Question

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:​

Answer 1

The p-value is
`p-value = 0.62578`

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{(\sigma )/(n_1) } + (\sigma )/(n_2) } }`

=>
`t = \frac{7.70 -7.80}{\sqrt{(0.5 )/(12) } + (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`