Question

15. A sample of 10 networking sites for a specific month has a mean number of visits of 26.1 and a standard deviation of 4.2. Find a 99 percent confidence interval of the true mean.

NOTE: be sure to take the square root of a given Variance in the word problems to obtain the standard deviation and use that for your figuring

a. 21.8, 30.4

b. 22.2, 28.4

c. 20.1, 33.2

d. 21.9, 36.8

Answer 1

A sample of 10 networking sites for a specific month has a mean number of visits of 26.1 and a standard deviation of 4.2. Now we want to find the 99% confidence interval of the true mean.

Here, n = sample size = 10, S = sample standard deviation = 4.2 , Sample mean = 26.1

Since population standard deviation is not given so we use t-distribution critical value to find the 99% confidence interval.

Calculation of level of significance

DF of t-distribution:-

Critical Value:-

[ Value are getting from t-distribution probability table with corresponding , DF = 9 and two tailed probability ]

99% Confidence Interval:-

[ Now we put the value of and we get,]

[ Round to one decimal places ]

So we get 99% confidence interval of the true mean is [ 21.8, 30.4 ]

Answer:- Correct answer is " Option a. 21.8, 30.4 "