Answer:
<u>Part a</u>:
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.
<u>Part b: </u>CI = (−3.637,−2.163)
Step-by-step explanation:
AS the sample sizes are sufficiently large (n1,n2> 30) we can use the sample standard deviations in place of population standard deviations.
<u>Part a</u>:
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.
<u>Part b: </u>
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)
