Question

A state legislator wants to determine whether his voters' performance rating (0 - 100) has changed from last year to this year. The following table shows the legislator's performance from the same ten randomly selected voters for last year and this year. Use this data to find the 98 % confidence interval for the true difference between the population means. Assume that the populations of voters' performance ratings are normally distributed for both this year and last year. Rating (last year) 74 72 41 | 69 57 95 85 76 79 88 Rating (this year) 85 | 48 49 | 45 84 86 57 72 67 92 Copy Data Step 2 of 4: Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places.

Step 2 of 4: Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places.

Step 3 of 4: Calculate the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Step 4 of 4: Construct the 98%98% confidence interval. Round your answers to one decimal place.

Answer 1

Answer:

Given,

Mean = 5.1

Standard deviation = 17.785450

degree of freedom = n -1 = 10 - 1 = 9

t(alpha/2,df) = t(0.02/2 , 9) = 2.821433 = 2.82

Margin of error = t*s/sqrt(n)

substitute values

= 2.82*17.785450/sqrt(10)

= 15.860394

98% CI = xbar +/- E

substitute values

= 5.1 +/- 15.860394

= (-10.8 , 21.0)