Answer:
The 99% confidence interval is [tex]0.3003 < I < 0.3997[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 500[/tex]
The the number that are parents x = 175
The proportion of parents is mathematically represented as
[tex]\r p = \frac{x}{n}[/tex]
substituting values
[tex]\r p = \frac{175}{500}[/tex]
[tex]\r p = 0.35[/tex]
The level of confidence is given as 99% which implies that the level of significance is
[tex]\alpha = 100 - 99[/tex]
[tex]\alpha =[/tex]1%
[tex]\alpha = 0.01[/tex]
The critical value for this level of significance is obtained from the table of critical value as
[tex]t_{x, \alpha } = t_{175, 0.05} = 2.33[/tex]
Generally the margin of error is mathematically evaluated as
[tex]M =\frac{ t_{175, 0.01 } * \sqrt{\r p (1-\r p)} }{\sqrt{n} }[/tex]
substituting values
[tex]M =\frac{ 2.33 * \sqrt{\r 0.35 (1-0.35)} }{\sqrt{500} }[/tex]
[tex]M = 0.0497[/tex]
Generally the 99% confidence interval is mathematically represented as
[tex]I = \r p \pm M[/tex]
[tex]\r p -M < I < \r p + M[/tex]
substituting values
[tex]0.35 -0.0497 < I < 0.35 + 0.0497[/tex]
[tex]0.3003 < I < 0.3997[/tex]
