Answer to Question #267147 in Statistics and Probability for Audz

Question

The Minister of Information proposed that Itheku newspaper should increase its news coverage. This was proposed to enhance free flow of information to the rural areas. Also, the newspaper would be carrying information about the Covid-19 pandemic to enlighten the public on the severity of the virus. An Itheku daily newspaper selected a simple random sample of 1500 readers from their list of 20000 subscribers. They asked whether the paper should increase its coverage of local news. Thirty-five percent of the sample did not support the idea of more local news coverage. What is the 98.5% confidence interval for the proportion of readers who would like more coverage of local news?

Answer 1

The confidence interval for proportion can be estimated using the next formula:

$p{\scriptscriptstyle *}=p±Cr* \sqrt{{\frac {p(1-p)} n}}$ , where p - proportion obtained from data, Cr - critical value, n - sample size

Since we have large sample size, we can use Z-score(two-tailed) as critical value, then:

$P(Z>Cr)={\frac {1+0.985} 2}=0.9925\implies Cr = 2.43$

$p={\frac {1500} {20000}}=0.075$

So, we have

$p{\scriptscriptstyle *}=p±Cr* \sqrt{{\frac {p(1-p)} n}}=0.075±2.43*\sqrt{{\frac {0.075(1-0.075)} {20000}}}=0.075±2.43*0.00186=0.075±0.00452$

The sought confidence interval is $p{\scriptscriptstyle *}\in (0.07048;0.07952)$