# Solved-Charla is the new statistician at a cola company. She wants to estimate the proportion of the population who enjoy their latest idea

#### ByDr. Raju Chaudhari

Oct 4, 2020

SolCharla is the new statistician at a cola company. She wants to estimate the proportion of the population who enjoy their latest idea for a flavour enough to make it a successful product. Charla wants to obtain a 99 percent confidence level estimate of the population proportion and she wants the estimate to be within
0.02 of the true proportion.

a. Using only the information given above, how large a sample is required? Each of your answers should be rounded up to a whole number.
b. A quick preliminary estimate of the proportion obtained by her predecessor was 0.30. Knowing this, how large a sample is required?

#### Solution

(a) If no prior estimate of the proportion is available, then we assume it as 0.50.

Given that the proportion is $p =0.5$, margin of error $E =0.02$. The confidence coefficient is $0.99$, so $\alpha = 0.01$.

The critical value of $Z$ is $Z_{\alpha/2} =Z_{0.005}= 2.5758$. (From Normal Table)

The minimum sample size required to estimate the proportion is

 \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.5(1-0.5)\bigg(\frac{2.5758}{0.02}\bigg)^2\\ &=4146.716\\ &\approx 4147. \end{aligned}

Thus, the sample of size $n=4147$ will ensure that the $99$% confidence interval for the proportion will have a margin of error $0.02$.

(b) A quick preliminary estimate of the proportion obtained by her predecessor was 0.30.

Given that the proportion is $p =0.3$, margin of error $E =0.02$. The confidence coefficient is $0.99$, so $\alpha = 0.01$.

The critical value of $Z$ is $Z_{\alpha/2} =Z_{0.005}= 2.5758$. (From Normal Table)

The minimum sample size required to estimate the proportion is

 \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.3(1-0.3)\bigg(\frac{2.5758}{0.02}\bigg)^2\\ &=3483.2415\\ &\approx 3483. \end{aligned}

Thus, the sample of size $n=3484$ will ensure that the $99$% confidence interval for the proportion will have a margin of error $0.02$.