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Someone was interested in determining if there is a difference in the proportion of US adults that working hard was the most important thing for a child to learn in 2012 and 2018. They found data on this in the General Social Survey. In 2018, out of 1542 respondents, 415 said it was the most important. In 2012, out of 1323 respondents, 326 said it was the most important. Find the margin of error with 95% confidence.

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Answer:

0.01521

Step-by-step explanation:

The Margin of Error for Difference in Proportions is given as:

z × √p1 × (1 - p1)/n1 - p2 × (1 - p2)/n2

z = z score for 95% confidence interval

= 1.96

In 2018, out of 1542 respondents, 415 said it was the most important.

p1 = x1/n1

x1= 326 n1 = 1323

p1 = 326/1323

0.246409675

Approximately = 0.246

In 2012, out of 1323 respondents, 326 said it was the most important.

p2 = x2/n2

x2 = 415, n2 = 1542

p2= 415/1942

p2 = 0.2136972194

Approximately = 0.214

Margin of Error =

1.96 × √0.246× (1 - 0.246)/1323- 0.214 × (1 - 0.214)/1542

1.96 × √0.0005699169 - 0.0005097276

1.96 × √(0.0000601893)

1.96 × 0.0077581763

= 0.0152060256

Approximately = 0.01521

Therefore, Margin of Error = 0.01521

You can use the formula for margin of error for difference in proportion to find the margin of error with 95% confidence.

The margin of error with 95% confidence for the given data is given by

MOE = 0.32

How to calculate margin of error for difference in proportion to find the margin of error ?

Suppose we've two samples with sample sizes [tex]n_1[/tex] and [tex]n_2[/tex] respectively. Let the interested quantities found out in those samples out of their sample sizes are  [tex]x_1[/tex] and [tex]x_2[/tex] respectively. Then we have the margin of error for difference in proportion to find the margin of error as:

[tex]MOE = Z_{\alpha/2} \times \sqrt{\dfrac{p_1(1-p_1)}{n_1} + \dfrac{p_2(1-p_2)}{n_2}}[/tex]

where we have:

[tex]p_1 = \dfrac{x_1}{n_1}\\\\p_2 = \dfrac{x_2}{n_2}[/tex]

and

[tex]\alpha = \text{Level of significance}[/tex]

Using the given data to get the margin of error

[tex]Year = 2015\\n_1 = 1542\\x_1 = 415\\\\Year = 2012\\n_2 = 1323\\x_2 = 326\\\\\alpha = 100-95\% = 1-0.95 = 0.05[/tex]

For 95% confidence, we have Z score as 1.96

The proportions are calculated as:

[tex]p_1 = \dfrac{415}{1542} = 0.269\\\\p_2 = \dfrac{326}{1323} = 0.246[/tex]

Thus, margin of error is calculated as

[tex]MOE = 1.96 \times \sqrt{\dfrac{0.269 \times 0.731}{1542} + \dfrac{0.246 \times 0.754}{1323}} = 1.96 \times \sqrt{0.000127 + 0.00014}\\\\MOE = 1.96 \times 0.01634 = 0.032[/tex]

The margin of error with 95% confidence for the given data is given by

MOE = 0.32

Learn more about margin of error here:

https://brainly.com/question/4757147

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