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 99% 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) 48 52 85 83 60 92 78 90 83 71 Rating (this year) 64 63 66 88 44 84 67 80 93 81 Step 3 of 4 : Calculate the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Answer 1

From the Data: = -1.2 and Sd = 12.865

= 0.01, the degrees of freedom = n – 1 = 9

The critical value = 3.25

The Confidence interval is given by

________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

#	Difference	Mean	(X-Mean)2
1	16	-1.2	295.84
2	11	-1.2	148.84
3	-19	-1.2	316.84
4	5	-1.2	38.44
5	-16	-1.2	219.04
6	-8	-1.2	46.24
7	-11	-1.2	96.04
8	-10	-1.2	77.44
9	10	-1.2	125.44
10	10	-1.2	125.44