Question 1170482: A consumer advocacy group suspects that a local supermarket’s 500 grams of sugar actually weigh less than 500 grams. The group took a random sample of 20 such packages, weighed each one, and found the mean weight for the sample to be 496 grams with a standard deviation of 8 grams. Using 10% significance level, would you conclude that the mean weight is less than 500 grams? Also, find the 90% confidence interval of the true mean.
Hypothesis:
H_0:
H_1:
The level of significance is a = _____, df = ____, and the critical region is t = ____.
Compute for the value of one sample t-test.
Decision rule:
Conclusion:
Compute for the confidence interval.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's perform the hypothesis test and calculate the confidence interval.
**1. Define Hypotheses**
* **Null Hypothesis (H0):** The mean weight of sugar packages is 500 grams (μ = 500).
* **Alternative Hypothesis (H1):** The mean weight of sugar packages is less than 500 grams (μ < 500).
**2. Set Significance Level (α)**
* α = 0.10
**3. Determine Degrees of Freedom (df)**
* Sample size (n) = 20
* df = n - 1 = 20 - 1 = 19
**4. Find the Critical t-value**
* This is a left-tailed test since H1: μ < 500.
* Using a t-distribution table or calculator with α = 0.10 and df = 19, the critical t-value is approximately -1.729.
**5. Compute the t-statistic**
* Sample mean (x̄) = 496 grams
* Sample standard deviation (s) = 8 grams
* Population mean (μ) = 500 grams
* t = (x̄ - μ) / (s / √n)
* t = (496 - 500) / (8 / √20)
* t = -4 / (8 / 4.472)
* t = -4 / 1.789
* t ≈ -2.236
**6. Decision Rule**
* Reject H0 if the calculated t-statistic is less than the critical t-value (-1.729).
**7. Conclusion**
* Since the calculated t-statistic (-2.236) is less than the critical t-value (-1.729), we reject the null hypothesis.
* Therefore, we conclude that there is sufficient evidence at the 10% significance level to suggest that the mean weight of the sugar packages is less than 500 grams.
**8. Compute the 90% Confidence Interval**
* Confidence level = 90%
* α = 1 - 0.90 = 0.10
* α / 2 = 0.05
* Using a t-distribution table or calculator with α/2 = 0.05 and df = 19, the t-value is approximately 1.729.
* Confidence interval = x̄ ± t * (s / √n)
* Confidence interval = 496 ± 1.729 * (8 / √20)
* Confidence interval = 496 ± 1.729 * 1.789
* Confidence interval = 496 ± 3.093
* Confidence interval = (492.907, 499.093)
**Summary**
* **Hypothesis:**
* H0: μ = 500
* H1: μ < 500
* **Level of significance:** α = 0.10
* **Degrees of freedom:** df = 19
* **Critical region:** t < -1.729
* **t-statistic:** t ≈ -2.236
* **Decision rule:** Reject H0 if t < -1.729.
* **Conclusion:** Reject H0. The mean weight is less than 500 grams.
* **90% Confidence Interval:** (492.907, 499.093)
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