Exam 9: Fundamentals of Hypothesis Testing: One-Sample Tests

SCENARIO 9-6 The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tail test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test. -Referring to Scenario 9-6, if the test is performed with a level of significance of 0.05, the null hypothesis would be rejected.

A is a numerical quantity computed from the data of a sample and is used in reaching a decision on whether or not to reject the null hypothesis.

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant. A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 cases: $n = 46$ ; Arithmetic Mean $= 28.00$ ; Standard Deviation $= 25.92$ ; Standard Error $= 3.82$ ; Null Hypothesis: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ Test Statistic $= 2.09$ ; One-Tail Test Upper Critical Value $= 1.3006 ; p$ -value $= 0.021 ;$ Decision $=$ Reject. -Suppose, in testing a hypothesis about a mean, the p-value is computed to be 0.034. The null hypothesis should be rejected if the chosen level of significance is 0.01.

SCENARIO 9-7 A major home improvement store conducted its biggest brand recognition campaign in the company's history. A series of new television advertisements featuring well-known entertainers and sports figures were launched. A key metric for the success of television advertisements is the proportion of viewers who "like the ads a lot". A study of 1,189 adults who viewed the ads reported that 230 indicated that they "like the ads a lot." The percentage of a typical television advertisement receiving the "like the ads a lot" score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i.e. if there is evidence that the population proportion of "like the ads a lot" for the company's ads is less than 0.22) at a 0.01 level of significance. -Referring to Scenario 9-7, the null hypothesis would be rejected.

SCENARIO 9-11-A You are the quality control manager of a water bottles company. One of the biggest complaints in the past years has been the breakage and, hence, the concern on the durability of the connector between the lid and the bottle which many users use as a handle for the bottles. To collect evidence before implementing any modification to the production process, your department has subjected 50 water bottles to a durability test and the following data on the number of times the handles have been used to lift the bottles before they break are contained in the file Scenario9-11-DataA.XLSX. 1495 1499 1502 1500 1491 1498 1498 1495 1488 1516 1513 1486 1504 1503 1493 1504 1489 1500 1495 1499 1501 1507 1511 1496 1486 1497 1510 1504 1493 1482 1511 1502 1520 1514 1486 1514 1500 1505 1512 1500 1504 1498 1503 1514 1474 1489 1488 1506 1517 1490 Assume that the number of times the handles have been used to lift the bottles before they break follows a normal distribution. You want to test to see if there is enough evidence that the mean number of times the handles have been used to lift the bottles before they break is more than 1500. -Referring to Scenario 9-11-A, the parameter of interest is

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant. A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 cases: $n = 46$ ; Arithmetic Mean $= 28.00$ ; Standard Deviation $= 25.92$ ; Standard Error $= 3.82$ ; Null Hypothesis: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ Test Statistic $= 2.09$ ; One-Tail Test Upper Critical Value $= 1.3006 ; p$ -value $= 0.021 ;$ Decision $=$ Reject. -In instances in which there is insufficient evidence to reject the null hypothesis, you must make it clear that this has proven that the null hypothesis is true.

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant. A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 cases: $n = 46$ ; Arithmetic Mean $= 28.00$ ; Standard Deviation $= 25.92$ ; Standard Error $= 3.82$ ; Null Hypothesis: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ Test Statistic $= 2.09$ ; One-Tail Test Upper Critical Value $= 1.3006 ; p$ -value $= 0.021 ;$ Decision $=$ Reject. -Suppose, in testing a hypothesis about a mean, the Z test statistic is computed to be 2.04. The null hypothesis should be rejected if the chosen level of significance is 0.01 and a two- tail test is used.

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant. A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 cases: $n = 46$ ; Arithmetic Mean $= 28.00$ ; Standard Deviation $= 25.92$ ; Standard Error $= 3.82$ ; Null Hypothesis: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ Test Statistic $= 2.09$ ; One-Tail Test Upper Critical Value $= 1.3006 ; p$ -value $= 0.021 ;$ Decision $=$ Reject. -A proper methodology in performing hypothesis tests is to ask whether a random sample can be selected from the population of interest.

SCENARIO 9-9 The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years but their mean SAT score is lower than previous years. He took a sample of 20 of this year's entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university's record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. -Referring to Scenario 9-9, the null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed.

