Question

In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface finish that is rougher than the specifications allow.
(a) Find a point estimate for the true proportion of crankshaft bearings that exceeds the roughness specification.
(b) Find a 95% two-sided confidence interval for the proportion of bearing in the population that exceeds the roughness specification.
(c) What is the margin of error?

Answer 1

Solution :

Given that,

n = 85

x = 10

Point estimate = sample proportion = = x / n = 10/85=0.118

1 -   = 1- 0.118 =0.882

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2   * ((( * (1 - )) / n)

= 1.96 (((0.118*0.882) / 85)

E = 0.0686

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.118- 0.0686< p < 0.118+0.0686

0.0494< p < 0.1866

The 95% confidence interval for the population proportion p is : 0.0494, 0.1866