Question

There are 480 welders employed at the Weller Shipyards Corporation. A sample of 45 welders revealed that 27 had graduated from a registered welding course. Construct the 99% confidence interval for the proportion of all welders who graduated from a registered welding course. (Round the final answers to 3 decimal places.)

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 27 / 45 = 0.600

1 - $\hat p$ = 1 - 0.600 = 0.4

Z_$\alpha$/2 = 2.576

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 2.576 * ($\sqrt$((0.600 * 0.4) / 45)

Margin of error = E = 0.188

A 99% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.600 - 0.188 < p < 0.600 + 0.188

0.412 < p < 0.788

The 99% confidence interval for the population proportion p is : 0.412 , 0.788