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Suppose the sample in Exercise 7.64 has produced \(\hat p = .83\) and we wish to test \({H_0}:P = 0.9\) against the alternative \({H_a}:p < .9\)

a. Calculate the value of the z-statistic for this test.

Short Answer

Expert verified
  1. The z-statistic is -2.333.

Step by step solution

01

Given Information

The number of sample size is 100.

The hypothesis are given by

\(\begin{aligned}{H_0}:p = 0.9\\{H_a}:p < 0.9\end{aligned}\)

02

z-statistics test

When the variations are known as well as the sampling size is high, a z-test is used to assess if two population means vary. The Z test is a statistical test performed on data that roughly follows a normally distributed. For hypothesis testing, the z test can be used to one sample, samples collected, as well as percentages.

03

Compute the z-statistic

The z-statistic is computed as

\(\begin{aligned}z &= \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}{q_0}}}{n}} }}\\ &= \frac{{0.83 - 0.9}}{{\sqrt {\frac{{0.9 \times 0.1}}{{100}}} }}\\ &= \frac{{ - 0.07}}{{0.03}}\\ &= - 2.333\end{aligned}\)

Therefore, the z-statistic is -2.333.

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Most popular questions from this chapter

The opinions of employees at Asia-Pacific firms regarding fraud, bribery, and corruption in the workplace were elicited in the 2015 AsiaPacific (APAC) Fraud Survey. Interviews were conducted with a sample 1,508 employees of large APAC companies. One question concerned whether or not the employeeโ€™s company had implemented a โ€œwhistle-blowerโ€ hotlineโ€”that is, a phone number that an employee can call to report fraud or other types of misconduct without fear of retribution. The 2015 survey found that 680 of the 1,508 respondentsโ€™ companies had not implemented a whistle-blower hotline. In comparison, 436 of the 681 respondents in the 2013 survey reported their companies had not implemented a whistle-blower hotline.

a. TestH0:p=.5 against Ha:p<.5using data from the 2015 survey and ฮฑ=.05Give a practical interpretation of the results.

b. Test H0:p=.5against Ha:p>.5using data from the 2013 survey andฮฑ=.05 Give a practical interpretation of the results.

Refer to Exercise 7.99.

a. Find b for each of the following values of the population mean: 74, 72, 70, 68, and 66.

b. Plot each value of b you obtained in part a against its associated population mean. Show b on the vertical axis and m on the horizontal axis. Draw a curve through the five points on your graph.

c. Use your graph of part b to find the approximate probability that the hypothesis test will lead to a Type II error when m = 73.

d. Convert each of the b values you calculated in part a to the power of the test at the specified value of m. Plot the power on the vertical axis against m on the horizontal axis. Compare the graph of part b with the power curve of this part.

e. Examine the graphs of parts b and d. Explain what they reveal about the relationships among the distance between the true mean m and the null hypothesized mean m0, the value of b, and the power.

A random sample of 64 observations produced the following summary statistics: \(\bar x = 0.323\) and \({s^2} = 0.034\).

a. Test the null hypothesis that\(\mu = 0.36\)against the alternative hypothesis that\(\mu < 0.36\)using\(\alpha = 0.10\).

b. Test the null hypothesis that \(\mu = 0.36\) against the alternative hypothesis that \(\mu \ne 0.36\) using \(\alpha = 0.10\). Interpret the result.

Hotelsโ€™ use of ecolabels. Refer to the Journal of Vacation Marketing (January 2016) study of travelersโ€™ familiarity with ecolabels used by hotels, Exercise 2.64 (p. 104). Recall that adult travelers were shown a list of 6 different ecolabels, and asked, โ€œSuppose the response is measured on a continuous scale from 10 (not familiar at all) to 50 (very familiar).โ€ The mean and standard deviation for the Energy Star ecolabel are 44 and 1.5, respectively. Assume the distribution of the responses is approximately normally distributed.

a. Find the probability that a response to Energy Star exceeds 43.

b. Find the probability that a response to Energy Star falls between 42 and 45. c. If you observe a response of 35 to an ecolabel, do you think it is likely that the ecolabel was Energy Star? Explain.

Shopping vehicle and judgment. Refer to the Journal of Marketing Research (December 2011) study of grocery store shoppersโ€™ judgments, Exercise 2.85 (p. 112). For one part of the study, 11 consumers were told to put their arm in a flex position (similar to carrying a shopping basket) and then each consumer was offered several choices between a vice product and a virtue product (e.g., a movie ticket vs. a shopping coupon, pay later with a larger amount vs. pay now). Based on these choices, a vice choice score was determined on a scale of 0 to 100 (where higher scores indicate a greater preference for vice options). The data in the next table are (simulated) choice scores for the 11 consumers. Suppose that the average choice score for consumers with an extended arm position (similar to pushing a shopping cart) is known to be \(\mu = 50\) . The researchers theorize that the mean choice score for consumers shopping with a flexed arm will be higher than 43 (reflecting their higher propensity to select a vice product) Test the theory at \(\alpha = 0.05\)

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