Answer:
The value is [tex]E =19.6 [/tex]
Step-by-step explanation:
From the question we are told that
The mean of Jo cows is [tex]\= x_1 = 1350[/tex]
The standard deviation of Jo cow is [tex]s_1 = 50[/tex]
The mean of Val cows is [tex]\= x_2 = 1420 [/tex]
The standard deviation of Val cows is [tex]s_2 = 50[/tex]
The sample size for both Val and Jo is n = 25
Let assume that the level of significance is [tex]\alpha = 0.05[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
Hence margin of error for Jo is
=> [tex]E = 1.96 * \frac{50}{\sqrt{25} }[/tex]
=> [tex]E =19.6 [/tex]
