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Complete Question
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Answer:
The lower bound is [tex]0.0234[/tex]
The upper bound is [tex]0.100[/tex]
So from the value obtained the solution to the question are
1 Does not include
2 sufficient
3 not different
Step-by-step explanation:
From the question we are told that
The sample size of individuals who earned in excess of $100,000 per year is [tex]n_ 1 = 1205[/tex]
The number of individuals who earned in excess of $100,000 per year that said yes is
[tex]w = 712[/tex]
The sample size individuals who earned less than $100,000 per year is [tex]n_2 = 1310[/tex]
The number of individuals who earned less than $100,000 per year that said yes is
[tex]v= 693[/tex]
The sample proportion of individuals who earned in excess of $100,000 per year that said yes is
[tex]\r p _ 1 = \frac{w}{n_1 }[/tex]
substituting values
[tex]\r p _ 1 = \frac{712}{1205}[/tex]
[tex]\r p _ 1 =0.5909[/tex]
The sample proportion of individuals who earned less than $100,000 per year that said yes is
[tex]\r p _ 1 = \frac{v}{n_2 }[/tex]
substituting values
[tex]\r p _ 1 = \frac{693 }{1310}[/tex]
[tex]\r p _ 1 = 0.529[/tex]
Given that the confidence level is 95% then the level of significance is mathematically represented as
[tex]\alpha = 1 -0.95[/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]
Generally the margin of error is
[tex]E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{ \r p _1 (1- \r p_1 )}{n_1} + \frac{ \r p _2 (1- \r p_2 )}{n_2} } }[/tex]
substituting values
[tex]E = 1.96 * \sqrt{ \frac{ 0.5909 (1- 0.5909 )}{1205} + \frac{ 0.592 (1- 0.6592 )}{1310} } }[/tex]
[tex]E =0.03846[/tex]
Generally the 95% confidence interval is
[tex](\r p_1 - \r p_2) - E < p_1 - p_2 <( \r p_1 - \r p_2 ) + E[/tex]
substituting values
[tex](0.5909 - 0.529 ) - 0.03846 < p_1 - p_2 < (0.5909 - 0.529 ) + 0.03846[/tex]
[tex]0.02344 < p_1 - p_2 < 0.10036[/tex]
The lower bound is [tex]0.0234[/tex]
The upper bound is [tex]0.100[/tex]
So from the value obtained the solution to the question are
1 Does not include
2 sufficient
3 not different
The lower bound is 0.0234 and the upper bound is 0.100. Then the 95% confidence interval is (0.0234, 0.100)
What is the margin of error?
The probability or the chances of error while choosing or calculating a sample in a survey is called the margin of error.
The research group asked the following question of individuals who earned in excess of $100,000 per year and those who earned less than $100,000 per year.
The sample size of individuals who earned in excess of $100,000 per year will be
[tex]\rm n_1 =1205[/tex]
The sample size of individuals who earned less than $100,000 per year will be
[tex]\rm n_1 =1205[/tex]
The number of individuals who earn an excess of $100,000 per year that said yes will be
[tex]\rm w = 712[/tex]
The number of individuals who earn less than $100,000 per year that said yes will be
[tex]\rm v= 693[/tex]
Then the sample proportion of individuals who earned in excess of $100,000 per year that said yes will be
[tex]\rm \hat{p}_1=\dfrac{w}{n_1}\\\\\hat{p}_1=\dfrac{712}{1205}\\\\\hat{p}_1= 0.5909[/tex]
Then the sample proportion of individuals who earned less than $100,000 per year that said yes will be
[tex]\rm \hat{p}_2=\dfrac{v}{n_2}\\\\\hat{p}_2=\dfrac{693}{1310}\\\\\hat{p}_2= 0.529[/tex]
The confidence level is 95% then the level of significance is mathematically represented as
[tex]\alpha =1-0.95\\\\\alpha =0.05[/tex]
Then the critical value of α/2 from the normal distribution table. Then the value of z is 1.96, then the error of margin will be
[tex]E = z_{\alpha /2} \times \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\\\\E = 1.96 \times \sqrt{\dfrac{05909(1-0.5909)}{1205} + \dfrac{0.529(1-0529)}{1310}}\\\\E = 0.03846[/tex]
The 95% confidence interval will be
[tex]\begin{aligned} (\hat{p}_1-\hat{p}_2)-E & < p_1-p_2 < (\hat{p}_1-\hat{p}_2) + E\\\\(0.5909 - 0.529) - 0.03846 & < p_1-p_2 < (0.5909 - 0.529) + 0.03846\\\\0.02344 & < p_1-p_2 < 0.10036 \end{aligned}[/tex]
More about the margin of error link is given below.
https://brainly.com/question/6979326
