Answer:
The test statistic is ![z = -2.1](https://tex.z-dn.net/?f=z%20%3D%20-2.1)
The p-value of the test is 0.0358 > 0.01, which means that the defective rates of the two suppliers are not significant different at the 1% significance level.
Step-by-step explanation:
Before testing the hypothesis, we need to understand the central limit theorem and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
and standard deviation ![s = \sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
A simple random sample of 400 parts from supplier 1 finds 20 defective.
This means that:
![p_1 = \frac{20}{400} = 0.05, s_1 = \sqrt{\frac{0.05*0.95}{400}} = 0.0109](https://tex.z-dn.net/?f=p_1%20%3D%20%5Cfrac%7B20%7D%7B400%7D%20%3D%200.05%2C%20s_1%20%3D%20%5Csqrt%7B%5Cfrac%7B0.05%2A0.95%7D%7B400%7D%7D%20%3D%200.0109)
A simple random sample of 200 parts from supplier 2 finds 20 defective.
This means that:
![p_2 = \frac{20}{200} = 0.1, s_2 = \sqrt{\frac{0.1*0.9}{200}} = 0.0212](https://tex.z-dn.net/?f=p_2%20%3D%20%5Cfrac%7B20%7D%7B200%7D%20%3D%200.1%2C%20s_2%20%3D%20%5Csqrt%7B%5Cfrac%7B0.1%2A0.9%7D%7B200%7D%7D%20%3D%200.0212)
Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. Test whether the defective rates of the parts from two suppliers are significant different at the 1% significance level.
At the null hypothesis, we test if they are equal, that is, if the subtraction of the proportions is 0. So
![H_0: p_1 - p_2 = 0](https://tex.z-dn.net/?f=H_0%3A%20p_1%20-%20p_2%20%3D%200)
At the alternate of the null hypothesis, we test if they are different, that is, if the subtraction of the proportions is different of 0. So
![H_a: p_1 - p_2 \neq 0](https://tex.z-dn.net/?f=H_a%3A%20p_1%20-%20p_2%20%5Cneq%200)
The test statistic is:
![z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
In which X is the sample mean,
is the value tested at the null hypothesis and s is the standard error.
0 is tested at the null hypothesis:
This means that ![\mu = 0](https://tex.z-dn.net/?f=%5Cmu%20%3D%200)
From the sample proportions:
![X = p_1 - p_2 = 0.05 - 0.1 = -0.05](https://tex.z-dn.net/?f=X%20%3D%20p_1%20-%20p_2%20%3D%200.05%20-%200.1%20%3D%20-0.05)
![s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0109^2 + 0.0212^2} = 0.0238](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7Bs_1%5E2%2Bs_2%5E2%7D%20%3D%20%5Csqrt%7B0.0109%5E2%20%2B%200.0212%5E2%7D%20%3D%200.0238)
Value of the test statistic:
![z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![z = \frac{-0.05 - 0}{0.0238}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B-0.05%20-%200%7D%7B0.0238%7D)
![z = -2.1](https://tex.z-dn.net/?f=z%20%3D%20-2.1)
P-value of the test and decision:
The p-value of the test is the probability that the sample proportion differs from 0 by at least 0.05, that is, P(|z| < 2.1), which is 2 multiplied by the p-value of Z = -2.1.
Z = -2.1 has a p-value of 0.0179
2*0.0179 = 0.0358.
The p-value of the test is 0.0358 > 0.01, which means that the defective rates of the two suppliers are not significant different at the 1% significance level.