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

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.

Instruction 9-7 Microsoft Excel was used on a set of data involving the number of parasites found in a random sample of 46 Blue Tiger butterflies captured in Kakadu National Park. A biologist wants to know if the mean number of parasites per butterfly is greater than 20. 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 butterflies: $n = 46$ ; Arithmetic Mean $= 28.00$ ; Standard Deviation $= 25.92$ ; Standard Error $= 3.82$ ; Null Hypothesis: $H _ { 0 } : \mu \leq 20.000 ; a = 0.10 ; \mathrm { df } = 45 ; T$ Test Statistic $= 2.09$ ; One-Tailed Test Upper CriticalValue $= 1.3006 ; p$ value $= 0.021$ ; Decision $=$ Reject. -Referring to Instruction 9-7,the lowest level of significance at which the null hypothesis can be rejected is______.

A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches.To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90,a random sample of 100 doctors was selected.Suppose you reject the null hypothesis.What conclusion can you draw?

Which of the following statements is NOT true about the level of significance in a hypothesis test?

Instruction 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-tailed 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 Instruction 9-6,if the test is performed with a level of significance of 0.10,the null hypothesis would be rejected.

You have created a 95% confidence interval for $\mu$ with the result (10,15). -What decision will you make if you test H₀: $\mu$ = 16 versus H₁: $\mu$ $\neq$ 16 at $\alpha$ = 0.05?

In hypothesis testing,it is not relevant whether the test is a one-tail or two-tail test.

Instruction 9-9 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 at a 0.01 level of significance. -Referring to Instruction 9-9,state the alternative hypothesis for this study.

A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches.To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90,a random sample of 100 doctors was selected.Suppose that the test statistic is -2.20.Can we conclude that H₀ should be rejected at the (α = 0.10,(α = 0.05,and (α = 0.01 level of Type I error?

You should report only the results of hypothesis tests that show statistical significance and omit those for which there is insufficient evidence in the findings.

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

In conducting research,you should document both good and bad results.

Instruction 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 Instruction 9-3,the p-value of the test is ________.

Instruction 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 Instruction 9-3,the appropriate hypotheses to determine if the manufacturer's claim appears reasonable are

Instruction 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-tailed 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 Instruction 9-6,suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650.What would be the p-value of this one-tailed test?

Instruction 9-4 A drug company is considering marketing a new local anaesthetic. The effective time of the anaesthetic 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 anaesthetic is such that the effective time should be normal with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anaesthetic; otherwise, it will continue to produce the older drug. 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 Instruction 9-4,if the level of significance had been chosen as 0.05,the company would market the new anaesthetic.

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 rejected at a level of significance of 0.05.

A Type II error is committed when

Instruction 9-1 A student claims that she can correctly identify whether a person is studying a business or science major by the way the person dresses. Suppose in actuality that if someone is studying a business major, she can correctly identify that person as a business student 87% of the time. When a person is studying a science major, the student will incorrectly identify that person as a business student 16% of the time. Presented with one person and asked to identify the area of study of this person (who is studying either a business or science major), she considers this to be a hypothesis test with the null hypothesis being that the person is a business student and the alternative that the person is a science student. -Referring to Instruction 9-1,what would be a Type I error?

Instruction 9-7 Microsoft Excel was used on a set of data involving the number of parasites found in a random sample of 46 Blue Tiger butterflies captured in Kakadu National Park. A biologist wants to know if the mean number of parasites per butterfly is greater than 20. 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 butterflies: $n = 46$ ; Arithmetic Mean $= 28.00$ ; Standard Deviation $= 25.92$ ; Standard Error $= 3.82$ ; Null Hypothesis: $H _ { 0 } : \mu \leq 20.000 ; a = 0.10 ; \mathrm { df } = 45 ; T$ Test Statistic $= 2.09$ ; One-Tailed Test Upper CriticalValue $= 1.3006 ; p$ value $= 0.021$ ; Decision $=$ Reject. -Referring to Instruction 9-7,the biologist can conclude that there is sufficient evidence to show that the mean number of parasites per butterfly is greater than 20 with no more than a 5% probability of incorrectly rejecting the true null hypothesis.