Question

In a random sample of 400 items where 82 were found to be​ defective, the null hypothesis that 20​% of the items in the population are defective produced ZSTAT=+0.25. Suppose someone is testing the null hypothesis H0​: π=0.20 against the​ two-tail alternative hypothesis H1​: π≠0.20 and they choose the level of significance α=0.01. What is their statistical​ decision? What is the statistical​ decision? Determine the​ p-value. The​ p-value for the given ZSTAT is ​p-value=nothing. ​(Type an integer or a decimal. Round to three decimal places as​ needed.)

Answer 1

Z - statistic = 0.25

P-value

P(Z=0.25) = 0.802587

The result is not significant at p<0.01

Explanation:

Step(i):-

Given the random sample size 'n'' = 400

A random sample of 400 items where 82 were found to be​ defective

Sample proportion

`p = (x)/(n) = (82)/(400) = 0.205`

Given Population proportion

P = 0.20

Null Hypothesis : P = 0.20

Alternative Hypothesis : P≠ 0.20

Step(ii):-

Test statistic

`Z = \frac{p-P}{\sqrt{(PQ)/(n) } }`

`Z = \frac{0.205-0.20}{\sqrt{(0.20X0.80)/(400) } }= (0.005)/(0.02) =0.25`

Z - statistic = 0.25

Level of significance =0.01

P-value

P(Z=0.25) = 0.802587

Conclusion:-

The result is not significant at p<0.01