SOLUTION: Please help me solve this: Test the indicated claim about the means of two populations. Assume that the two samples are independent simple random samples selected from normally di

Question 1198711: Please help me solve this:
Test the indicated claim about the means of two populations. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal.
Women: xbar1=12.5hrs, s1=3.9hrs, n1=14
Men: xbar2=13.8hrs, s2=5.2hrs, n2=17
Use a 0.05 significance level to test the claim that the mean amount of time spent watching television by women is smaller than the mean amount of time spent watching television by men.
Include(a) the null and alternative hypotheses, (b) test statistic, (c) critical value(s) or P-value (or range of P-value) as appropriate, (d) conclusion about the null hypotheses, and (e) conclusion about the claim in your answer.
Note: Value of Test statistic t≈-0.795
Answer by textot(100) (Show Source): You can put this solution on YOUR website!
**a) Hypotheses:**
* **Null Hypothesis (H0):** μ₁ - μ₂ ≥ 0
* This translates to: The mean amount of time spent watching television by women is greater than or equal to the mean amount of time spent watching television by men.
* **Alternative Hypothesis (H1):** μ₁ - μ₂ < 0
* This translates to: The mean amount of time spent watching television by women is smaller than the mean amount of time spent watching television by men.
**b) Test Statistic**
* Given: t ≈ -0.795
**c) Critical Value**
* **Degrees of Freedom:**
* Using the Welch-Satterthwaite equation for unequal variances:
* df ≈ [(s₁²/n₁)²/(n₁-1) + (s₂²/n₂)²/(n₂-1)]²
/ [(s₁²/n₁)²/(n₁-1)²/(n₁-1) + (s₂²/n₂)²/(n₂-1)²/(n₂-1)]
* df ≈ [(3.9²/14)²/(14-1) + (5.2²/17)²/(17-1)]²
/ [(3.9²/14)²/(14-1)²/(14-1) + (5.2²/17)²/(17-1)²/(17-1)]
* df ≈ 27.83
* We'll use df = 27 for the t-distribution table.
* **Critical Value (One-tailed test at α = 0.05):**
* From the t-distribution table with 27 degrees of freedom and α = 0.05, the critical value is approximately -1.703.
**d) Conclusion about the Null Hypothesis**
* **Compare the test statistic to the critical value:**
* Calculated t-statistic (-0.795) > Critical value (-1.703)
* **Decision:** Since the calculated t-statistic is greater than the critical value, we **fail to reject the null hypothesis**.
**e) Conclusion about the Claim**
* **There is not enough evidence at the 0.05 significance level to support the claim that the mean amount of time spent watching television by women is smaller than the mean amount of time spent watching television by men.**
**In summary:**
* The test statistic does not fall in the rejection region.
* We do not have sufficient evidence to conclude that women watch less television than men.

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