Question 1199646
**1. Set up Hypotheses**

* **Null Hypothesis (H0):** μ = 1.6 (The mean number of text-speech instances in current emails is the same as in 2001)
* **Alternative Hypothesis (H1):** μ > 1.6 (The mean number of text-speech instances in current emails is greater than in 2001)

**2. Calculate the Test Statistic (z-score)**

* Since we know the population standard deviation (σ = 1.25), we can use the z-test:
   * z = (x̄ - μ) / (σ / √n) 
   * where:
      * x̄ = Sample mean (1.98)
      * μ = Population mean (1.6)
      * σ = Population standard deviation (1.25)
      * n = Sample size (121)

   * z = (1.98 - 1.6) / (1.25 / √121)
   * z = 0.38 / (1.25 / 11)
   * z = 0.38 / 0.1136 
   * z ≈ 3.34

**3. Determine Critical Value**

* **Significance Level (α) = 0.05**
* This is a one-tailed test (since H1 is μ > 1.6).
* Find the critical z-value for α = 0.05 in a standard normal distribution table.
* The critical z-value is approximately 1.645.

**4. Make a Decision**

* **Calculated z-score (3.34) is greater than the critical value (1.645).**
* **Therefore, we reject the null hypothesis.**

**5. Conclusion**

* At the 0.05 significance level, there is sufficient evidence to conclude that the mean number of text-speech instances in current emails is significantly higher than the mean number of text-speech instances in emails from 2001.

**In summary:**

* The calculated z-score is 3.34.
* We reject the null hypothesis at the 0.05 significance level. 
* This suggests that the use of text-speech in emails has significantly increased since 2001.