Statistics - 95% Confidence Level

[KingBigness]

Homework Statement

Two different types of injection–moulding machines are used to form plastic parts. A part is considered defective if it has excessive shrinkage or is discoloured. Two random samples, each of size 300, are selected and 15 defective parts are found in the sample from machine 1 whereas 8 defective parts are found in the sample from machine 2.

a) Test, at the 5% level of significance, if both machines produce the same fraction of defective parts. State your null and alternative hypotheses and the conclusion of the test in plain language.

b)Find a 95% confidence interval for the difference in the fraction of defective parts produced by the two machines. How does this confidence interval confirm the conclusion reached in part (a) above?

Homework Equations

The Attempt at a Solution

a) I found that there is no significant difference in the fraction of defective parts produced by the machine at the 5% level of significance.
I would put the working up here but it will take a while to type out...if you really want to see it I'll type it up.

Part b is where I am confused. Do I just do the same working as part a, but instead of using α=0.05 I use α=0.95?

Thank you.

 

[mighty2000]

Thank you for your post. I can provide some insight and guidance on your question.

a) The null hypothesis (H0) in this case would be that both machines produce the same fraction of defective parts, while the alternative hypothesis (Ha) would be that there is a significant difference in the fraction of defective parts produced by the two machines. Using a significance level of 5%, we can test this by calculating the p-value and comparing it to the significance level. If the p-value is less than 0.05, we can reject the null hypothesis and conclude that there is a significant difference in the fraction of defective parts produced by the two machines. If the p-value is greater than 0.05, we fail to reject the null hypothesis and conclude that there is no significant difference in the fraction of defective parts produced by the two machines. Therefore, it seems that your conclusion is correct.

b) In order to find a 95% confidence interval for the difference in the fraction of defective parts produced by the two machines, we can use the same data and perform a two-sample t-test. This will give us a confidence interval for the difference in means, which can then be converted to a confidence interval for the difference in proportions. The confidence interval will provide a range of values within which the true difference in proportions is likely to lie. If this interval includes 0, it supports the conclusion from part a that there is no significant difference in the fraction of defective parts produced by the two machines. If the interval does not include 0, it suggests that there is a significant difference in the fraction of defective parts produced by the two machines.

I hope this helps. Let me know if you have any further questions or if you would like to see my calculations. Keep up the good work!