A random sample of 85 sixth-graders in a large city take a course designed to improve scores on a reading comprehension test. Based on this sample, a 90% confidence interval for the mean improvement in test scores for all sixth-graders in the city taking this course is found to be (12.6, 14.8). Which of the following are the sample mean and margin of error on which this interval is based?A. Sample Mean = 13.7; Margin of Error = 1.1B. Sample mean = 13.7; Margin of Error = 2.2C. Sample mean is unknown; Margin of Error = 2.2

For this case we can find the margin of error like this:

Since we need to divide the width of the interval by 2.

And now with the margin of error we can find the sample mean with any of the two following ways:

So for this case the correct answer would be:

A. Sample Mean = 13.7; Margin of Error = 1.1

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

(1)  

In order to calculate the critical value we need to find first the degrees of freedom, given by:  

We know that the confidence interval is given by:

For this case we can find the margin of error like this:

Since we need to divide the width of the interval by 2.

And now with the margin of error we can find the sample mean with any of the two following ways:

So for this case the correct answer would be:

A. Sample Mean = 13.7; Margin of Error = 1.1

ƒ = 0.03b

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step-by-step explanation: