Answer to Question #105328 in Statistics and Probability for Katenda

Question

QUESTIONS 5, 6 AND 15 ARE BASED ON THE FOLLOWING INFORMATION.

A Social Scientist took a random sample of 30 adults with Autism Spectrum Disorder (ASD) and found their reading time to be normally distributed with sample mean and sample standard deviation of 90 wpm and 18 wpm respectively.

Question 5

What is the 99% confidence interval estimate of the population mean?
(1) (80.9626; 99.0374)
(2) (80.9429; 99.0571)
(3) (81.9090; 98.0910)
(4) (81.5213; 98.4787)
(5) (82.3428; 97.6572)


Question 6

What is the upper limit for the 95% confidence interval estimate of the population mean?
(1) 96.7206
(2) 96.4412
(3) 83.5588
(4) 83.2794
(5) None of the above.

Answer 1

n=30, mean=90,sd=18

Question 5

What is the 99% confidence interval estimate of the population mean?

Confidence limits=mean ±(critical value) (SE)

=$\bar x ±Z_\alpha*\frac {SD} {\sqrt n}$

The critical value for 99% is 2.58 from the z-table.

=90±2.58*$\frac {18}{\sqrt {30}}$

=90±8.4787

=(81. 5213,98.4787)

The answer is (4)

Question 6

What is the upper limit for the 95% confidence interval estimate of the population mean?

Upper limit =mean+(critical value) (SE)

=$\bar x+Z_\alpha *\frac {SD} {\sqrt n}$

The critical value of 95% from the z-table is 1.96.

$=90+1.96*\frac {18}{\sqrt {30 }}$

=90+6.4412

=96.4412

The answer is (2)