Question

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2: Suppose a sample of 5733 floppy disks is drawn. Of these disks, 4988 were not defective. Using the data, construct the 98 % confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places. Answer/How to Enter) 2 Points Keypad Lower endpoint Upper endpoint

Answer 1

Solution :

Given that,

n = 5733

x = 4988

Point estimate = sample proportion = $\hat p$ = x / n = 4988 /5733 =0.870

1 - $\hat p$ = 1 - 0.870 = 0.13

At 98% confidence level

$\alpha$ = 1 - 98%

$\alpha$ =1 - 0.98 =0.02

$\alpha$/2 = 0.01

Z$\alpha$/2 = Z_{0.01 = 2.326}

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

= 2.326 ($\sqrt$((0.870 * 0.13) / 5733 )

= 0.010

A 98% confidence interval for population proportion p is ,

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

0.870 - 0.010 < p < 0.870 +0.010

0.86 < p < 0.88

Lower endpoint = 0.86

Upper endpoint = 0.88