SOLUTION: A random sample of 30 cheese bars manufactured by Tasty Diary showed a mean weight of 189g with a standard deviation of 18g. If the average weight of the cheese bar is 200g, is the

Question 1171065: A random sample of 30 cheese bars manufactured by Tasty Diary showed a mean weight of 189g with a standard deviation of 18g. If the average weight of the cheese bar is 200g, is the manufacturer meeting their standard at 0.05 (two-tailed) level?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to conduct the hypothesis test to determine if the manufacturer is meeting their standard.
**1. State the Hypotheses**
* **Null Hypothesis (H₀):** The average weight of the cheese bars is 200g.
* H₀: μ = 200
* **Alternative Hypothesis (H₁):** The average weight of the cheese bars is different from 200g.
* H₁: μ ≠ 200 (two-tailed test)
**2. Determine the Test Statistic**
Since we have a sample mean, sample standard deviation, and a sample size, and the population standard deviation is unknown, we will use a t-test.
The formula for the t-statistic is:
t = (x̄ - μ) / (s / √n)
Where:
* x̄ = sample mean
* μ = population mean
* s = sample standard deviation
* n = sample size
**3. Calculate the Test Statistic**
* x̄ = 189g
* μ = 200g
* s = 18g
* n = 30
t = (189 - 200) / (18 / √30)
t = -11 / (18 / 5.477)
t = -11 / 3.286
t ≈ -3.347
**4. Determine the Critical Value or P-value**
* Significance level (α) = 0.05
* Degrees of freedom (df) = n - 1 = 30 - 1 = 29
* Type of test: Two-tailed
* **Critical Value Approach:**
* Using a t-table or calculator, we find the critical t-values for a two-tailed test with α = 0.05 and df = 29.
* The critical t-values are approximately ±2.045.
* **P-value Approach:**
* Using a t-table or calculator, we find the p-value associated with t = -3.347 and df = 29.
* p-value ≈ 0.0022
**5. Make a Decision**
* **Critical Value Approach:**
* The calculated t-statistic is -3.347.
* The critical t-values are ±2.045.
* Since |-3.347| > 2.045, we reject the null hypothesis.
* **P-value Approach:**
* The p-value is approximately 0.0022.
* The significance level is 0.05.
* Since the p-value (0.0022) is less than the significance level (0.05), we reject the null hypothesis.
**6. Conclusion**
There is sufficient evidence at the 0.05 significance level to conclude that the average weight of the cheese bars is different from 200g. Therefore, the manufacturer is not meeting their standard.

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