Question 1179607
To construct a 98% confidence interval for the difference between the two population means, we'll use a two-sample t-test. Here's how we can approach this problem:

**1. Define the given information:**

* Sample 1 (with daycare):
    * n1 = 50 (sample size)
    * x1 = 6.4 (sample mean)
    * s1 = 1.20 (sample standard deviation)
* Sample 2 (without daycare):
    * n2 = 50 (sample size)
    * x2 = 9.3 (sample mean)
    * s2 = 1.83 (sample standard deviation)
* Confidence level = 98% (which means alpha = 1 - 0.98 = 0.02)

**2. Calculate the degrees of freedom (df):**

* df = n1 + n2 - 2 = 50 + 50 - 2 = 98

**3. Calculate the pooled standard deviation (sp):**

* Since we have two independent samples, we can calculate the pooled standard deviation:
    * sp = √[((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / df]
    * sp = √[((49 * 1.20^2) + (49 * 1.83^2)) / 98]
    * sp ≈ √[(70.56 + 163.6329) / 98]
    * sp ≈ √(234.1929 / 98)
    * sp ≈ √2.389723469
    * sp ≈ 1.545873

**4. Calculate the standard error (SE):**

* SE = sp * √((1/n1) + (1/n2))
* SE = 1.545873 * √((1/50) + (1/50))
* SE = 1.545873 * √(0.02 + 0.02)
* SE = 1.545873 * √0.04
* SE = 1.545873 * 0.2
* SE ≈ 0.3091746

**5. Find the t-value:**

* For a 98% confidence interval and 98 degrees of freedom, we need to find the t-value that corresponds to an alpha of 0.02. This means we are looking for the t-value that leaves 0.01 in each tail.
* Using a t-table or calculator, the t-value for df = 98 and alpha/2 = 0.01 is approximately 2.365.

**6. Calculate the margin of error (ME):**

* ME = t-value * SE
* ME = 2.365 * 0.3091746
* ME ≈ 0.7311187

**7. Calculate the confidence interval:**

* Confidence interval = (x1 - x2) ± ME
* Confidence interval = (6.4 - 9.3) ± 0.7311187
* Confidence interval = -2.9 ± 0.7311187
* Lower bound = -2.9 - 0.7311187 = -3.6311187
* Upper bound = -2.9 + 0.7311187 = -2.1688813

**8. Round the results:**

* The 98% confidence interval is approximately (-3.63, -2.17).

**Conclusion:**

The 98% confidence interval for the difference between the two population means is approximately (-3.63, -2.17). This suggests that mothers working in companies that provide daycare facilities on premises miss significantly fewer days of work compared to those who don't.