3. Measurements on the percentage 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

a. Test the Hypothesis H0: µ = 2.95 versus H1: µ 0 ≠ 2.95, and draw appropriate conclusions. Use the P-value approach.
b. Find a 99% two-sided CI on the mean percentage of enrichment. Are you confortable with the statement that the mean percentage of enrichment is 2.95%? Why?
c. What would you use to check the normality assumption of the data?

Here is a hint for part a):

Try a one-sample t-test. You will need to calculate mean and standard deviation to plug into the formula. Calculate, then determine the p-value from a t-table. The p-value is the actual level of the test statistic you calculate.

I'll let you take it from here.

a. To test the hypothesis H0: µ = 2.95 versus H1: µ ≠ 2.95, we can use the P-value approach.

Step 1: State the null and alternative hypotheses:
- Null Hypothesis (H0): The mean percentage of enrichment is equal to 2.95% (µ = 2.95)
- Alternative Hypothesis (H1): The mean percentage of enrichment is not equal to 2.95% (µ ≠ 2.95)

Step 2: Calculate the test statistic:
- First, calculate the sample mean (x̄) and sample standard deviation (s) of the data.
- Next, calculate the standard error (SE) using the formula SE = s / √n, where n is the sample size.
- Then, calculate the t-statistic using the formula t = (x̄ - µ) / SE, where x̄ is the sample mean and µ is the hypothesized population mean (2.95).

Step 3: Determine the p-value:
- Find the corresponding p-value using the t-distribution table or software.
- For a two-sided test, find the p-value for both tails.
- The p-value is the probability of obtaining a test statistic as extreme as the observed test statistic, assuming the null hypothesis is true.

Step 4: Compare the p-value with the significance level (α):
- If the p-value is less than the significance level (α), reject the null hypothesis.
- If the p-value is greater than or equal to the significance level, fail to reject the null hypothesis.

b. To find the 99% two-sided confidence interval (CI) on the mean percentage of enrichment, we need to calculate the margin of error and construct the interval.

Step 1: Calculate the sample mean (x̄) and sample standard deviation (s).
Step 2: Determine the critical value based on the desired confidence level (in this case, 99% two-sided). The critical value corresponds to the appropriate t-score or z-score depending on the sample size and knowledge of the population standard deviation.
Step 3: Calculate the margin of error using the formula ME = critical value * (s / √n), where ME is the margin of error.
Step 4: Construct the confidence interval by subtracting the margin of error from the sample mean for the lower limit and adding the margin of error to the sample mean for the upper limit.

c. To check the normality assumption of the data, we can use graphical methods, such as a histogram or a normal probability plot, or statistical tests, such as the Shapiro-Wilk test or the Anderson-Darling test. These methods can help assess if the data follows a normal distribution or not.

To test the hypothesis using the P-value approach, we follow these steps:

Step 1: State the null and alternative hypotheses:
H0: µ = 2.95 (the mean percentage of enrichment is equal to 2.95%)
H1: µ ≠ 2.95 (the mean percentage of enrichment is not equal to 2.95%)

Step 2: Select the significance level, α. Let's assume α = 0.05.

Step 3: Calculate the test statistic:
For a one-sample t-test, we can calculate the t-test statistic using the formula:

t = (x̄ - µ) / (s / √n)

where x̄ is the sample mean, µ is the hypothesized population mean (2.95 in this case), s is the sample standard deviation, and n is the sample size.

In this case, the sample mean x̄ is the average of the measurements: (3.11 + 2.88 + 3.08 + 3.01 + 2.84 + 2.86 + 3.04 + 3.09 + 3.08 + 2.89 + 3.12 + 2.98) / 12 = 3.00.

The sample standard deviation s is calculated using the formula:

s = √[(Σ(xi - x̄)²) / (n - 1)]

where xi represents each measurement.

Using the given measurements, we can calculate the sample standard deviation s as:

s = √[(Σ(xi - x̄)²) / (n - 1)] = √[(0.0104 + 0.0728 + 0.0004 + 0.0001 + 0.0791 + 0.0601 + 0.0016 + 0.0081 + 0.0004 + 0.0716 + 0.0109 + 0.0001) / (12 - 1)]
= √[0.4052 / 11] = √0.0369 = 0.192

Now we can calculate the test statistic:
t = (x̄ - µ) / (s / √n) = (3.00 - 2.95) / (0.192 / √12) = 0.05 / 0.0555 ≈ 0.9009

Step 4: Calculate the p-value:
The p-value is the probability of obtaining a test statistic more extreme than the observed value, assuming the null hypothesis is true. Since the alternative hypothesis is two-sided, we want to find the p-value for both tails.

To find the p-value, we can use the t-distribution table or a statistical software. Using the t-distribution table for a two-sided test, at α = 0.05 level of significance and degrees of freedom (n - 1) = 11, the critical value is approximately 2.201.

The p-value is defined as P(|t| > 0.9009) = P(t > 0.9009) + P(t < -0.9009). From the t-distribution table, we find that P(t > 0.9009) ≈ 0.198 and P(t < -0.9009) ≈ 0.198. Thus, the p-value is approximately 0.198 + 0.198 = 0.396.

Step 5: Make a decision and draw appropriate conclusions:
If the p-value is less than the significance level (α), we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

In this case, the p-value (0.396) is greater than the significance level (0.05). Therefore, we fail to reject the null hypothesis. We do not have sufficient evidence to conclude that the mean percentage of enrichment is different from 2.95%.

b. To find a 99% two-sided confidence interval (CI) on the mean percentage of enrichment, we can use the following formula:

CI = x̄ ± t * (s / √n)

where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and t is the critical value from the t-distribution table for a given confidence level and degrees of freedom (n - 1).

For a 99% confidence level and degrees of freedom (12 - 1 = 11), the critical value is approximately 3.106.

Using the same values as before, we can calculate the confidence interval as:

CI = 3.00 ± 3.106 * (0.192 / √12) ≈ 3.00 ± 0.644

The 99% confidence interval is approximately (2.356, 3.644).

Based on this confidence interval, we can see that the mean percentage of enrichment of 2.95% is within the confidence interval. Therefore, we can say that we are comfortable with the statement that the mean percentage of enrichment is 2.95%.

c. To check the normality assumption of the data, we can use graphical methods such as a histogram, a normal probability plot, or a box plot. These methods allow us to visually examine the distribution of the data and assess its normality.

For example, we can create a histogram of the measurements or use a normal probability plot to see if the data points follow a roughly linear pattern. If the histogram appears bell-shaped or the points on the normal probability plot approximately follow a straight line, it suggests that the data is normally distributed.

Alternatively, we can perform a formal statistical test for normality, such as the Shapiro-Wilk test. This test calculates a test statistic and p-value to determine whether the data significantly deviates from normality.