Question

A researcher wishes to see if the average weights of newborn male infants are higher than the

Answer 1

Answer:

The  p-value is  $p-value = 0.62578$

Step-by-step explanation:

From the question we are told that    

    The  sample size of male infant is  $n_1 = 12$

     The  sample size of female infant is $n_2= 9$

     The sample mean of male infant is  $\= x_1 = 7.70 \ lb$

      The sample mean of female infant is  $\= x_2 = 7.80 \ lb$

     The population standard deviation is  $\sigma = 0.5$

       The significance level is  $\alpha = 0.05$

The null hypothesis is  $H_o : \mu_ 1 = \mu_2$

The  alternative hypothesis is  $H_1 : \mu_1 > \mu_2$

The  test statistics is mathematically represented as

            $t =\frac{\= x_1 - \= x_2 }{\sqrt{\frac{\sigma }{n_1} } + \frac{\sigma }{n_2} } }$

=>          $t = \frac{7.70 -7.80}{\sqrt{\frac{0.5 }{12} } + \frac{0.5 }{9} } }$

=>        $t = -0.3207$

From the z-table  the p-value is  obtained, the value is  

     $p-value = P(Z > -0.3207) = 0.62578$

     $p-value = 0.62578$