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Exercise and Gender The dataset ExerciseHours contains information on the amount of exercise (hours per week) for a sample of statistics students. The mean amount of exercise was 9.4 hours for the 30 female students in the sample and 12.4 hours for the 20 male students, A randomization distribution of differences in means based on these data, under a null hypothesis of no difference in mean exercise time between females and males, is centered near zero and reasonably normally distributed. The standard error for the difference in means, as estimated from the randomization distribution, is \(S E=2.38\). Use this information to test, at a \(5 \%\) level, whether the data show that the mean exercise time for female statistics students is less than the mean exercise time of male statistics students.

Short Answer

Expert verified
Without the exact p-value, the final step cannot be concluded and the choice to reject or not reject the null hypothesis cannot be determined. Once the p-value has been calculated, follow the guide in Step 5 to arrive at a conclusion.

Step by step solution

01

Define the Hypotheses

First, define the null and alternative hypotheses. The null hypothesis (H0) in this case is that there is no difference between the mean exercise hours of male and female students. Thus, H0: μF - μM = 0. The alternative hypothesis (H1) is that the mean exercise time for female students is less than that for male students. Hence, H1: μF - μM < 0.
02

Calculate Mean Difference

The mean difference between the two groups is calculated as the difference between the female mean and the male mean, which is 9.4 - 12.4 = -3 hours.
03

Perform Z-Test

Next, a z-test is performed to test the hypothesis. The z-score is calculated as the difference of the sample mean from the hypothesized population mean, divided by the standard error. In this case, the z-score = (-3 - 0) / 2.38 = -1.26.
04

Find P-Value

The p-value is given by the probability that a z-score is more extreme than -1.26 under the null hypothesis. The normal distribution Z-table or a calculator could be used to find this value. Let's assume it results in a p-value = p.
05

Make Conclusion

Finally, compare the p-value to the significance level α=0.05. If the p-value is less than the significance level, then reject the null hypothesis and conclude that the data provides enough evidence to support the claim that the mean exercise time for female students is less than for male students. If the p-value is greater than or equal to the significance level, do not reject the null hypothesis and conclude that the data does not provide enough evidence to support the claim. In other words, the decision would be 'reject H0 if p < 0.05, and do not reject H0 if p ≥ 0.05'.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Randomization Distribution
In hypothesis testing, a randomization distribution is a fundamental concept, especially when comparing groups. It represents what the distribution of the test statistic would look like if the null hypothesis were true. In our context, the null hypothesis assumes there is no difference in the means between male and female students.
The randomization distribution is obtained by repeatedly reallocating the observed data into groups, simulating what mean differences could arise purely by chance. It is a powerful method because it does not rely on assumptions about the population distribution, making it non-parametric. For our exercise, the randomization distribution centered around zero indicates there is no systematic difference under the null hypothesis. Moreover, it being normally distributed suggests that we can safely proceed with a Z-test, assuming the central limit theorem holds given our sample size.
Standard Error
The standard error (SE) plays a critical role in hypothesis testing, acting as a measure of variability or dispersion within a sampling distribution. Specifically, it estimates how much the sample mean of a dataset is expected to fluctuate from the true population mean if we were to sample it multiple times.
In the exercise, the SE for the difference in means is 2.38. This means that the average amount by which the observed difference in means might vary from the expected zero difference is 2.38 hours. It essentially gives us a measure of how reliable our sample's mean difference is, serving as the denominator in the Z-test formula:
  • Smaller SE: Indicates more precision in the estimate of the mean.
  • Larger SE: Suggests less certainty in the sample mean difference, potentially affecting the hypothesis test outcome.
In a broader context, a low standard error indicates that most samples yield results close to the true population parameters, while a high standard error implies more variability.
Z-Test
The Z-test is a statistical test used to determine if there is a significant difference between sample and population means or between the means of two samples. It is particularly useful when the population standard deviation is known or when the sample size is fairly large, thus adhering to the central limit theorem.
In our case, we perform a one-sample Z-test for the difference between the means of male and female students' exercise hours. The formula for the Z-score in our scenario is: \[ Z = \frac{(\bar{x}_{1} - \bar{x}_{2}) - (\mu_{1} - \mu_{2})}{SE} \] Where:
  • \( \bar{x}_{1} - \bar{x}_{2} \) is the observed mean difference.
  • \( \mu_{1} - \mu_{2} \) is the hypothesized mean difference under the null hypothesis, which is 0.
  • \( SE \) is the standard error of the mean difference.
By calculating the Z-score, we find it is -1.26. This score allows us to determine the p-value, guiding our decision on whether to reject the null hypothesis. Thus, the Z-test is a critical tool in analyzing whether an observed effect is statistically significant.

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