Answer to Question #113720 in Statistics and Probability for Katenda

Question

Africa Check is interested in the activity of fake news tweets. From a sample of 50 tweets, there
are 100 impressions on average. Assume fake news tweet activity is normally distributed and the
population standard deviation is 25 impressions. Use this information to answer questions 1
and 2.

Question 1

What is the 90% confidence interval estimate for the population mean?

(1) (94.1840; 105.8160)
(2) (93.0704; 106.9296)
(3) (90.8783; 109.1217)
(4) (94.0709; 105.9291)
(5) None of the above.

Question 2

Suppose the sample size increases to 100 (while the mean and standard deviation remain the
same). What is the 90% confidence interval estimate for the population mean?

(1) (94.1840; 105.8160)
(2) (93.0704; 106.9296)
(3) (90.8783; 109.1217)
(4) (95.8875; 104.1125)
(5) (95.8075; 104.1925)

Answer 1

Question 1.

$\overline{x}=100\text{ --- point estimate of population mean}.\\ \mu\text{ --- population mean}.\\ \sigma=25\text{ --- population standard deviation}.\\ N=50\text{ --- size of sample}.\\ \alpha=0.1.\\ \mu=\overline{x}\pm z_{\alpha/2}[\frac{\sigma}{\sqrt{N}}]\\ \Phi(z_{0.05})=0.5-0.05=0.45\\ z_{0.05}\approx 1.645 \text{ (we used Laplace table}).\\ (94.1840;105.8160)\text{ --- our confidence interval}.\\ \text{The answer is (1)}.$

Question 2.

$N=100\\ \overline{x}=100\\ \sigma=25\\ \alpha=0.1\\ (95.8875;104.1125)\text{ --- our confidence interval}.\\ \text{The answer is (4)}.$