1144 Words5 Pages

Answers to Week Two Homework Assignment ( Assignment 1)
Chapter 7 Exercise 13
You should choose a sample size of 1000 people. Although the law of averages tells you that the average of 1000 people is very likely to be close to 60 inches -- and thus under 65 inches tall -- here you are winning a prize for each person over 65 inches tall, not when the average of all the people is over 65 inches tall. Having a larger sample will give you more chances at getting the prize. The law of averages pertains to the variability of a sample average -- not variability in individual data values themselves. Here, we are concerned with individual data values.
Chapter 7 Exercise 18
No, we can say we are quite confident that the average wage is between $25*…show more content…*

(b) One correct answer to this part is as follows: We can say that we are 95 percent confident the average balance for all overdue bills is in the range from $480 to $520. Since the standard error SE for the sample mean should equal 100/square root of 100 = 10, the margin of error would be twice this value, or $20. This means $480 to $520 is a 95 percent confidence interval for the average of all overdue balances. Another correct answer to this part is as follows: Instead of the average balance, we can find the percentage of all balances that fall in this interval. When we think of all balances, $480 is 0.2 standard deviations below the average (480 -500)/100 = - 0.20), and $520 is 0.2 standard deviations above the average (520 -500)/100 = 0.20). Looking in the Normal table, we see that the area up to z = 0.20 is 58%, and the area up to z = - 0.20 is 42%. The area between them is 16%, so we can say that 16% of all balances fall between $480 and*…show more content…*

If they want to cut this by a factor of three to get it down to $4000, they need to multiply the sample size by 3^2=9, and get a sample size of 25×9= 225. Here is how we can calculate it more directly. We want ME = 4000, and we know ME = 2×SE. Therefore, SE = ME/2 =2000, and also, SE = SD/Square root of sample size. So, 2000 = 30000/Square root of sample size. Solving for the Square root of sample size, we get Square root of sample size = 30000/2000 = 15. Taking its square, the sample size is found as 225. Chapter 9 Exercise 1 No it is not a good defense. If you choose 40 random employees from the corporation, the standard error would equal 6/Square root of 40 = .95 days. The 12 days in this department corresponds to (12-8.2)/.95 = 4 standard errors above the corporation average of 8.2. This is much higher than two or three standard errors, and it appears to be beyond chance variation. Chapter 9 Exercise 3 The p- value tells you how likely it would be to get results at least as extreme as this if there was no difference in the taste and only chance variation was operating. In this problem, p-value of 0.02 means that, if there is no difference in taste, then there is only 2% chance that 70% or more people would declare one drink better than the

(b) One correct answer to this part is as follows: We can say that we are 95 percent confident the average balance for all overdue bills is in the range from $480 to $520. Since the standard error SE for the sample mean should equal 100/square root of 100 = 10, the margin of error would be twice this value, or $20. This means $480 to $520 is a 95 percent confidence interval for the average of all overdue balances. Another correct answer to this part is as follows: Instead of the average balance, we can find the percentage of all balances that fall in this interval. When we think of all balances, $480 is 0.2 standard deviations below the average (480 -500)/100 = - 0.20), and $520 is 0.2 standard deviations above the average (520 -500)/100 = 0.20). Looking in the Normal table, we see that the area up to z = 0.20 is 58%, and the area up to z = - 0.20 is 42%. The area between them is 16%, so we can say that 16% of all balances fall between $480 and

If they want to cut this by a factor of three to get it down to $4000, they need to multiply the sample size by 3^2=9, and get a sample size of 25×9= 225. Here is how we can calculate it more directly. We want ME = 4000, and we know ME = 2×SE. Therefore, SE = ME/2 =2000, and also, SE = SD/Square root of sample size. So, 2000 = 30000/Square root of sample size. Solving for the Square root of sample size, we get Square root of sample size = 30000/2000 = 15. Taking its square, the sample size is found as 225. Chapter 9 Exercise 1 No it is not a good defense. If you choose 40 random employees from the corporation, the standard error would equal 6/Square root of 40 = .95 days. The 12 days in this department corresponds to (12-8.2)/.95 = 4 standard errors above the corporation average of 8.2. This is much higher than two or three standard errors, and it appears to be beyond chance variation. Chapter 9 Exercise 3 The p- value tells you how likely it would be to get results at least as extreme as this if there was no difference in the taste and only chance variation was operating. In this problem, p-value of 0.02 means that, if there is no difference in taste, then there is only 2% chance that 70% or more people would declare one drink better than the

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