I have a question on a quiz and I don't even know where to begin on figuring out the question... please help:

Measurements on the percentave of enrichment of 12 fuel rods used in a nuclear reactor were reported as follows:

3.11
2.88
3.08
3.01
2.84
2.86
3.04
3.09
3.08
2.89
3.12
2.98

Is there an indication that the rods do not meet a target of 2.95%? Assume alpha=0.05.

Also is there sufficient evidence to suggest that the variance is greater than 0.085? Assume alpha=.01.

First, calculate the mean (M), variance (V), and standard deviation (SD) of the observed sample.

Next, test whether the observed mean is significantly different from 2.95. Since this is a small sample (n=12), you will need to use a Students t distribution table (Your stats book probably has such a table). Your question implies a 2-tailed test with alpha=.05, so find the t-value which accounts for 97.5% of sample with (n-1) degrees of freedom. In my table, i get 2.201. Is your observed mean M within 2.201*SD of the target value of 2.95?

2) I get a variance (V) of 0.011, which of course, is less than .085. So no, there is not sufficient evidence to suggest the actual variance is greater than .085. That said, such a test would use a Chi-squared distribution table. I believe the chi-squared statistic is ((n-1)*V/.085)

To answer the first question about the target percentage of 2.95%, you need to calculate the mean, variance, and standard deviation of the observed sample. Here's how you can do that:

1. Calculate the mean (M):
- Add up all the values: 3.11 + 2.88 + 3.08 + 3.01 + 2.84 + 2.86 + 3.04 + 3.09 + 3.08 + 2.89 + 3.12 + 2.98 = 35.98.
- Divide the sum by the number of values (n=12): M = 35.98 / 12 = 2.999.

2. Calculate the variance (V):
- For each value, subtract the mean, square the result, and then sum all the squared differences.
- (3.11 - 2.999)^2 + (2.88 - 2.999)^2 + ... + (2.98 - 2.999)^2 = 0.0345.
- Divide the sum by (n-1): V = 0.0345 / (12-1) = 0.00305.

3. Calculate the standard deviation (SD):
- Take the square root of the variance: SD = sqrt(0.00305) ≈ 0.055.

Now that you have the mean (M) and standard deviation (SD), you can test if the observed mean is significantly different from 2.95% using a t-test. Since this is a small sample size (n=12) and a 2-tailed test with alpha=0.05, you need to find the t-value that accounts for 97.5% of the sample with (n-1) degrees of freedom. According to your table, the t-value is 2.201.

Next, calculate the range within which the observed mean M falls: M ± 2.201 * SD.
In this case, it is 2.999 ± 2.201 * 0.055 = 2.999 ± 0.121.
Therefore, the range is (2.878, 3.12).

Since the target value of 2.95% falls within this range, the observed mean does not differ significantly from the target value of 2.95%.

For the second question regarding the variance, you calculated a variance (V) of 0.00305, which is less than 0.085. This suggests that there is not sufficient evidence to suggest the actual variance is greater than 0.085.

A test for variance would typically use a Chi-squared distribution table, and the chi-squared statistic can be calculated as ((n-1) * V) / 0.085.

Remember that these calculations assume that the data follows a normal distribution and the sample is representative of the population.