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From a random sample of 60 new games, the mean price for a new game is $51.98. Assume the population has a standard deviation of $47.19. This creates a 95% confidence interval for the population mean of (40.04, 63.92). Does it seem likely that the population mean could be greater than $66? Explain. Choose the correct answer below. A. No, it does not seem likely because $66 is greater than the right endpoint of the confidence interval. B. Yes, it seems likely because $66 is greater than the left endpoint of the confidence interval. C. Yes, it seems likely because $66 is between the endpoints of the confidence interval. D. No, it does not seem likely because $66 is between the endpoints of the confidence interval. 

          From a random sample of 60 new games, the mean price for a new game is $51.98. Assume the population has a standard deviation of $47.19. This creates a 95% confidence interval for the population mean of (40.04, 63.92). Does it seem likely that the population mean could be greater than $66? Explain.
Choose the correct answer below.
A. No, it does not seem likely because $66 is greater than the right endpoint of the confidence interval.
B. Yes, it seems likely because $66 is greater than the left endpoint of the confidence interval.
C. Yes, it seems likely because $66 is between the endpoints of the confidence interval.
D. No, it does not seem likely because $66 is between the endpoints of the confidence interval.
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From a random sample of 60 new games, the mean price for a new game is 51.98. Assume the population has a standard deviation of47.19. This creates a 95% confidence interval for the population mean of (40.04, 63.92). Does it seem likely that the population mean could be greater than $66? Explain. Choose the correct answer below. A. No, it does not seem likely because 66 is greater than the right endpoint of the confidence interval. B. Yes, it seems likely because66 is greater than the left endpoint of the confidence interval. C. Yes, it seems likely because 66 is between the endpoints of the confidence interval. D. No, it does not seem likely because66 is between the endpoints of the confidence interval.

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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From a random sample of 60 new games, the mean price for a new game is $51.98. Assume the population has a standard deviation of $47.19. This creates a 95% confidence interval for the population mean of (40.04,63.92). Does it seem likely that the population mean could be greater than $66 ? Explain. Choose the correct answer below. No, it does not seem likely because $66 is greater than the right endpoint of the confidence interval. B. Yes, it seems likely because $66 is greater than the left endpoint of the confidence interval. C. Yes, it seems likely because $66 is between the endpoints of the confidence interval. D. No, it does not seem likely because $66 is between the endpoints of the confidence interval. From a random sample of 60 new games, the mean price for a new game is $51.98. Assume the population has a standard deviation of $47.19. This creates a 95% confidence interval for the population mean of 40.04,63.92.Does it seem likely that the population mean could be greater than $66?Explain. Choose the correct answer below No,it does not seem likely because S66 is greater than the right endpoint of the confidence interval B.Yes,it seems likely because $66 is greater than the left endpoint of the confidence interval O C.Yes,it seems likely because S66 is between the endpoints of the confidence interval O D. No,it does not seem likely because $66 is between the endpoints of the confidence interval
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Transcript

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00:01 Hello, here we have to solve the following problem.
00:02 Sample has a size of 60 items and that is n1.
00:09 Mean of the sample is 90 and the population standard deviation sigma is 15.
00:19 First in question a we have to find 95 confidence interval.
00:29 Let's do this.
00:31 So here this confidence interval can be found as following that is mean plus minus t coefficient times sigma over square root of n.
00:43 Here we need to determine number of degrees of freedom which is n1 minus 1 that is 59.
00:55 And now let's use either calculator or a table to determine t.
01:00 Let's do this.
01:02 That is 2.
01:05 Now once it's calculated let's determine the confidence interval.
01:32 Then this confidence interval is from 86 .13 to 93 .87.
01:44 That is answer to question one.
01:47 Now let's move on to question two, question b...
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