In: Statistics and Probability
A state legislator wants to determine whether his voters' performance rating (00 - 100100) 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%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) | 8383 | 8989 | 9191 | 7171 | 5656 | 7575 | 5959 | 4343 | 9292 | 7878 |
---|---|---|---|---|---|---|---|---|---|---|
Rating (this year) | 5454 | 8383 | 6868 | 7575 | 4949 | 8989 | 8080 | 6969 | 7272 | 5656 |
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Step 1 of 4:
Find the point estimate for the population mean of the paired differences. Let x1x1 be the rating from last year and x2x2 be the rating from this year and use the formula d=x2−x1d=x2−x1 to calculate the paired differences. Round your answer to one decimal place.
A state legislator wants to determine whether his voters' performance rating (00 - 100100) 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%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) | 8383 | 8989 | 9191 | 7171 | 5656 | 7575 | 5959 | 4343 | 9292 | 7878 |
---|---|---|---|---|---|---|---|---|---|---|
Rating (this year) | 5454 | 8383 | 6868 | 7575 | 4949 | 8989 | 8080 | 6969 | 7272 | 5656 |
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Step 2 of 4:
Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places.
from above:
Step 1 of 4:
point estimate for the population mean of the paired differences =-4.2
Step 2 of 4:
sample standard deviation =19.696587
step 3:
for 99% CI; and 9 degree of freedom, value of t= | 3.250 | ||
therefore confidence interval=sample mean -/+ t*std error | |||
margin of errror =t*std error= | 20.242975 |
step 4:
lower confidence limit = | -24.44 | |
upper confidence limit = | 16.04 |
(try -24.4 , 16.0 if required to one decimals)