Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Answer:
The test statistic is [tex]z = -2.1[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
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:
[tex]p_1 = \frac{20}{400} = 0.05, s_1 = \sqrt{\frac{0.05*0.95}{400}} = 0.0109[/tex]
A simple random sample of 200 parts from supplier 2 finds 20 defective.
This means that:
[tex]p_2 = \frac{20}{200} = 0.1, s_2 = \sqrt{\frac{0.1*0.9}{200}} = 0.0212[/tex]
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
[tex]H_0: p_1 - p_2 = 0[/tex]
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
[tex]H_a: p_1 - p_2 \neq 0[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{s}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis and s is the standard error.
0 is tested at the null hypothesis:
This means that [tex]\mu = 0[/tex]
From the sample proportions:
[tex]X = p_1 - p_2 = 0.05 - 0.1 = -0.05[/tex]
[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0109^2 + 0.0212^2} = 0.0238[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{s}[/tex]
[tex]z = \frac{-0.05 - 0}{0.0238}[/tex]
[tex]z = -2.1[/tex]
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.
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
