Question

Question 18 9 pts Based on a sample of size 19, a 98% confidence interval for the mean score of all students, p. on an aptitude test is from 79.2 to 94.8. Find the margin of error Question 19 9 pts A group of 75 randomly selected students have a mean score of 25.2 on a placement test. Nationwide, the standard deviation of all students taking the test is 27. This sample data is used to calculate a 90 percent confidence interval for the mean score of all students taking the test. What is the margin of error?

Answer 1

Answer 18)

Margin of error = (Upper limit – lower limit) / 2

Margin of error = (94.8 - 79.2) / 2

Margin of Error, E = 7.8

19)

Given that,

Sample mean = $\bar x$ = 25.2

Population standard deviation =   $\sigma$ = 2.7
Sample size = n =76

At 90% confidence level the z is ,

$\alpha$  = 1 - 90% = 1 - 0.90 = 0.10

$\alpha$ / 2 = 0.10/ 2 = 0.05

Z$\alpha$/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z$\alpha$/2* ($\sigma$ /$\sqrt$n)

E = 1.645* (2.7 / $\sqrt$ 75)

E = 0.5129

90% confidence interval estimate of the population mean is,

$\bar x$- E < $\mu$ < $\bar x$ + E

25.2- 0.5129< $\mu$ < 25.2+0.5129

24.6871< $\mu$ < 25.7129

(24.6871, 25.7129)

Margin of Error, E = 0.5129