Question

a random sample of 25 people the mean commute time to work is 34.5 min and the standard deviation was 7.3 min. use t distribution to construct an 80% confidence interval for the population mean. what is the margin of error

Answer 1

Step 1: Subtract 1 from your sample size. 25 – 1 = 24. This gives you degrees of freedom,

Step 2: Subtract the confidence level from 1, then divide by two.

(1 – .80) / 2 = .10

Step 3: Look up your answers to step 1 and 2 in the t-distribution table. For 24 degrees of freedom (df) and ? = 0.10, my result is 1.318

Step 4: Divide your sample standard deviation by the square root of your sample size.

7.3 / ?(25) = 1.46

Step 5: Multiply step 3 by step 4.

1.318 × 1.46 = 1.92428

Step 6: For the lower end of the range, subtract step 5 from the sample mean.

34.5 – 1.92428 = 32.57572

Step 7: For the upper end of the range, add step 5 to the sample mean.

34.5 + 1.92428 = 36.42428

80 confidence interval lower bound is 32.57572 and upper bound is 36.42428

MArgin of error ?

find the margin of error for the sample size n=25, standard deviation s=7.3, and confidence level 80.0% using t-distribution.

First, find the critical value: t=1.31783593367315

Next, find the standard error of the mean: SE = S/sqrt(n) =73/50

Finally, the margin of error is ME= t?SE = 1.31783593367315 * (73/50) ? 1.9240404631628

Answer: ME = 1.9240404631628