A random ample of400 teleconnectors manufactured by a certai

A random ample of400 teleconnectors manufactured by a certain process are tested, and 30 are found be defective. Let p represent the proportion of components manufactured by this process that are defective Find a 95 confidence interval for p How many components must be sampled so that the 95% confidence interval will specify the proportion defective to within plusminus 0.02.

Solution

Given that n = 400      x = 30     P =0.02   Q =   0.98   p = x/n = 30/400 = 0.0750 q = 1- 0.0750 = 0.9250

The null hypothesis is

H₀ : the proportion of defective is within ± 0.02 i.e., P = ± 0.02

Against the alternative hypothesis

H₁ : the proportion of defective is not within ± 0.02 i.e., P ± 0.02

The test statistic is

Z = (p- P_0­ / P₀Q₀/n ) N(0,1)

Z = (0.0750 – 0.02/ (0.02)(0.98)/400 ) N(0,1)

Z = (0.0550/ 0.000049 ) N(0,1)

Z = (0.0550 / 0.007) N(0,1)

Z = 7.8571

Here Z_tab = 1.96

Therefore Z_cal   > Z_tab we reject the null hypothesis

Hence the proportion of defective is not within ±0.02

The 95% confident interval for P are given by

p±1.96pq/n

0.0750 ±1.96(0.0750)(0.9250)/400

0.0750 ±1.960.000173

0.0750±1.96* 0.013170

[0.0492 , 0.1008]