Answer:
The 95% confidence interval is [tex]0.503 < p < 0.535[/tex]
The interpretation is that there is 95% confidence that the true population proportion lie within the confidence interval
Step-by-step explanation:
From the question we are told that
The sample size is n = 1003
The number that indicated television are a luxury is k = 521
Generally the sample mean is mathematically represented as
[tex]\r p = \frac{k}{n}[/tex]
[tex]\r p = \frac{521}{1003}[/tex]
[tex]\r p = 0.519[/tex]
Given the confidence level is 95% then the level of significance is mathematically evaluated 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} } = 1.96[/tex]
The margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p )}{n} }[/tex]
=> [tex]E = 1.96 * \sqrt{ \frac{ 0.519 (1- 0.519 )}{1003} }[/tex]
=> [tex]E = 0.016[/tex]
The 95% confidence interval is mathematically represented as
[tex]\r p -E < p < \r p +E[/tex]
=> [tex]0.519 - 0.016 < p < 0.519 + 0.016[/tex]
=> [tex]0.503 < p < 0.535[/tex]
