Question 1193375
**1. Set up Hypotheses**

* **Null Hypothesis (H0):** The mean weight of the cement bags is equal to the expected population mean. 
    * μ = 45 kg

* **Alternative Hypothesis (H1):** The mean weight of the cement bags is significantly greater than the expected population mean.
    * μ > 45 kg

**2. Determine Test Statistic**

* Since the population variance is unknown and the sample size is large (n = 36), we will use a **t-test**.

* **Calculate the t-statistic:**

   * t = (sample mean - population mean) / (sample standard deviation / √sample size)
   * t = (44 - 45) / (√1.25 / √36) 
   * t = -1 / (1.118 / 6) 
   * t = -5.37

**3. Determine Degrees of Freedom**

* Degrees of freedom (df) = sample size - 1 = 36 - 1 = 35

**4. Find the Critical Value**

* Using a t-distribution table or a statistical software (like Python's scipy library), find the critical value for a one-tailed t-test with 35 degrees of freedom and a significance level of 0.05. 
    * The critical value for a one-tailed t-test at 0.05 significance level with 35 degrees of freedom is approximately 1.690.

**5. Compare Test Statistic to Critical Value**

* Calculated t-statistic (-5.37) < Critical value (1.690)

**6. Make a Decision**

* Since the calculated t-statistic is less than the critical value, we **fail to reject the null hypothesis**.

**7. Conclusion**

* There is **not enough evidence** to support the sales manager's claim that the mean weight of the cement bags is significantly greater than the expected population mean at a 0.05 significance level.

**In Summary:**

The statistical analysis does not support the sales manager's claim. The sample mean of 44 kg is lower than the expected population mean of 45 kg, and the t-test does not provide sufficient evidence to conclude that the difference is statistically significant.