Question

A hospital wishes to justify the benefits of nutrition programs for pregnant women using birth weight data from newborns. The hospital hopes to show that the mean birth weight for newborns from mothers who complete the program is higher than the birth weight for newborns from mothers who do not complete the program. A group of 10 pregnant women were randomly divided into two groups; the first group received the nutrition program and the second group did not receive the program. The resulting weights (in grams) of the newborn babies from each group are shown below. Assume normality.

Answer 1

Hypothesis tests are used to verify or refute a claim about a given dataset. From the given that, we have the following results.

• The p value is: $p = 0.3157$

• The test statistic is $t = 4.6503$

• The p value associated to the test statistic is: $p = 0.0048$

Given

$\bar x_1 =2520.2$              $\bar x_2=2446$

$\sigma_1=143.468$            $\sigma_2=111.101$

$n_1 = n_2 = 5$

(1) Calculate the p-value.

Start by calculating the F-test statistic;

$F =(\frac{\sigma_1}{\sigma_2})^2$

$F =(\frac{143.468}{111.01})^2$

$F =1.670$

Calculate the degrees of freedom using:

$df = n - 1$

So, we have:

$df_1 =n_1 - 1 = 5 - 1 =4$

$df_2 =n_2 - 1 = 5 - 1 =4$

The p-value of two tailed F-test at

$F =1.670$

$df_1 = df_2 = 4$

is

$p = 0.3157$

Hence, the p value is 0.3157

(2) Calculate the test statistic

First, we calculate the pooled standard deviation using

$s_{pooled} = \sqrt{\frac{(n_1 - 1)^2\times \sigma_1 + (n_2 - 1)^2\times \sigma_2}{n_1 + n_2 - 2}}$

So, we have:

$s_{pooled} = \sqrt{\frac{(6 - 1)^2\times 143.468 + (6 - 1)^2\times 111.10}{6+6 - 2}}$

$s_{pooled} = \sqrt{\frac{6364.2}{10}}$

$s_{pooled} = \sqrt{636.42}$

$s_{pooled} = 25.227$

The test statistic (t) is then calculated using:

$t = \frac{(\bar x_1 - \bar x_2)}{s_{pooled} \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

$t = \frac{(2520.2 - 2446)}{25.227 \times \sqrt{\frac{1}{5} + \frac{1}{5}}}$

$t = \frac{(2520.2 - 2446)}{25.227 \times \sqrt{\frac{2}{5}}}$

$t = \frac{74.2}{25.227 \times 0.6325}}$

$t = \frac{74.2}{15.956}}$

$t = 4.6503$

Hence, the test statistic is 4.6503

(3) Calculate the p-value associated to the above test statistic.

First, calculate the degrees of freedom (df)

$df = n_1 + n_2 - 2 = 5 + 5 -2 = 8$

The p-value for the right-tailed t-test at  

$t = 4.6503$

$df= 8$

is

$p = 0.0048$

Hence, the p value associated to the test statistic is 0.0048

Summarily:

The p value is: $p = 0.3157$

The test statistic is $t = 4.6503$

The p value associated to the test statistic is: $p = 0.0048$

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