SOLUTION: In a study of the effect of predictability on stress, subjects were either given a public speaking assignment with an hours advanced notice or they were instructed to give the spee

Question 1168569: In a study of the effect of predictability on stress, subjects were either given a public speaking assignment with an hours advanced notice or they were instructed to give the speech immediately. Stress experienced right before beginning the speech was measured. They collected data from a small sample of 14 people with 7 in each group, and found that the group that was given notice of the speech had a mean stress level of 4.00 while the group who did not have advanced notice had a mean stress level of 7.71. The standard error of the mean difference was .71. Test the hypothesis that giving warning of an upcoming stressful event reduces the stress experienced by individuals.
Question:
1. Statistical Hypotheses H0: µ1 - µ2 = H1: µ1 - µ2 ≠
2. Decision Rule
3. Reject the null hypothesis at the .05 level of significance if t equals or is greater than ___ or if t equals or is less than__ given df = __
4. Calculate the appropriate t test
5. Decision: ____ the null hypothesis at the .05 level of significance because the obtained t of ____ is __ than the critical t of ___
6. How would you interpret your decision?

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this hypothesis test step-by-step.
**1. Statistical Hypotheses**
* μ1 = mean stress level of the group given notice.
* μ2 = mean stress level of the group without notice.
* **H₀: μ1 - μ2 = 0** (There is no difference in mean stress levels.)
* **H₁: μ1 - μ2 < 0** (The group given notice has a lower mean stress level.)
* This is a left-tailed test, as we're testing if the notice *reduces* stress.
**2. Decision Rule**
* Significance level (α) = 0.05
* Degrees of freedom (df) = n1 + n2 - 2 = 7 + 7 - 2 = 12
**3. Reject the Null Hypothesis**
* We need to find the critical t-value for a one-tailed test with α = 0.05 and df = 12.
* Using a t-table or calculator:
* t(0.05, 12) = -1.782 (This is the critical value for a left-tailed test)
* Reject H₀ if t ≤ -1.782.
**4. Calculate the Appropriate t-test**
* Mean difference (x̄1 - x̄2) = 4.00 - 7.71 = -3.71
* Standard error of the mean difference (SE) = 0.71
* t = (x̄1 - x̄2) / SE
* t = -3.71 / 0.71
* t ≈ -5.225
**5. Decision**
* Decision: **Reject** the null hypothesis at the 0.05 level of significance because the obtained t of **-5.225** is **less** than the critical t of **-1.782**.
**6. Interpretation**
* The results of the t-test indicate that there is a statistically significant difference in stress levels between the two groups.
* Specifically, the group that was given notice of the public speaking assignment experienced significantly lower stress levels compared to the group that was not given notice.
* This suggests that providing warning of an upcoming stressful event can effectively reduce the amount of stress experienced.

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