Answer:
The 95% confidence interval is [tex]0.3795 < p < 0.4405[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 1000[/tex]
The number of approved loan is k = 410
Generally the sample proportion is mathematically represented as
[tex]\r p = \frac{k}{n}[/tex]
substituting values
[tex]\r p = \frac{410}{1000}[/tex]
[tex]\r p = 0.41[/tex]
Given that the confidence level is 95% then the level of significance is mathematically represented as
[tex]\alpha = 100 - 95[/tex]
[tex]\alpha = 5\%[/tex]
[tex]\alpha = 0.05[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table,the value is
[tex]Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }= 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r p(1- \r p)}{n} }[/tex]
substituting values
[tex]E = 1.96 * \sqrt{\frac{ 0.41(1- 0.41)}{1000} }[/tex]
[tex]E = 0.03048[/tex]
The 95% confidence interval for p is mathematically represented as
[tex]\r p - E < p < \r p + E[/tex]
substituting values
[tex]0.41 - 0.03048 < p < 0.41 + 0.03048[/tex]
[tex]0.3795 < p < 0.4405[/tex]
