To determine if the instructor can prove that the mean score is higher than 66, we conduct a one-sample t-test with the following setup:

1. State the Hypotheses

• Null Hypothesis ($H_0$): The mean score is at most 66 ($\mu \leq 66$).
• Alternative Hypothesis ($H_a$): The mean score is higher than 66 ($\mu > 66$).

2. Set the Significance Level

Let the significance level be $\alpha = 0.05$.

3. Calculate the Test Statistic

The test statistic for a one-sample t-test is given by:

$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$

Where:

• $\bar{x}$: Sample mean
• $\mu_0$: Hypothesized mean ($66$)
• $s$: Sample standard deviation
• $n$: Sample size

Step 1: Compute Sample Statistics

The given data is:

$61.9, 85.5, 62.7, 69.2, 66.5, 73.5, 61.9, 88, 75.6, 85.5, 73.5, 73.5, 64.3, 68.4, 64.3, 66.5, 66.5$

We will compute:

• Sample mean ($\bar{x}$)
• Sample standard deviation ($s$)
• Sample size ($n = 17$)

Let me compute these values. numpy as np

Data provided

data = [61.9, 85.5, 62.7, 69.2, 66.5, 73.5, 61.9, 88, 75.6, 85.5, 73.5, 73.5, 64.3, 68.4, 64.3, 66.5, 66.5]

Compute statistics

sample_mean = np.mean(data) sample_std = np.std(data, ddof=1) # Use ddof=1 for sample standard deviation sample_size = len(data)

sample_mean, sample_std, sample_sizeFrom the data:

• Sample Mean ($\bar{x}$): $71.02$
• Sample Standard Deviation ($s$): $8.46$
• Sample Size ($n$): $17$

Step 2: Compute the Test Statistic

Using the formula: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ Substituting:

• $\bar{x} = 71.02$
• $\mu_0 = 66$
• $s = 8.46$
• $n = 17$

Let me calculate this. Hypothesized mean mu_0 = 66

Compute t-statistic

t_statistic = (sample_mean - mu_0) / (sample_std / np.sqrt(sample_size)) t_statisticThe calculated $t$-statistic is $t = 2.45$.

4. Determine the Critical Value

For a one-tailed t-test at $\alpha = 0.05$ with $df = n - 1 = 16$, we find the critical t-value ($t_{critical}$) from the t-distribution table or software.

Let me calculate the critical value. scipy.stats import t

Degrees of freedom

df = sample_size - 1

Critical t-value for one-tailed test at alpha = 0.05

alpha = 0.05 t_critical = t.ppf(1 - alpha, df) t_criticalThe critical $t$-value for a one-tailed test at $\alpha = 0.05$ with $df = 16$ is $t_{critical} = 1.75$.

5. Decision Rule

• If $t > t_{critical}$, reject the null hypothesis ($H_0$).
• Otherwise, do not reject $H_0$.

Since $t = 2.45 > 1.75$, we reject the null hypothesis.

6. Conclusion

There is sufficient evidence to conclude that the mean score on the first statistics test is higher than 66.

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Do you have any questions or want further clarification?

Related Questions:

1. How is the critical $t$-value determined for different confidence levels?
2. What would happen if the significance level was changed to $\alpha = 0.01$?
3. Can you explain how the degrees of freedom affect the test statistic?
4. How would this test differ if the population standard deviation were known?
5. What are the assumptions of the one-sample t-test?

Tip:

When conducting hypothesis tests, always check the assumptions, such as the normality of the data, especially for small sample sizes.