Stats unknown variance hypothesis testing

[Punchlinegirl]

Homework Statement

A machine produces metal rods used in an automobile suspension system. A random sample of 12 rods is selected and the diameter is measred.The sample mean is 8.28. and the significance level is 0.05. Is there strong evidence to indicate that mean rod diameter is not 8.20 mm using a fixed level test?

Homework Equations

The Attempt at a Solution

I tried finding the variance by using E=x-$$\mu$$ = 8.28-8.20 = .08
and n = (z _$$\alpha$$/2)$$\sigma$$/ E)^2 to get $$\sigma$$=.141. Then I plugged this into Z_0 = x-$$\mu$$/$$\sigma$$/$$\sqrt{n}$$. Then I found Z_0 to be 1.97 which would mean the mean 8.20 would be rejected, but I don't think this is right. Can someone tell me what I'm doing wrong and help me out?

 

[mighty2000]

I would first like to clarify the significance level and the hypothesis being tested. The significance level of 0.05 indicates that there is a 5% chance of rejecting the null hypothesis when it is actually true. The null hypothesis in this case would be that the mean rod diameter is 8.20 mm. The alternative hypothesis would be that the mean rod diameter is not 8.20 mm.

To conduct a hypothesis test, we need to calculate the standard deviation of the sample mean. This can be done using the formula \sigma/\sqrt{n}, where \sigma is the population standard deviation and n is the sample size. In this case, we do not have the population standard deviation, so we will use the sample standard deviation as an estimate. This can be calculated using the formula \sqrt{\frac{\sum(x_i-\bar{x})^2}{n-1}}, where x_i is the individual measurements and \bar{x} is the sample mean.

Plugging in the values, we get \sigma/\sqrt{n} = \frac{0.08}{\sqrt{12}} = 0.0231. This is the standard error of the sample mean.

Next, we need to calculate the test statistic, which is the number of standard errors between the sample mean and the hypothesized mean. This can be calculated using the formula (x-\mu)/\sigma/\sqrt{n}. Plugging in the values, we get (8.28-8.20)/0.0231 = 3.46.

Now, we can use a Z-test to determine the p-value. The p-value is the probability of obtaining a test statistic at least as extreme as the one we calculated, assuming that the null hypothesis is true. The p-value can be calculated using a Z-table or a statistical software. In this case, the p-value is very small (less than 0.0001), indicating strong evidence against the null hypothesis.

Finally, we can compare the p-value to the significance level of 0.05. Since the p-value is smaller than the significance level, we can reject the null hypothesis and conclude that there is strong evidence to indicate that the mean rod diameter is not 8.20 mm. However, it is important to note that this conclusion is based on the assumption that the sample is representative of the population and that the measurements are accurate. Further studies and experiments may be needed to confirm these results