Deck 9: Hypothesis Testing
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Deck 9: Hypothesis Testing
1
A professional employee in a large corporation receives an average of = 42.4 e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 30 employees showed that they were receiving an average of x = 33.6 e-mails per day. The computer server through which the e-mails are routed showed that = 19.4. Has the new policy had any effect? Use a 5% level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee. What is the value of the test statistic?
A)0.454
B)0.083
C)-0.454
D)-2.485
E)-0.083
A)0.454
B)0.083
C)-0.454
D)-2.485
E)-0.083
-2.485
2
A professional employee in a large corporation receives an average of =44.2 e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 49 employees showed that they were receiving an average of x = 39.1 e-mails per day. The computer server through which the e-mails are routed showed that = 16.7. Has the new policy had any effect? Use a 1% level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee. Find (or estimate) the P-value.
A)0.984
B)0.033
C)0.967
D)0.008
E)0.016
A)0.984
B)0.033
C)0.967
D)0.008
E)0.016
0.033
3
Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with = 2.3%. A random sample of 15 Australian bank stocks has a sample mean of x = 8.71%. For the entire Australian stock market, the mean dividend yield is = 5.5%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 5.5%? Use = 0.05. What is the value of the test statistic?
A)-1.396
B)-0.360
C)5.405
D)0.360
E)-5.405
A)-1.396
B)-0.360
C)5.405
D)0.360
E)-5.405
5.405
4
A severe storm has an average peak wave height of 16.4 feet for waves hitting the shore. Suppose that a storm is in progress with a severe storm class rating. Let us say that we want to set up a statistical test to see if the wave action (i.e., height) is dying down or getting worse. If you wanted to test the hypothesis that the waves are dying down, what would you use for the alternate hypothesis? Is the P-value area on the left, right, or on both sides of the mean?
A)H1 : is less than 16.4 feet; the P-value area is on the right of the mean
B)H1 : is not equal to 16.4 feet; the P-value area is on the right of the mean
C)H1 : is not equal to 16.4 feet; the P-value area is on the left of the mean
D)H1 : is greater than 16.4 feet; the P-value area is on the right of the mean
E)H1 : is less than 16.4 feet; the P-value area is on the left of the mean
A)H1 : is less than 16.4 feet; the P-value area is on the right of the mean
B)H1 : is not equal to 16.4 feet; the P-value area is on the right of the mean
C)H1 : is not equal to 16.4 feet; the P-value area is on the left of the mean
D)H1 : is greater than 16.4 feet; the P-value area is on the right of the mean
E)H1 : is less than 16.4 feet; the P-value area is on the left of the mean
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5
Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with = 2.3%. A random sample of 11 Australian bank stocks has a mean x = 9.89%. For the entire Australian stock market, the mean dividend yield is = 7.9%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 7.9%? Use = 0.05. What is the level of significance?
A)0.025
B)0.050
C)0.100
D)0.900
E)0.975
A)0.025
B)0.050
C)0.100
D)0.900
E)0.975
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6
Suppose that the mean time for a certain car to go from 0 to 60 miles per hour was 8.1 seconds. Suppose that you want to test the claim that the average time to accelerate from 0 to 60 miles per hour is less than 8.1 seconds. What would you use for the alternative hypothesis?
A)H1 : <8.1 seconds
B)H1 : 8.1 seconds
C)H1 : 8.1 seconds
D)H1 : 8.1 seconds
E)H1 : = 8.1 seconds
A)H1 : <8.1 seconds
B)H1 : 8.1 seconds
C)H1 : 8.1 seconds
D)H1 : 8.1 seconds
E)H1 : = 8.1 seconds
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7
Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with = 2.9%. A random sample of 14 Australian bank stocks has a sample mean of x = 6.35%. For the entire Australian stock market, the mean dividend yield is = 5.2%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 5.2%? Use = 0.05. Find (or estimate) the P-value.
