Question

Construct two confidence intervals for the population mean: a 95% confidence interval and a 99% confidence interval. Assume that your data is normally distributed and the population standard deviation is unknown Here is the data set of 60 numbers. 69 35 60 55 49 60 72 70 70 73 68 72 74 69 46 48 70 55 49 60 72 70 76 56 59 64 71 69 55 61 70 55 45 69 54 48 60 61 50 59 60 62 63 53 64 50 69 52 68 70 69 59 58 69 65 61 59 71 71 68

Answer 1

First, we need to find the mean and sample standard deviation

Mean = (sum of all values)/(total number of values)

setting the values, we get

Mean = (69+35+...+71+68)/(60) here  (69+35+...+71+68) shows sum of all values

we get

Mean = 3709/60 = 61.82 =

and standard deviation =

where xi are the given values and x(bar) is mean, n is the sample size = 60

setting the values, we get

standard deviation =

this gives us

Standard deviation (s) = 8.92

Now, we have unknown population standard deviation, so we will use t distribution. We need to calculate t critical values for the 95% and 99% confidence levels.

For 95% confidence interval, significance level = 1-0.95 = 0.05

degree of freedom = n-1 = 60-1 = 59

Using t distribution table, we get t critical = 2.00

Confidence interval =

setting the given values, we get

CI(95) =

Similarly, for 99% confidence interval, significance level = 1-0.99 = 0.01

degree of freedom = n-1 = 60-1 = 59

Using t distribution table, we get t critical = 2.66

Confidence interval =

setting the given values, we get

CI(99) =