Question

The birth weights for two groups of babies were compared in a study. In one group the mothers took a zinc supplement during pregnancy. In another group, the mothers took a placebo. A sample of 3535 babies in the zinc group had a mean birth weight of 32893289 grams. A sample of 3131 babies in the placebo group had a mean birth weight of 30243024 grams. Assume that the population standard deviation for the zinc group is 755755 grams, while the population standard deviation for the placebo group is 620620 grams. Determine the 98%98% confidence interval for the true difference between the mean birth weights for "zinc" babies versus "placebo" babies.

Step 2 of 3 :  

Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.

Answer 1

Here, we have given that,

The birth weights for two groups of babies were compared in this study,

1st group= Mothers took a zinc supplement during pregnancy

2nd group= Mothers took Placebo

n1= number of babies in the zinc group=169

= Mean birth weight of babies in zinc group=3019 grams

=Population standard deviation of babies in zinc group=813 grams

n2= number of babies in the placebo group=158

= Mean birth weight of babies in placebo group=3084 grams

=Population standard deviation of babies in placebo group=648 grams

Here, Population standard deviation is not equal.

Now, we want to find the 80% confidence interval for the true difference between the mean birth weight for zinc babies versus the placebo babies.

Formula for CI is as follows

\bar d - t\frac{\alpha}{2} *\frac{Sd}{\sqrt(n)} \leq \mu d \leq \bar d + t\frac{\alpha}{2} *\frac{Sd}{\sqrt(n)}

Degrees of freedom = n-1 = 15-1 = 14

\alpha = level of significance= 1-c=1-0.95=0.05

We get

Tc= 2.010   ( using t table)

Now,

Margin of error=

first we find the Sp=

=

=544516

Now, for finding the margin of error we need to find the Z critical value

c=confidence level =0.80

= level of significance=c-1= 0.80-1=0.20

= =   = -1.282 using Excel =NORMSINV(PROB=0.10)

we get the margin of error as follows,

Margin of error=

=

= -77223.37

Now, we get the 80% CI as follows,

Interpretation:

This confidence interval shows that we are 95% confident that the difference in the population mean will falls under this interval.