We can calculate power before collect data to ensure that we are able to detect a differences or change of defective. Example: Customer feedback 2.62% of defective rate (11 reject units out of total quantity of 420) due to dented cup for product A2019. They request supplier to perform sampling inspection check on 60K inventory. What is the 95% lower bound of proportion and power of 0.9 to detect the change of defective rate from lower bound to 0.026 or more?

**Test and CI for One Proportion **

Test of p = 0.026 vs p > 0.026

95% Lower

Sample X N Sample p Bound Z-Value P-Value

1 11 420 0.026190 0.013373 0.02 0.490

Using the normal approximation.

You are hoping that sample evidence will demonstrate the feasibility of your theory (stated as alternative hypothesis) by demonstrating the unlikeliness of the truth of the null hypothesis.

The alternative hypothesis is referred to as research hypothesis since it represents what the analyst hopes will be found to be true from the research conducted.

The p-value of 0.490 suggests that the data are consistent with the null hypothesis (H0: p = 0.026), that is, the proportion of defective is not greater than the 0.026.The 95% lower bound is 0.013373.

Next is to calculate sample size required to detect the change of defective from 0.013 to 0.026 or more with power of 0.9.

1 Choose **Stat > Power and Sample Size > 1 Proportion**.

2 In **Comparison proportions**, enter *0.013*.

3 In **Power values**, enter 0.9

4 In Hypothesized proportion, enter 0.026

5 Click **Options**.

6 Under **Alternative Hypothesis**, choose **Less Than**.

7 Click **OK**in each dialog box.

**Power and Sample Size **

Test for One Proportion

Testing p = 0.026 (versus < 0.026)

Alpha = 0.05

Sample Target

Comparison p Size Power Actual Power

0.013 980 0.9 0.900068

In order to detect the change of defective% from 1.3% to 2.6% or more with power of 0.9, we require sample size of 980.

(b)Suppose that *comparison p=0.05*

Determine the sample size needed to detect this difference with a probability of at least 0.95. Use α= 0.05.

**Power and Sample Size **

Test for One Proportion

Testing p = 0.026 (versus > 0.026)

Alpha = 0.05

Sample Target

Comparison p Size Power Actual Power

0.05 668 0.95 0.950026

The sample size needed to detect this difference with a probability of at least 0.95 is 668.

(c)Determine the power value if sample size of 300,400,500,600,700 and 800 were being used. Use α= 0.05.

**Power and Sample Size **

Test for One Proportion

Testing p = 0.026 (versus < 0.026)

Alpha = 0.05

Sample

Comparison p Size Power

0.013 300 0.373348

0.013 400 0.493823

0.013 500 0.600809

0.013 600 0.691595

0.013 700 0.765963

0.013 800 0.825175

The result shows that power is reduced when sample size decreased from 800 to 300.

Conclusion:

As sample size increases, α decreases (assuming you hold power and the minimum difference constant).

As sample size increases, power increases (assuming you hold α and the minimum difference constant).

As sample size increases, the minimum detectable difference decreases (assuming you fix α and power constant)