Question 1193539
Certainly, let's analyze both scenarios.

**1. Cement Bag Weights**

**a) Set up Hypotheses**

* **Null Hypothesis (H0):** The mean weight of the cement bags is less than or equal to the expected mean weight. 
    * μ ≤ 45 kg 
* **Alternative Hypothesis (H1):** The mean weight of the cement bags is significantly greater than the expected mean weight.
    * μ > 45 kg

**b) Choose the Test Statistic**

* Since the population variance is unknown and the sample size is greater than 30, we can use the **one-sample t-test**.

**c) Calculate the Test Statistic**

* **Given:**
    * Sample size (n): 36
    * Sample mean (x̄): 44 kg
    * Sample variance (s²): 1.25 kg²
    * Population mean (μ₀): 45 kg
* **Calculate the standard error:**
    * Standard Error (SE) = s / √n 
        * SE = √(1.25) / √36 
        * SE ≈ 0.2041 
* **Calculate the t-score:**
    * t = (x̄ - μ₀) / SE 
        * t = (44 - 45) / 0.2041 
        * t ≈ -4.898

**d) Determine Critical Value**

* **Significance Level:** α = 0.05
* **Degrees of Freedom (df):** n - 1 = 36 - 1 = 35
* **One-tailed test (right-tailed):** Find the critical t-value from a t-distribution table. 
    * t_critical ≈ 1.690

**e) Decision Rule**

* If the calculated t-score is greater than the critical t-value, reject the null hypothesis.
* If the calculated t-score is less than or equal to the critical t-value, fail to reject the null hypothesis.

**f) Make a Decision**

* Our calculated t-score (-4.898) is less than the critical t-value (1.690).

* **Conclusion:** We fail to reject the null hypothesis.

**Interpretation**

The evidence does not support the sales manager's claim that the mean weight of the cement bags is significantly greater than the expected mean weight.

**2. Washing Machine Motor Time Between Failures**

**a) Set up Hypotheses**

* **Null Hypothesis (H0):** The mean time between failures is less than or equal to 90 days.
    * μ ≤ 90 days 
* **Alternative Hypothesis (H1):** The mean time between failures is greater than 90 days.
    * μ > 90 days

**b) Choose the Test Statistic**

* Similar to the first scenario, we can use the **one-sample t-test** since the population variance is unknown.

**c) Calculate the Test Statistic**

* **Given:**
    * Sample size (n): 25
    * Sample mean (x̄): 93 days
    * Sample variance (s²): 16 days²
    * Population mean (μ₀): 90 days
* **Calculate the standard error:**
    * SE = s / √n 
        * SE = √(16) / √25 
        * SE = 4 / 5 
        * SE = 0.8 days
* **Calculate the t-score:**
    * t = (x̄ - μ₀) / SE 
        * t = (93 - 90) / 0.8 
        * t = 3 / 0.8 
        * t = 3.75

**d) Determine Critical Value**

* **Significance Level:** α = 0.05
* **Degrees of Freedom (df):** n - 1 = 25 - 1 = 24
* **One-tailed test (right-tailed):** Find the critical t-value from a t-distribution table. 
    * t_critical ≈ 1.711

**e) Decision Rule**

* If the calculated t-score is greater than the critical t-value, reject the null hypothesis.
* If the calculated t-score is less than or equal to the critical t-value, fail to reject the null hypothesis.

**f) Make a Decision**

* Our calculated t-score (3.75) is greater than the critical t-value (1.711).

* **Conclusion:** We reject the null hypothesis.

**Interpretation**

The evidence supports the quality manager's intuition that the mean time between failures of the motors is significantly greater than 90 days.