Question

Calculate a 90onfidence interval for a sample mean of 15 with a sample standard deviation of 5 and a sample size of 25. the answer should be accurate to the nearest decimal. given t = 1.711.

Answer 1

The confidence interval for the given sample will be 15±1.711.

Answer 2

The confidence interval for the given sample will be 15±1.711.

To find the answer, we have to know about the confidence interval.

How to find the confidence interval for a sample?

• A confidence interval in statistics is a range of values that is established using observed data and computed at a chosen confidence level and may contain the actual value of the parameter under study.
• In order to calculate a confidence interval, the sample mean (X) and, if applicable, the standard deviation d must be known.

• We have the expression for confidence interval as,

                           $C= X+t (\frac{d}{\sqrt{n} } )\\ or \\C= X-t (\frac{d}{\sqrt{n} } )$

t-value for the chosen confidence level, X is the sample mean, d is the standard deviation, and n is the sample size.

• Thus, the confidence interval for the given sample will be,

                        $C=15+1.711(\frac{5}{\sqrt{25} } )=15+1.711\\or\\C=15-1.711$

Thus, the confidence interval for the given sample will be 15±1.711.

Learn more about the confidence interval here:

https://brainly.com/question/27630001

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