Question

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random samples of 5 babies' weights (pounds) for each gender showed the following:
Boys: 8.0 4.7 7.3 6.2 3.4
Girls: 5.3 2.8 6.4 6.8 7.4
To test the researcher's hypothesis, we should use the:

a) paired (dependent) samples t-test
b) independent samples t-test
c) large-sample z-test
d) t-test for correlation

Answer 1

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$  

$p_v =2*P(t_{8}>0.154) =0.881$  

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).  

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: $\mu_1 = \mu_2$  

Alternative hypothesis: $\mu_1 \neq \mu_2$  

Or equivalently:  

Null hypothesis: $\mu_1 - \mu_2 = 0$  

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$  

Our notation on this case :  

$n_1 =5$ represent the sample size for group 1  

$n_2 =5$ represent the sample size for group 2  

We can calculate the mean and deviation for eaach group with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =5.92$ represent the sample mean for the group 1  

$\bar X_2 =5.74$ represent the sample mean for the group 2  

$s_1=1.88$ represent the sample standard deviation for group 1  

$s_2=1.81$ represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:  

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$  

And now we can calculate the statistic:  

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$  

The degrees of freedom are given by:  

$df=5+5-2=8$

And now we can calculate the p value using the altenative hypothesis:  

$p_v =2*P(t_{8}>0.154) =0.881$  

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).