Question

Can i get a step by step, i am having difficulty with this question, I understand that the sample size is 500 and the sample mean is 190

A cellphone provider has the business objective of wanting to estimate the proportion of subscribers who would upgrade to a new cellphone with improved features if it were made available at a substantially reduced cost. Data are collected from a random sample of 500 subscribers. The results indicate that 190 of the subscribers would upgrade to a new cellphone at a reduced cost. Complete parts (a) and (b) below. a. Construct a 99% confidence interval estimate for the population proportion of subscribers that would upgrade to a new cellphone at a reduced cost. (Round to four decimal places as needed.)

Answer 1

Given that, data are collected from a random sample of 500 subscribers. The results indicate that 190 of the subscribers would upgrade to a new cellphone at a reduce cost.

That is, n = 500 and x = 190

=> sample proportion $(\hat p)$ = 190/500 = 0.38

a) A 99% confidence level has significance level of 0.01 and critical value is, $Z_{\alpha/2} = 2.575$

The margin of error (E) is,

$E = Z_{\alpha/2} *\sqrt {\frac {\hat p *(1-\hat p)}{n}} = 2.575 *\sqrt {\frac {0.38*(1-0.38)}{500}} = 0.0559$

A 99% confidence interval estimate for the population proportion of subscribers that would upgrade to a new cellphone at a reduced cost is,

P-E<A Sp+E

$0.38 - 0.0559 \leq \pi \leq0.38 + 0.0559$

$0.3241 \leq \pi \leq 0.4359$

Answer : 0.3241 ≤ π ≤ 0.4359