 0.05, should we reject H0or should we fail to reject H0?

The test statistic of z = 2.01 is obtained when testing the claim that p0.345.

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