Chapter 9 Quiz Practice Name: Other

Chapter 9 Quiz Practice Name: ___________________________________
Other
1. As a construction engineer for a city, you are responsible for ensuring that the company that is providing gravel for a new road puts as much gravel in each truckload as they claim to. It has been estimated that it will take 500 truckloads of gravel to complete this road, so you plan to measure the volume of gravel in an SRS of 25 trucks to make sure that the company isn’t delivering less gravel per truckload than they claim. Each truckload is supposed to have 20 m³ of gravel, so you will test the hypotheses versus at the a = 0.05 level.
(a) Describe what a Type I error would be in this context.

(b) What is the probability of making a Type I error when performing this test?

(c) Describe what a Type II error would be in this context.

(d) Which error—Type I or Type II—is a more serious problem for the city? Explain.

2. A certain cigarette brand advertises that the mean nicotine content of their cigarettes is 1.5 mg, but you are suspicious and plan to investigate the advertised claim by testing the hypotheses versus at the significance level. You will do so by measuring the nicotine content of 30 randomly selected cigarettes of this brand.
(a) Describe what a Type I error would be in this context.

(b) Describe what a Type II error would be in this context.

(c) From the perspective of public health, which error—Type I or Type II—is more serious? Explain.
(d) Explain why you know that the probability of making a Type I error when performing this test is 0.05.
3. Nationally, the proportion of red cars on the road is 0.12. A statistically-minded fan of the Philadelphia Phillies (whose team color is red) wonders if fans who park at Citizens Bank Park (the Phillies home field) are more likely to drive red cars. One day during a home game, he takes an SRS of 210 cars parked in the lot while a game is being played, and counts 35 red cars. (There are 21,000 parking spaces.)
(a) Is this convincing evidence that Phillies fans prefer red cars more than the general population? Support your conclusion with a test of significance.

4. LeRoy, a starting player for a major college basketball team, made only 40% of his free throws last season. During the summer, he worked on developing a softer shot in hopes of improving his free throw accuracy. In the first eight games of this season, LeRoy made 25 free throws in 40 attempts. You want to investigate whether LeRoy’s work over the summer will result in a higher proportion of free-throw successes this season. What conclusion would you draw at the = 0.01 level about LeRoy’s free throw shooting? Justify your answer with a complete significance test.

5. Sweet corn of a certain variety is known to produce individual ears of corn with a mean weight of 8 ounces. A farmer is testing a new fertilizer designed to produce larger ears of corn, as measured by their weight. He finds that 32 randomly-selected ears of corn grown with this fertilizer have a mean weight of 8.23 ounces and a standard deviation of 0.8 ounces. There are no outliers in the data.
(a) Do these samples provide convincing evidence at the a= 0.05 level that the fertilizer had a positive impact on the weight of the corn ears? Justify your answer.

6. Does too much sleep impair intellectual performance? Researchers examined this commonly held belief by comparing the performance of subjects on the mornings following A. two normal night’s sleep and B. two nights of “extended sleep.” The order of these two treatments was determined randomly. In the morning they were given a number of tests of ability to think quickly and clearly. One test was for vigilance where the lower the score, the more vigilant the subject, (vigilance = alertness). The following data were collected:

	Vigilance Score
Subject	A	B	C	D	E	F	G	H	I	J
Normal Sleep	8	9	14	4	12	11	3	26	3	8
Extended Sleep	8	9	17	2	21	16	9	38	10	0

Carry out an appropriate test to help answer the researchers’ question.

Chapter 9 Quiz Practice

Answer Section
OTHER
1. ANS:

A. Type I error: concluding that the mean volume of gravel per truck is below 20 m³ when it is equal to (or greater than) 20 m³. B. P(Type I error) = = 0.05 C. Type II error: not concluding that the mean volume of gravel per truck is less than 20 m³ when it is. D. A Type II error is probably more problematic for the city, since it means they would be paying full price for underweight trucks.

