To test the hypothesis that "the mean time for students at this college to complete the entrance exam is greater than 82 minutes," we perform a one-sample t-test. Here's the process:

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Hypotheses:

• Null Hypothesis ($H_0$): $\mu \leq 82$ (The mean completion time is less than or equal to 82 minutes.)
• Alternative Hypothesis ($H_a$): $\mu > 82$ (The mean completion time is greater than 82 minutes.)

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Significance Level:

$\alpha = 0.05$

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Given Data:

Sample times: $111, 98, 46, 77, 98, 112, 112, 105$
Sample size ($n$) = 8

We calculate the sample mean ($\bar{x}$) and standard deviation ($s$).

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Test Statistic:

The test statistic is calculated using the t-test formula:

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

Where:

• $\bar{x}$: Sample mean
• $\mu_0$: Hypothesized population mean ($82$)
• $s$: Sample standard deviation
• $n$: Sample size

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I will calculate these values now.import numpy as np from scipy.stats import t

Given data

data = np.array([111, 98, 46, 77, 98, 112, 112, 105]) mu_0 = 82 # Hypothesized population mean alpha = 0.05 # Significance level

Sample statistics

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

Test statistic

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

Critical t-value for a one-tailed test

df = n - 1 # Degrees of freedom t_critical = t.ppf(1 - alpha, df)

p-value

p_value = 1 - t.cdf(t_statistic, df)

sample_mean, sample_std, t_statistic, t_critical, p_value(94.875, 22.91872285147794, 1.588919219761797, 1.894578605061305, 0.07805124416847908)(94.875, 22.91872285147794, 1.588919219761797, 1.894578605061305, 0.07805124416847908)success### Results:

1. Sample Mean ($\bar{x}$): $94.88$
2. Sample Standard Deviation ($s$): $22.92$
3. Test Statistic ($t$): $1.59$
4. Critical t-value ($t_{critical}$): $1.89$
5. p-value: $0.078$

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Conclusion:

Since the test statistic ($1.59$) is less than the critical t-value ($1.89$) and the p-value ($0.078$) is greater than the significance level ($0.05$), we fail to reject the null hypothesis. There is insufficient evidence to conclude that the mean time for students to complete the entrance exam is greater than 82 minutes.

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Would you like further explanation or assistance with interpreting this test?

1. Would changing the significance level to $0.10$ affect the result?
2. What assumptions are made in this test?
3. How is the p-value calculated in a t-test?
4. Can we use a Z-test here instead of a t-test? Why or why not?
5. What is the power of this test for detecting a mean of $90$ minutes?

Tip: Always check the normality of the data distribution when performing t-tests for small samples.