Question

After a month-long survey, you collect 146 responses. You found that 83% of the customers indicated the life of the motor is greater than 10 years. You would like to understand what proportion of the motors last longer than 10 years. How many surveys would you need to collect in total in order to obtain a margin of error of /- 5% with 89.9% confidence

Answer 1

Answer:

The number of surveys needed to be collected is 152.

Step-by-step explanation:

The (1 - α)% confidence interval for population proportion p is:

$CI=\hat p \pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error for this interval is:

$MOE= z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The information provided is:

$\hat p=0.83\\MOE=0.05\\(1-\alpha )\%=89.9\%$

Compute the critical value of z for 89.9% confidence level as follows:

$z_{\alpha/2}=z_{0.101/2}=z_{0.0505}=1.64$

*Use a z-table.

Compute the sample size value as follows:

$MOE= z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

       $n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)}}{MOE}]^{2}$

          $=[\frac{1.64\times \sqrt{0.83(1-0.83)}}{0.05}]^{2}$

          $=151.801024\\\approx 152$

Thus, the number of surveys needed to be collected is 152.