To solve this hypothesis test, we proceed as follows:

Step 1: State the hypotheses

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

This is a one-tailed t-test because we are testing if the mean is greater than 66.

Step 2: Calculate the sample statistics

The sample scores are: $61.9, 85.5, 62.7, 69.2, 66.5, 73.5, 61.9, 88, 75.6, 85.5, 73.5, 73.5, 64.3, 68.4, 64.3, 66.5, 66.5$ Let $n = 17$ (sample size).

1. Sample mean ($\bar{x}$): $\bar{x} = \frac{\text{Sum of all values}}{n}$

2. Sample standard deviation ($s$): $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

Step 3: Compute the test statistic ($t$)

The test statistic is calculated using: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ where:

• $\mu_0 = 66$ (mean under $H_0$),
• $s$ is the sample standard deviation,
• $n = 17$ is the sample size.

Step 4: Determine the critical value

For a one-tailed test with $\alpha = 0.01$ and $n - 1 = 16$ degrees of freedom, use a t-table or calculator to find the critical value.

Now, let's compute the values. numpy as np from scipy.stats import t

Data

scores = np.array([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])

Hypothesized mean

mu_0 = 66

Sample size

n = len(scores)

Sample mean and standard deviation

sample_mean = np.mean(scores) sample_std = np.std(scores, ddof=1) # ddof=1 for sample standard deviation

Test statistic (t)

t_statistic = (sample_mean - mu_0) / (sample_std / np.sqrt(n))

Degrees of freedom

df = n - 1

Critical value for one-tailed test at alpha = 0.01

alpha = 0.01 critical_value = t.ppf(1 - alpha, df)

sample_mean, sample_std, t_statistic, critical_value### Results:

1. Sample mean ($\bar{x}$): $71.018$
2. Sample standard deviation ($s$): $8.457$
3. Test statistic ($t$): $2.446$
4. **Critical value at $\alpha = 0.01$ (one-tailed, $df = 16$): $2.583$

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Conclusion:

Since the test statistic $t = 2.446$ is less than the critical value $2.583$, we fail to reject the null hypothesis. There is insufficient evidence at the 1% significance level to conclude that the mean score is greater than 66.

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Would you like further details or explanations?
Here are five related questions to consider:

1. How would the conclusion change if the significance level was 5% instead of 1%?
2. What assumptions must be satisfied for this test to be valid?
3. How would the result change if the sample size was larger?
4. Can you explain the difference between a one-tailed and a two-tailed test?
5. How is the critical value determined for a given significance level and degrees of freedom?

Tip: Always check the normality assumption when performing t-tests, especially for small sample sizes!