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 90% 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): 87 67 68 75 59 60 50 41 75 72

Rating (this year): 85 52 51 53 50 52 80 44 48 57

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=x2âx1 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 90%90% confidence interval. Round your answers to one decimal place.

Answer 1

Answer:

Step 1 of 4

Point estimate for the population mean of the paired differences = -8.2

Step 2 of 4

Sample standard deviation of the paired differences = 16.116244

Step 3 of 4

Margin of Error = ±9.326419

Step 4 of 4

90% Confidence interval = (-17.5, 1.1)

Step-by-step explanation:

The ratings from last year and this year are given in table as

Rating (last year) | x1 | 87 67 68 75 59 60 50 41 75 72

Rating (this year) | x2| 85 52 51 53 50 52 80 44 48 57

Difference | x2 - x1 | -2 -15 -17 -22 -9 -8 30 3 -27 -15

Step 1 of 4

Mean = (Σx)/N = (-82/10) = -8.2 to 1 d.p.

Step 2 of 4

Standard deviation for the sample

= √{[Σ(x - xbar)²]/(N-1)} = 16.116244392951 = 16.116244 to 6 d.p.

Step 3 of 4

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = -8.2

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 10 - 1 = 9.

Significance level for 90% confidence interval

= (100% - 90%)/2 = 5% = 0.05

t (0.05, 9) = 1.83 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 16.116244

n = sample size = 10

σₓ = (16.116244/√10) = 5.0964038367

Margin of Error = (Critical value) × (standard Error of the mean) = 1.83 × 5.0964038367 = 9.3264190212 = 9.326419 to 6 d.p.

Step 4 of 4

90% Confidence Interval = (Sample mean) ± (Margin of Error)

CI = -8.2 ± (9.326419)

90% CI = (-17.5264190212, 1.1264190212)

90% Confidence interval = (-17.5, 1.1)

Hope this Helps!!!