For a sample group of 22, the mean is 86 with a standard deviation of 5. What is the 90% confidence interval of the true mean of the sample? Can you please explain the steps you used to get the answer. Thanks.

Formula:

CI90 = mean + or - 1.645(sd/√n)

Substituting your data:

CI90 = 86 + or - 1.645(5/√22)

Finish the calculations for your confidence interval.

Thank you so much!

To calculate the 90% confidence interval of the true mean of a sample group, we can use the formula:

Confidence interval = Mean ± (Critical value * Standard deviation/square root of sample size)

1. Find the critical value:

The critical value is based on the desired level of confidence and the sample size. In this case, we want a 90% confidence level.

To find the critical value, we can use a z-table or a calculator. Since the sample size is greater than 30, we can use the standard normal distribution and find the z-value for a 90% confidence level.

The z-value for the 90% confidence level is approximately 1.645.

2. Calculate the confidence interval:

Now, we can plug in the values into the formula:

Confidence interval = 86 ± (1.645 * 5 / √22)

To calculate the square root of the sample size (√22), take the square root of 22, which is approximately 4.69.

Confidence interval = 86 ± (1.645 * 5 / 4.69)

Confidence interval = 86 ± 1.75

So, the 90% confidence interval of the true mean of the sample is (84.25, 87.75).

This means we are 90% confident that the true mean of the population lies between 84.25 and 87.75 based on this sample.