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Question 2381 random elementary schools were asked for their average exam scores (sample mean = 535, sample standard deviation = 7). Calculate the 98% confidence interval.1 point(528.00, 542.00)(533.15, 536.85)Not possible to calculate based on this information.(533.41, 536.59)

Question

Question 2381 random elementary schools were asked for their average exam scores (sample mean = 535, sample standard deviation = 7). Calculate the 98% confidence interval.1 point(528.00, 542.00)(533.15, 536.85)Not possible to calculate based on this information.(533.41, 536.59)

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Solution

To calculate the 98% confidence interval, we first need to find the standard error (SE). The formula for the standard error when the population standard deviation is unknown is s/√n, where s is the sample standard deviation and n is the sample size.

In this case, s = 7 and n = 81.

SE = 7/√81 = 7/9 = 0.78

Next, we need to find the z-score that corresponds to a 98% confidence level. The z-score for a 98% confidence level is approximately 2.33 (you can find this value in a standard z-table or using a calculator).

The margin of error (ME) is then calculated as the z-score times the standard error.

ME = 2.33 * 0.78 = 1.82

Finally, we calculate the confidence interval by subtracting the margin of error from the sample mean for the lower limit and adding the margin of error to the sample mean for the upper limit.

Lower limit = 535 - 1.82 = 533.18 Upper limit = 535 + 1.82 = 536.82

So, the 98% confidence interval for the mean score is approximately (533.18, 536.82).

So, the closest answer among the options provided is "(533.15, 536.85)".

This problem has been solved

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