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

The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is Greater than 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no Entertainment changes will be made. Suppose she found that the sample mean was 30.45 years And the sample standard deviation was 5 years. If she wants to have a level of significance at 0.01 What conclusion can she make?

SCENARIO 9-11-B 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 bottle to 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-DataB.XLSX. 1493 1506 1515 1491 1500 1505 1517 1510 1506 1503 1503 1491 1495 1496 1496 1505 1493 1486 1504 1483 1514 1494 1497 1501 1493 1490 1510 1494 1494 1495 1494 1486 1495 1506 1506 1507 1502 1498 1510 1501 1500 1505 1492 1486 1501 1496 1501 1521 1510 1498 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-B, you can conclude that there is enough evidence that the mean number of times the handles have been used to lift the bottles before they break is no more than 1500 when allowing for a 1% probability of committing a Type I error.

It is possible to directly compare the results of a confidence interval estimate to the results obtained by testing a null hypothesis if

The symbol for the level of significance of a statistical test is a) $\alpha$ . b) $1 - \alpha$ . c) $\beta$ . d) $1 - \beta$ .

The symbol for the probability of committing a Type I error of a statistical test is a) $\alpha$ . b) $1 - \alpha$ . c) $\beta$ . d) $1 - \beta$ .

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 engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650.

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. -Referring to Scenario 9-1, the manager can conclude that there is sufficient evidence to show that the mean number of defective bulbs per case is greater than 20 during the morning shift with no more than a 5% probability of incorrectly rejecting the true null hypothesis.

SCENARIO 9-8 One of the biggest issues facing e-retailers is the ability to turn browsers into buyers. This is measured by the conversion rate, the percentage of browsers who buy something in their visit to a site. The conversion rate for a company's website was 10.1%. The website at the company was redesigned in an attempt to increase its conversion rates. A sample of 200 browsers at the redesigned site was selected. Suppose that 24 browsers made a purchase. The company officials would like to know if there is evidence of an increase in conversion rate at the 5% level of significance. -Referring to Scenario 9-8, the company officials can conclude that there is sufficient evidence that the conversion rate at the company's website has increased using a level of significance of 0.05.

If a researcher rejects a false null hypothesis, she has made a _______decision.

You know that the probability of committing a Type II error (β) is 5%, you can tell that the power of the test 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. -A sample is used to obtain a 95% confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be accepted at a level of significance of 0.01.

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. -Referring to Scenario 9-1, the null hypothesis would be rejected.

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. What would be the p-value of this one-tail test?

SCENARIO 9-5 A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45 years, versus an alternative that the mean age is greater than 45 years. They take a sample and calculate a p-value of 0.0202. -Referring to Scenario 9-5, the null hypothesis would be rejected at a significance level of $\alpha$ = 0.01.

If the p-value is less than α in a two-tail test,

SCENARIO 9-5 A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45 years, versus an alternative that the mean age is greater than 45 years. They take a sample and calculate a p-value of 0.0202. -Referring to Scenario 9-5, the bank can conclude that the mean age is greater than 45 at a significance level of $\alpha$ = 0.01.

The symbol for the probability of committing a Type II error of a statistical test is a) $\alpha$ . b) $1 - \alpha$ . c) $\beta$ . d) $1 - \beta$ .

SCENARIO 9-11-B 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 bottle to 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-DataB.XLSX. 1493 1506 1515 1491 1500 1505 1517 1510 1506 1503 1503 1491 1495 1496 1496 1505 1493 1486 1504 1483 1514 1494 1497 1501 1493 1490 1510 1494 1494 1495 1494 1486 1495 1506 1506 1507 1502 1498 1510 1501 1500 1505 1492 1486 1501 1496 1501 1521 1510 1498 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-B, the population of interest is

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 president can conclude that there is sufficient evidence to show that the mean SAT score of the entering class this year is lower than previous years with no more than a 5% probability of incorrectly rejecting the true 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. -In instances in which there is insufficient evidence to reject the null hypothesis, you must make it clear that this does not prove that the null hypothesis is true.