A researcher wishes to see if the average weights of newborn male infants are higher than the
average weights of newborn female infants. She selects a random sample of 12 male infants and
finds the mean weight is 7.70 pounds. She selects a random sample of 9 female infants and finds
that the mean Leight is 7.80 pounds. Assume that the variables are normally distributed and the
population standard deviation is 0.5 for each group.
Using alpha=0.05 to test if the mean weight of the males is higher than the mean weight of the
females, the pvalue of the test is:​

Respuesta :

Answer:

The  p-value is  [tex]p-value = 0.62578[/tex]

Step-by-step explanation:

From the question we are told that    

    The  sample size of male infant is  [tex]n_1 = 12[/tex]

     The  sample size of female infant is [tex]n_2= 9[/tex]

     The sample mean of male infant is  [tex]\= x_1 = 7.70 \ lb[/tex]

      The sample mean of female infant is  [tex]\= x_2 = 7.80 \ lb[/tex]

     The population standard deviation is  [tex]\sigma = 0.5[/tex]

       The significance level is  [tex]\alpha = 0.05[/tex]

The null hypothesis is  [tex]H_o : \mu_ 1 = \mu_2[/tex]

The  alternative hypothesis is  [tex]H_1 : \mu_1 > \mu_2[/tex]

The  test statistics is mathematically represented as

            [tex]t =\frac{\= x_1 - \= x_2 }{\sqrt{\frac{\sigma }{n_1} } + \frac{\sigma }{n_2} } }[/tex]

=>          [tex]t = \frac{7.70 -7.80}{\sqrt{\frac{0.5 }{12} } + \frac{0.5 }{9} } }[/tex]

=>        [tex]t = -0.3207[/tex]

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

     [tex]p-value = P(Z > -0.3207) = 0.62578[/tex]

     [tex]p-value = 0.62578[/tex]