Question

Identify the P-VALUE used in a hypothesis test of the following claim and sample data:

Claim: "The proportion of defective tablets manufactured in this factory is less than 6%."

A random sample of 500 tablets from this factory is selected, and it is found that 20 of them were defective. Test the claim at the 0.05 significance level.

Answer 1

Answer:

The calculated value Z = 2 > 1.96 at 0.05 level of significance

Alternative Hypothesis is accepted

The proportion of defective tablets manufactured in this factory is less than 6%."

Step-by-step explanation:

Step(i):-

Given Population proportion P = 0.06

Sample size 'n' = 500

A random sample of 500 tablets from this factory is selected, and it is found that 20 of them were defective.

Sample proportion

                                  $p^{-} = \frac{x}{n} = \frac{20}{500} =0.04$

Null hypothesis :H₀: P = 0.06

Alternative Hypothesis :H₁:P<0.06

Level of significance = 0.05

Z₀.₀₅ = 1.96

Step(ii):-

       Test statistic

                          $Z = \frac{p^{-} -P}{\sqrt{\frac{P Q}{n} } }$

                         $Z = \frac{0.04 -0.06}{\sqrt{\frac{0.06 X 0.94}{500} } }$

                        Z =  - 2

                    |Z|= |-2| = 2

Step(iii):-

The calculated value Z = 2 > 1.96 at 0.05 level of significance

Null hypothesis is rejected

Alternative Hypothesis is accepted

The proportion of defective tablets manufactured in this factory is less than 6%."