SCENARIO 9-6 The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tail test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test. -Referring to Scenario 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. Then if the test is performed with a level of significance of 0.05, the null hypothesis would be rejected.

The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a GPS navigation system if it had been available For an additional cost of $300. The manager believes from previous information that the proportion is 0)30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they Would have purchased the GPS navigation system. If you were to conduct a test to determine whether There is evidence that the proportion is different from 0.30, which test would you use?

For a given level of significance, if the sample size is increased but the summary statistics remain the same, the probability of committing a Type II error will increase.

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant. A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 cases: $n = 46$ ; Arithmetic Mean $= 28.00$ ; Standard Deviation $= 25.92$ ; Standard Error $= 3.82$ ; Null Hypothesis: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ Test Statistic $= 2.09$ ; One-Tail Test Upper Critical Value $= 1.3006 ; p$ -value $= 0.021 ;$ Decision $=$ Reject. -The larger the p-value, the more likely you are to reject the null hypothesis.

SCENARIO 9-9 The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years but their mean SAT score is lower than previous years. He took a sample of 20 of this year's entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university's record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. -Referring to Scenario 9-9, the null hypothesis would be rejected.

SCENARIO 9-4 A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with a mean of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normally distributed with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision. -Referring to Scenario 9-4, for a test with a level of significance of 0.10, the critical value would be ________.

SCENARIO 9-7 A major home improvement store conducted its biggest brand recognition campaign in the company's history. A series of new television advertisements featuring well-known entertainers and sports figures were launched. A key metric for the success of television advertisements is the proportion of viewers who "like the ads a lot". A study of 1,189 adults who viewed the ads reported that 230 indicated that they "like the ads a lot." The percentage of a typical television advertisement receiving the "like the ads a lot" score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i.e. if there is evidence that the population proportion of "like the ads a lot" for the company's ads is less than 0.22) at a 0.01 level of significance. -Referring to Scenario 9-7, the company officials can conclude that there is sufficient evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.01.

SCENARIO 9-11-A You are the quality control manager of a water bottles company. One of the biggest complaints in the past years has been the breakage and, hence, the concern on the durability of the connector between the lid and the bottle which many users use as a handle for the bottles. To collect evidence before implementing any modification to the production process, your department has subjected 50 water bottles to a durability test and the following data on the number of times the handles have been used to lift the bottles before they break are contained in the file Scenario9-11-DataA.XLSX. 1495 1499 1502 1500 1491 1498 1498 1495 1488 1516 1513 1486 1504 1503 1493 1504 1489 1500 1495 1499 1501 1507 1511 1496 1486 1497 1510 1504 1493 1482 1511 1502 1520 1514 1486 1514 1500 1505 1512 1500 1504 1498 1503 1514 1474 1489 1488 1506 1517 1490 Assume that the number of times the handles have been used to lift the bottles before they break follows a normal distribution. You want to test to see if there is enough evidence that the mean number of times the handles have been used to lift the bottles before they break is more than 1500. -Referring to Scenario 9-11-A, for a test with a level of significance of 0.01, the critical value would be ________.

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant. A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 cases: $n = 46$ ; Arithmetic Mean $= 28.00$ ; Standard Deviation $= 25.92$ ; Standard Error $= 3.82$ ; Null Hypothesis: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ Test Statistic $= 2.09$ ; One-Tail Test Upper Critical Value $= 1.3006 ; p$ -value $= 0.021 ;$ Decision $=$ Reject. -Suppose, in testing a hypothesis about a mean, the p-value is computed to be 0.043. The null hypothesis should be rejected if the chosen level of significance is 0.05.

SCENARIO 9-4 A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with a mean of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normally distributed with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision. -Referring to Scenario 9-4, if the level of significance had been chosen as 0.05, the company would market the new anesthetic.

SCENARIO 9-3 An appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W. From previous studies, it is believed that power consumption for microwave ovens is normally distributed with a population standard deviation of 15 W. A consumer group has decided to try to discover if the claim appears true. They take a sample of 20 microwave ovens and find that they consume a mean of 257.3 W. -Referring to Scenario 9-3, for this test to be valid, it is necessary that the power consumption for microwave ovens has a normal distribution.