PTS: 1
2. ANS:

Type I error: Concluding that the mean nicotine content per cigarette is greater than 1.5 mg when it is equal to (or less than) 1.5 mg. B. Type II error: Not concluding that the mean nicotine level is greater than 1.5 mg per cigarette when it is. C. A Type II error would mean that you fail to discover that the cigarettes have a higher nicotine content that the company claims, which means people will be exposed to more nicotine that they expect. A Type I error might bring unwarranted negative publicity to the tobacco company, but that is not a public health issue! D. When we choose a significance level of _ = 0.05 we will reject the null hypothesis if we get a sample mean nicotine content that is far enough above 1.5 mg. so that it only occurs in 5% of samples when the true mean is 1.5 mg. In other words, if the null hypothesis is true, 5% of all possible sample means will cause us to (incorrectly) reject H₀.

PTS: 1
3. ANS:

A. State: We wish to test versus , where p = the true proportion of red cars in the parking lot. We will use a significance level of a = 0.05. Plan: The procedure is a one-sample z-test for a proportion. Conditions: Random: The fan took an SRS of 210 cars. 10%: 210 is clearly less than 10% of all the cars in the lot. Large counts: Assuming is true, , and Do: so ; P-value = 0.0187. Conclude: A P-value of 0.0187 is less than a = 0.05, so we reject H₀ and conclude that there is convincing evidence that the true proportion of red cars at Citizens Bank Park is greater than 0.12. B. If is true, there is a probability of 0.0187 of getting a sample proportion of red cars as far or farther above 0.12 as 0.1667 is.

PTS: 1
4. ANS:

State: We wish to test versus , where p = Leroy’s new free throw accuracy, expressed as the true proportion of all shots he will take this season that he’ll make. We will use a significance level of a = 0.01. Plan: The procedure is a one-sample z-test for a proportion. Conditions: Random: We will have to assume that LeRoy’s first 40 attempts this season are an SRS of all shots he might over the remainder of his career (and that he will take more than 400 shots) . Large counts: Assuming is true, , and Do: , so ; P-value = 0.0018. Conclude: A P-value of 0.0018 is less than a = 0.01, so we reject H₀ and conclude that there is convincing evidence that LeRoy’s free-throw shooting percentage has improved.

PTS: 1
5. ANS:

A. State: We wish to test versus , where m = mean weight (in ounces) of all corn ears of this variety grown with the new fertilizer. We are using a significance level of a = 0.05. Plan: The procedure is a one-sample t-test for a mean. Conditions: Random: The farmer randomly selected 32 ears of corn. 10%: It seems reasonable to assume that the farmer’s crop consists of more than 280 ears of corn. Normal/Large sample: n=32 is large enough as long as there are no outliers or strong skew in the sample. We are told there are no outliers, so it seems safe to proceed. Do: ; df = 31; P-value = 0.057. Conclude: A P-value of 0.057 is greater than a = 0.05, so we fail to reject H₀: we do not have convincing evidence that the corn of this variety grown with the new fertilizer weighs more than 8 oz., on average. B. If , then ; df = 31; P-value = 0.04985, and we would reject H₀. This points out that it may not be wise to attach to much importance to statistical significance, since a small change in mean corn weight can change our statistical conclusion.

PTS: 1
6. ANS:

State: We wish to test versus , where = the mean difference in vigilance score (extended sleep – normal sleep) for all possible subject of the experiment. Since lower values indicate more vigilance, we are looking for positive values of this difference. We will use a significance level of _ = 0.05. Plan: The procedure is a one-sample t-test on paired data. Conditions: Random: The treatments (extended and normal sleep) were assigned in random order. 10%: We aren’t sampling, so there’s no need to check this condition. NormalLarge Sample: The dotplot (right) does not indicate strong skew, and the extreme values are not outliers by the 1.5 x IQR standard. Do: ; ; df = 9; P-value = 0.0593. Conclude: A P-value of 0.0593 is greater than _ = 0.05, so we fail to reject H₀: we do not have convincing evidence that extended sleep reduces vigilance.
PTS: 1