Ch 9 ex 54 business Stats 435 Exercises (Cincinnati Multiple Listing Service. Ap
ID: 3358035 • Letter: C
Question
Ch 9 ex 54 business Stats 435 Exercises (Cincinnati Multiple Listing Service. April, 2012). Data for the sale of 40 houses in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Con- Hamilton County, Ohio, the mean number of days needed to sell a house is 86 days test to determine whether the mean number of days until a house is sold is different than the Hamilton County mean of 86 days in the nearby county. Use a - 05 for the level of significance, and state your conclusion. blow up 54 On December 25, 2009, an airline passenger was subdued while attempting to a Northwest Airlines flight headed for Detroit, Michigan. The passenger had smuggled explosives hidden in his underwear past a metal detector at an airport screening facility. As a result, the Transportation Security Administration (TSA) proposed installing full-body scan- ners to replace the metal detectors at the nation's largest airports. This proposal resulted in strong objections from privacy advocates, who considered the scanners an invasion of privacy On January 5-6, 2010, USA Today conducted a poll of 542 adults to learn what proportion of airline travelers approved of using full-body scanners (USA Today, January 11,2010). The pol results showed that 455 of the respondents felt that full-body scanners would improve airline security and 423 indicated that they approved of using the devices a. Conduct a hypothesis test to determine if the results of the poll justify concluding that over 80% of airline travelers feel that the use of full-body scanners will improve airline security. Use .05 b. Suppose the TSA will go forward with the installation and mandatory use of full-body scanners if over 75% of airline travelers approve of using the devices. You have been told to conduct a statistical analysis using the poll results to determine if the TSA should go forward with mandatory use of the full-body scanners. Because this is viewed as a very sensitive decision, use a-01. What is your recommendation? A recent article concerning bullish and bearish sentiment about the stock market reported that 41% of investors responding to an American Institute of Individual Investors (AAII) poll were bullish on the market and 26% were bearish (USA Today, January 11, 2010). The article also reported that the long-term average measure of bullishness is .39 or 39% Suppose the AAll poll used a sample size of 450. Using 39 (the long-term average) as the population proportion of investors who are bullish, conduct a hypothesis test to determine if the current proportion of investors who are bullish is significantly greater than the long- 55. term average proportion. a. State the appropriate hypotheses for your significance test. b. Use the sample results to compute the test statistic and the p-value. c. Using .10, what is your conclusion? 56. Virtual call centers are staffed by individuals working out of their homes. Most home ents earn $10 to $15 per hour without benefits versus $7 to $9 per hour with benefits at a I call center (BusinessWeek, January 23, 2006). Regional Airways is considering mploying home agents, but only if a level of customer satisfaction greater than 80% can be maintained. A test was conducted with home servise assats In aExplanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.80
Alternative hypothesis: P > 0.80
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.01577
z = (p - P) /
z = 2.50
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is more than 2.50. We use the Normal Distribution Calculator to find P(z > 2.50).
Thus, the P-value = 0.0062
Interpret results. Since the P-value (0.0062) is less than the significance level (0.05), we cannot accept the null hypothesis.
We can conclude that over 80% of the airline travellers feels that use of full body scanners will improve airline security.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.75
Alternative hypothesis: P > 0.75
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.01778
z = (p - P) /
z = 1.71
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is more than 171. We use the Normal Distribution Calculator to find P(z > 1.71).
Thus, the P-value = 0.0436
Interpret results. Since the P-value (0.0436) is less than the significance level (0.05), we cannot accept the null hypothesis.
We can conclude that over 75% of the airline travellers feels that use of full body scanners will iapprove of using the device.