Question

please answer all parts in detail thank u :)

A random sample of 120 parts was taken from a big lot and was found that 111 of them met specification (a) [5 points] Construct a 99% confidence interval for population proportion p of parts that meet specification. (b) [5 points] Suppose that quality engineer wants to estimate the true proportion of parts that meet specification to within 0.04 with 95% confidence. How many parts should this quality engineer take? Assume the engineer has no prior knowledge about the proportion. (Hint: Use p = 0.5)

Answer 1

Solution :

a ) Given that

n = 120

x = 111

$\hat p$ = x / n = 111 / 120 = 0.925

1 -$\hat p$ = 1 - 0.925 = 0.075

At 99% confidence level the z is ,

$\alpha$  = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

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

= 2.576 * ($\sqrt$((0.925 * 0.075) / 120)

= 0.062

A 99% confidence interval for population proportion p is ,

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

0.925 - 0.062 < p < 0.925 + 0.062

0.863 < p < 0.987

b ) Given that,

$\hat p$$\hatp$ = 0.5

1 - $\hat p$ = 1 - 0.5 = 0.5

margin of error = E = 4% = 0.04

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

Sample size = n = ((Z_{$\alpha$ / 2}) / E)² * $\hat p$ * (1 - $\hat p$ )

= (1.960 / 0.04)² * 0.5 * 0.5

= 600.23

= 600

n = sample size = 600