Question

A business consultant wanted to investigate if providing day-care facilities on premises by companies reduces the absentee rate of working mothers with children under six years old. She took a sample of 45 such mothers from companies that provide day-care facilities on premises. 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 1.85 days.

a. Using a 2.5% significance level, can you conclude that the mean number of days missed per year by mothers working for companies that provide day-care facilities on premises is less than the mean number of days missed per year by mothers working for companies that do not provide day-care facilities on premises?
b. Construct a 98% confidence interval for the difference between the two population means.
i. State the hypothesis
ii. State the rejection or nor rejection region
iii. Calculate the test statistics
iv. State the conclusion

Answer 1

Part a:

The null and alternate hypotheses are

i) H0: ud ≥ u vs Ha: ud < u

ii) As the significance level is ∝= 0.025 and it is one tailed test then the critical region R < - 1.96

iii) The test statistics is =−9.15

iv) We accept the alternate hypothesis.

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Part b: CI = (−3.637,−2.163)

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Explanation:

AS the sample sizes are sufficiently large (n1,n2> 30) we can use the sample standard deviations in place of population standard deviations.

Part a:

The null and alternate hypotheses are

i) H0: ud ≥ u vs Ha: ud < u

where

ud is the mean of the days missed per year by mothers working for companies that provide day-care facilities on premises

and

u is the mean number of days missed per year by mothers working for companies that do not provide day-care facilities on premises

ii) As the significance level is ∝= 0.025 and it is one tailed test then the critical region R < - 1.96

iii) The test statistics is

z= xˉ1 − xˉ2/√ s 1² /n 1 +s 2² /n 2

​= 6.4−9.3/ √1.2 ² /45+ 1.85² /50​

=−9.15

iv) The calculated z = −9.15 values lies in the critical region= R < - 1.96 and the null hypothesis is rejected. Hence the mean of absentees for companies that provide day-care facilities on premises is less than the

the mean number of absentees for companies that do not provide day-care facilities on premises.

We accept the alternate hypothesis.

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Part b:

The confidence interval can be calculated using the formula

CI = ( xˉ1 − xˉ2) ± z∝/2 √ s 1² /n 1 +s 2² /n 2

CI = (6.4−9.3) ± 2.326 * √1.2 ² /45+ 1.85² /50​

CI = (−3.637,−2.163)

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