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Answer:

To test the null hypothesis that the mean mark is equal to 100 against the alternative that the mean mark is greater than 100, we can use a one-sample t-test since the population variance is unknown. Here's how you can perform the test:

Step 1: State the null and alternative hypotheses:

- Null hypothesis (H₀): The mean mark is equal to 100.

- Alternative hypothesis (H₁): The mean mark is greater than 100.

Step 2: Set the significance level (α):

In this case, the significance level is given as 0.05 or 5%.

Step 3: Compute the test statistic:

The test statistic for a one-sample t-test is calculated using the formula:

t = (X - μ) / (s / √n)

where X is the sample mean, μ is the population mean under the null hypothesis, s is the sample standard deviation, and n is the sample size.

Given:

X = 110 (sample mean)

s = 8 (sample standard deviation)

n = 13 (sample size)

μ (population mean under the null hypothesis) = 100

Substituting the values into the formula, we get:

t = (110 - 100) / (8 / √13)

t = 10 / (8 / √13)

t ≈ 3.012

Step 4: Determine the critical value:

Since the alternative hypothesis is one-tailed (greater than), we need to find the critical value for a one-tailed test at a 5% significance level with (n - 1) degrees of freedom. In this case, the degrees of freedom are 13 - 1 = 12.

Using a t-distribution table or statistical software, the critical value at α = 0.05 and 12 degrees of freedom is approximately 1.782.

Step 5: Make a decision:

If the test statistic t is greater than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

In this case, the test statistic t is approximately 3.012, which is greater than the critical value of 1.782. Therefore, we reject the null hypothesis.

Step 6: State the conclusion:

Based on the sample data, there is sufficient evidence to support the claim that the mean mark is greater than 100.

Explanation:

User Kevin Meier
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