A)0.931
B)0.034
C)0.069
D)0.138
E)1.862
A)0.931
B)0.034
C)0.069
D)0.138
E)1.862
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8
A professional employee in a large corporation receives an average of = 38.9 e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 31 employees showed that they were receiving an average of x = 30.1 e-mails per day. The computer server through which the e-mails are routed showed that = 19.4. Has the new policy had any effect? Use a 10% level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee. Are the data statistically significant at level ? Based on your answers, will you reject or fail to reject the null hypothesis?
A)The P-value is greater than than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.
B)The P-value is greater than than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.
C)The P-value is less than than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.
D)The P-value is less than than the level of significance and so the data are not statistically significant. Thus, we fail to reject the null hypothesis.
E)The P-value is greater than than the level of significance and so the data are statistically significant. Thus, we fail to reject the null hypothesis.
A)The P-value is greater than than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.
B)The P-value is greater than than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.
C)The P-value is less than than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.
D)The P-value is less than than the level of significance and so the data are not statistically significant. Thus, we fail to reject the null hypothesis.
E)The P-value is greater than than the level of significance and so the data are statistically significant. Thus, we fail to reject the null hypothesis.
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9
A professional employee in a large corporation receives an average of = 40.8 e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 31 employees showed that they were receiving an average of x = 32.2 e-mails per day. The computer server through which the e-mails are routed showed that = 17.3. Has the new policy had any effect? Use a 5% level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee. What are the null and alternate hypotheses?
A)H0 : 40.8 e-mails; H1 : =40.8 e-mails
B)H0 : =40.8 e-mails; H1 : 40.8 e-mails
C)H0 : = 40.8 e-mails; H1 : <40.8 e-mails
D)H0 : = 40.8 e-mails; H1 : 40.8 e-mails
E)H0 : 40.8 e-mails; H1 : .8 e-mails
A)H0 : 40.8 e-mails; H1 : =40.8 e-mails
B)H0 : =40.8 e-mails; H1 : 40.8 e-mails
C)H0 : = 40.8 e-mails; H1 : <40.8 e-mails
D)H0 : = 40.8 e-mails; H1 : 40.8 e-mails
E)H0 : 40.8 e-mails; H1 : .8 e-mails
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10
Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n =276 numerical entries from the file and r = 95 of the entries had a first nonzero digit of 1. Let p represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that p is less than 0.301 by using = 0.1. What is the level of significance?
A)0.100
B)0.950
C)0.800
D)0.900
E)0.200
A)0.100
B)0.950
C)0.800
D)0.900
E)0.200
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11
Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with = 2.9%. A random sample of 8 Australian bank stocks has a sample mean of x = 6.74%. For the entire Australian stock market, the mean dividend yield is = 6.2%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 6.2%? Use = 0.05. Are the data statistically significant at the given level of significance? Based on your answers, will you reject or fail to reject the null hypothesis?
A)The P-value is less than than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.
B)The P-value is less than than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.
C)The P-value is greater than than the level of significance and so the data are statistically significant. Thus, we fail to reject the null hypothesis.
D)The P-value is greater than than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.
E)The P-value is greater than than the level of significance and so the data are not statistically significant. Thus, we fail to reject the null hypothesis.
A)The P-value is less than than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.
B)The P-value is less than than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.
C)The P-value is greater than than the level of significance and so the data are statistically significant. Thus, we fail to reject the null hypothesis.
D)The P-value is greater than than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.
E)The P-value is greater than than the level of significance and so the data are not statistically significant. Thus, we fail to reject the null hypothesis.
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12
Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 275 numerical entries from the file and r = 67 of the entries had a first nonzero digit of 1. Let p represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that p is less than 0.301 by using = 0.1. What does the area of the sampling distribution corresponding to your P-value look like?
A)The area not in the left tail and the right tail of the standard normal curve.
B)The area not including the left tail of the standard normal curve.
