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.