Question 1179607: A business consultant wanted to investigate if providing day care facilities on premises
by companies reduces the absentee rate of working mothers from companies that provide
day care facilities on premises. Sample of 50 mothers selected from the companies that
provide day care facilities was taken. These mothers missed an average of 6.4 days from
work last year with a standard deviation of 1.20 days. Another sample of 50 such mothers
taken from companies that do not provide day care facilities on premises showed that these
mothers missed an average of 9.3 days last year with a standard deviation of 1.83 days.
Construct a 98 % confidence interval for the difference between the two population means.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
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