Answer:
The value is [tex]n_2 = 260[/tex]
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 width of a confidence interval is dependent on the margin of error which is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{ \sigma}{\sqrt{n} }[/tex]
Here [tex]\sigma[/tex] is the standard deviation
Let assume that [tex]\sigma[/tex] and [tex]Z_{\frac{\alpha }{2} } [/tex] are constant for the width of the confidence interval when the sample size is n =65 and now that it has been divided to two
=> [tex]E = \frac{ K }{\sqrt{n} }[/tex]
Here K is a constant
=> [tex]E * \sqrt{n} = K[/tex]
=> [tex]E_1 * \sqrt{n}_1 = E_2 * \sqrt{n}_2[/tex]
Now let [tex]E_1 \ and \ n_1[/tex] be the margin of error and sample size before the reduction
So [tex]n_1 = 65[/tex]
and let [tex]E_2 \ and \ n_2[/tex] be the margin of error and sample size after the reduction
So [tex]E_2 = \frac{1}{2} E_1[/tex]
=> [tex]E_1 * \sqrt{65} = \frac{E_1}{2} * \sqrt{n}_2[/tex]
=> [tex]\sqrt{65} = \frac{1}{2} * \sqrt{n}_2[/tex]
=> [tex]n_2 = 260[/tex]
