Question 1192469: A recent article in a leading magazine claimed that the homemakers spend an average of 22 hours per week preparing meals. A director of marketing at a food-processing company surveyed that a recent sample of 28 homemakers resulted in an average time preparation of meals of 17.5 hours per week, with a standard deviation of 4.3 hours. Can the director conclude that the average preparation of meals is lesser? Use 0.05 significance level.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! **1. Set up the Hypotheses**
* **Null Hypothesis (H0):** The average time homemakers spend preparing meals is 22 hours per week (μ = 22).
* **Alternative Hypothesis (H1):** The average time homemakers spend preparing meals is less than 22 hours per week (μ < 22).
**2. Calculate the t-statistic**
* **Formula:**
t = (sample mean - population mean) / (sample standard deviation / √sample size)
t = (17.5 - 22) / (4.3 / √28)
t = -4.5 / 0.811
t ≈ -5.55
**3. Determine the Degrees of Freedom**
* Degrees of Freedom (df) = sample size - 1 = 28 - 1 = 27
**4. Find the Critical Value**
* Since this is a one-tailed test (we're only interested in whether the mean is *less* than 22 hours), we'll use a one-tailed t-distribution table.
* For a significance level of 0.05 and 27 degrees of freedom, the critical t-value is approximately -1.703.
**5. Compare t-statistic to Critical Value**
* The calculated t-statistic (-5.55) is less than the critical t-value (-1.703).
**6. Make a Decision**
* Since the t-statistic falls in the critical region, we **reject the null hypothesis**.
**Conclusion:**
At the 0.05 significance level, there is sufficient evidence to conclude that the average time homemakers spend preparing meals is less than 22 hours per week.
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