Answer:
The margin of error reduces to half of it original size
Step-by-step explanation:
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [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]
Let assume the standard deviation is [tex]\sigma = 0.4[/tex]
When sample size is n = 25
[tex]E = 1.96 * \frac{0.4 }{\sqrt{25} }[/tex]
[tex]E =0.1568[/tex]
When sample size is n = 100
[tex]E_1 = 1.96 * \frac{0.4 }{\sqrt{100} }[/tex]
[tex]E_1 = 0.0784 [/tex]
So
[tex]\frac{E_1}{E} = \frac{0.0784}{0.1568} = \frac{1}{2}[/tex]
Hence the margin of error reduces to half of it original size
