Question

A state legislator wants to determine whether his voters' performance rating (00 - 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) 83 89 91 71 56 75 59 43 92 78

Rating (this year) 54 83 68 75 49 89 80 69 72 56

Step 1 of 4 : Find the point estimate for the population mean of the paired differences. Let x1 be the rating from last year and x2 be the rating from this year and use the formula d=x2−x1 to calculate the paired differences. Round your answer to one decimal place.

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 99% confidence interval. Round your answers to one decimal place.

Answer 1

Number	This Year	Last Year	Difference	( di - d̅ )2
	54	83	-29	615.04
	83	89	-6	3.24
	68	91	-23	353.44
	75	71	4	67.24
	49	56	-7	7.84
	89	75	14	331.24
	80	59	21	635.04
	69	43	26	912.04
	72	92	-20	249.64
	56	78	-22	316.84
Total	695	737	-42	3491.6

Step 1

d̅ = Σdi/n = -42 / 10 = -4.2

Step 2

S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √( 3491.6 / 9) = 19.696587

Step 3

Margin of Error = t(α/2, n-1) Sd / √(n) = 20.242232

Step 4

Confidence Interval :-
d̅ ± t(α/2, n-1) Sd / √(n)
t(α/2) = t(0.01 /2) = 3.2
-4.2 ± t(0.01/2) * 19.6966/√(10)
Lower Limit = -4.2 - t(0.01/2) 19.6966/√(10)
Lower Limit = -24.4
Upper Limit = -4.2 + t(0.01/2}) 19.6966/√(10)
Upper Limit = 16.0
99% Confidence interval is ( -24.4 , 16.0 )