Question

A business consultant wanted to investigate if providing day care facilities on a company’s premises reduces the absentee rate of parents working for the company. The consultant took a sample of 52 parents not working for companies providing on site daycare. These parents missed an average of 8.2 days of work last year. Another sample of 46 parents working for companies providing on site daycare showed an average 6.5 missed days of work last year. Assume that the standard deviation for the two populations are 1.75 days and 1.3 days respectively.

a) Construct a 94% confidence interval for the difference between the population means. Interpret your confidence interval in the context of the question.

b) Using a 2.5% level of significance, can you conclude that the mean number of days missed per year by parents not working for the companies providing on site daycare is greater than the number of days missed per year by parents working for companies that do provide on site daycare? Conduct a hypoth- esis test and write a conclusion in the context of the question.

Answer 1

Let X1 denote the parents not working for the companies providing on site daycare

n1= 52

Mean of X1=x1bar=8.2

SD of population 1=Ơ1=1.75

Let X2 denote the parents working for the companies providing on site daycare

n2=46

Mean of X2=x2bar=6.5

SD of population 2=Ơ2=1.3

(a) For 94% CI, α=1-0.94=0.06

For known SD, The 94% CI for the difference of means is given by:

(x1bar-x2bar) ± Z(α/2)*√(Ơ1^2/n1+Ơ2^2/n2).....(1)

Z(α/2)=Z(0.06/2)=Z(0.03)=1.88

From(1), the 94% CI is

(8.2-6.5) ± 1.88*√{(1.75^2)/52+(1.3^2/46)}

i.e, 1.7 ± 1.88*0.3092

i.e, (1.1187, 2.2813)

So confidence interval is

1.1187≤ (µ1-µ2) ≤ 2.2813

Interpretation of CI: The confidence interval doesnot contain 0, from which we can conclude that we are 94% confident that there is significant difference between the two means.

(b) To test

H0:µ1=µ2 vs H1:µ1>µ2

The test statistic under H0 is

Z= (X1bar-X2bar)/√(Ơ1^2/n1+Ơ2^2/n2)}

=(8.2-6.5)/√{(1.75^2)/52+(1.3^2/46)}

= 1.7/0.3092

=5.4981

For the right one tailed test at 2.5% level of significance, i.e For α= 0.025, Z(0.025) =1.96 [From normal table]

The test statistic Z=5.4981 > critical value = Z(0.025) =1.96

Hence, we reject the null hypothesis at 2.5% level of significance and conclude that the alternative hypothesis is true i.e, the mean number of days missed per year by parents not working for the companies providing on site daycare is greater than the number of days missed per year by parents working for companies that do provide on site daycare.