C)The area not including the right tail of the standard normal curve.
D)The area in the left tail and the right tail of the standard normal curve.
E)The area in the left tail of the standard normal curve.
A)The area not in the left tail and the right tail of the standard normal curve.
B)The area not including the left tail of the standard normal curve.
C)The area not including the right tail of the standard normal curve.
D)The area in the left tail and the right tail of the standard normal curve.
E)The area in the left tail of the standard normal curve.
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13
Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 499 numerical entries from the file and r = 131 of the entries had a first nonzero digit of 1. Let p represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that p is less than 0.301 by using = 0.01. What is the value of the test statistic?
A)1.874
B)-41.856
C)-0.084
D)0.084
E)-1.874
A)1.874
B)-41.856
C)-0.084
D)0.084
E)-1.874
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14
Suppose that the mean time for a certain car to go from 0 to 60 miles per hour was 7.8 seconds. Suppose that you want to test the claim that the average time to accelerate from 0 to 60 miles per hour is longer than 7.8 seconds. What would you use for the alternative hypothesis?
A)H1 : < 7.8 seconds
B)H1 : 7.8 seconds
C)H1 : = 7.8 seconds
D)H1 : 7.8 seconds
E)H1 : 7.8 seconds
A)H1 : < 7.8 seconds
B)H1 : 7.8 seconds
C)H1 : = 7.8 seconds
D)H1 : 7.8 seconds
E)H1 : 7.8 seconds
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15
A professional employee in a large corporation receives an average of = 41.5 e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 37 employees showed that they were receiving an average of x = 33.8 e-mails per day. The computer server through which the e-mails are routed showed that = 19.5 Has the new policy had any effect? Use a 10% level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee. What is the level of significance?
A) = 0.95
B) = 0.90
C) = 0.80
D) = 0.20
E) = 0.10
A) = 0.95
B) = 0.90
C) = 0.80
D) = 0.20
E) = 0.10
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16
Suppose that the mean time for a certain car to go from 0 to 60 miles per hour was 8.8 seconds. Suppose that you want to set up a statistical test to challenge the claim of 8.8 seconds. What would you use for the null hypothesis?
A)H0 : 8.8 seconds
B)H0 : = 8.8 seconds
C)H0 : < 8.8 seconds
D)H0 : 8.8 seconds
E)H0 : 8.8 seconds
A)H0 : 8.8 seconds
B)H0 : = 8.8 seconds
C)H0 : < 8.8 seconds
D)H0 : 8.8 seconds
E)H0 : 8.8 seconds
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17
Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 287 numerical entries from the file and r = 73 of the entries had a first nonzero digit of 1. Let p represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that p is less than 0.301 by using = 0.1. Are the data statistically significant at the significance level? Based on your answers, will you reject or fail to reject the null hypothesis?
A)The P-value is less than the level of significance so the data are not statistically significant. Thus, we reject the null hypothesis.
B)The P-value is greater than the level of significance so the data are not statistically significant. Thus, we fail to reject the null hypothesis.
C)The P-value is less than the level of significance so the data are not statistically significant. Thus, we fail to reject the null hypothesis.
D)The P-value is less than the level of significance so the data are statistically significant. Thus, we fail to reject the null hypothesis.
E)The P-value is less than the level of significance so the data are statistically significant. Thus, we reject the null hypothesis.
A)The P-value is less than the level of significance so the data are not statistically significant. Thus, we reject the null hypothesis.
B)The P-value is greater than the level of significance so the data are not statistically significant. Thus, we fail to reject the null hypothesis.
C)The P-value is less than the level of significance so the data are not statistically significant. Thus, we fail to reject the null hypothesis.
D)The P-value is less than the level of significance so the data are statistically significant. Thus, we fail to reject the null hypothesis.
E)The P-value is less than the level of significance so the data are statistically significant. Thus, we reject the null hypothesis.
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