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: