Question

The birthweights for a sample of babies are summarized below. The sample means and standard deviations are given for babies whose mothers smoked during pregnancy and for those whose mothers did not smoke.

Mother did not smoke	x? = 7.69 lb	s1 = 1.148 lb	n1 = 20
Mother smoked	x? = 6.88 lb	s2 = 1.389 lb	n2 = 22

The researchers are looking for evidence that the mean birthweights are different.

a. The null hypothesis is H₀: ?₁ = ?₂. The alternative hypothesis is H_a: ?₁ __________ ?₂ . (Fill in the blank to show the appropriate relation between ?₁ and ?₂.)

b. The degrees of freedom for the test are ________.

c. The P-value for the test is _________.

d. The decision is to (reject or not reject) _________the null hypothesis, at the 0.05 level of significance.

Answer 1

Here the researchers are looking for evidence that the mean birthweights are different.

Therefore alternative hypothesis include not equal to inequality

a. The null hypothesis is H0: ?1 = ?2. The alternative hypothesis is Ha: ?1 ?2 . (Fill in the blank to show the appropriate relation between ?1 and ?2.)

b) first we need to test the equality of standard deviations

Let's use minitab:

Step 1) Click on Stat >>>Basic Statistics >>>2-Variances ...

Fill the necessary information and then click on Option again fill the necessary information.

Look the following image:

Then click on OK again Click on OK , so we get the following output:

p-value = 0.408

Since p-value > 0.05 we fail to reject the equality assumption of two populations.

So that the pooled procedure more appropriate in this situation

Now lets test the equality of two means using pooled t test:

Level of significance = = 0.05

therfore level of confidence = 100 - 5 = 95%

Let's used minitab:

Steps 1) Click on Stat>>>Basic Statistics>>>2-Sample t...

Steps 1) Click on summarized data and then fill the required information in the boxes : look the following picture.

step 3) Click on Option, Look the following image :

then click on OK again click on OK

So we get the following output

b) Degrees of freedom = 20 + 22 - 2 = 40

c ) From the above output p-value = 0.047

Test statistic = T value = 2.05

Decision rule: 1) If p-value <= level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.047 < 0.05 so we used first rule.

That is we reject null hypothesis

d ) The decision is to reject the null hypothesis, at the 0.05 level of significance.