Question
According to the U.S. Census Bureau, 44% of men who worked at home were college graduates. In a sample of 510 women who worked at home, 162 were college graduates. (a) Find a point estimate for the proportion of college graduates among women who work at home. Round the answer to at least three decimal places.
The answer to A is 0.318.
Construct a 99% confidence interval
The answer to A is 0.318.
Construct a 99% confidence interval
Answers
Answer
(c) Based on the confidence interval, is it reasonable to believe that the proportion of college graduates among women who work at home is the same as the proportion of college graduates among men who work at home? Explain.
Answered by
GPT-4o mini
To construct a 99% confidence interval for the proportion of college graduates among women who work at home, we first need to calculate the sample proportion \( \hat{p} \):
\[
\hat{p} = \frac{x}{n}
\]
where:
- \( x \) is the number of college graduates (162),
- \( n \) is the total sample size (510).
\[
\hat{p} = \frac{162}{510} \approx 0.318
\]
Next, we calculate the standard error (SE) for the sample proportion:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.318(1 - 0.318)}{510}}
\]
Calculating \( 1 - \hat{p} \):
\[
1 - \hat{p} = 1 - 0.318 = 0.682
\]
Now calculating the numerator:
\[
\hat{p}(1 - \hat{p}) = 0.318 \times 0.682 \approx 0.2176
\]
Now the standard error:
\[
SE = \sqrt{\frac{0.2176}{510}} \approx \sqrt{0.000426} \approx 0.02066
\]
Now, for a 99% confidence interval, we need the z-value for 99% confidence. The z-value that corresponds to 99% confidence is approximately 2.576 (for a two-tailed test).
Now we can calculate the margin of error (ME):
\[
ME = z \cdot SE \approx 2.576 \cdot 0.02066 \approx 0.0533
\]
Now, we can determine the confidence interval:
\[
(\hat{p} - ME, \hat{p} + ME) = (0.318 - 0.0533, 0.318 + 0.0533) \approx (0.2647, 0.3713)
\]
Thus, rounding to three decimal places, the 99% confidence interval for the proportion of college graduates among women who work at home is:
\[
(0.265, 0.371)
\]
\[
\hat{p} = \frac{x}{n}
\]
where:
- \( x \) is the number of college graduates (162),
- \( n \) is the total sample size (510).
\[
\hat{p} = \frac{162}{510} \approx 0.318
\]
Next, we calculate the standard error (SE) for the sample proportion:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.318(1 - 0.318)}{510}}
\]
Calculating \( 1 - \hat{p} \):
\[
1 - \hat{p} = 1 - 0.318 = 0.682
\]
Now calculating the numerator:
\[
\hat{p}(1 - \hat{p}) = 0.318 \times 0.682 \approx 0.2176
\]
Now the standard error:
\[
SE = \sqrt{\frac{0.2176}{510}} \approx \sqrt{0.000426} \approx 0.02066
\]
Now, for a 99% confidence interval, we need the z-value for 99% confidence. The z-value that corresponds to 99% confidence is approximately 2.576 (for a two-tailed test).
Now we can calculate the margin of error (ME):
\[
ME = z \cdot SE \approx 2.576 \cdot 0.02066 \approx 0.0533
\]
Now, we can determine the confidence interval:
\[
(\hat{p} - ME, \hat{p} + ME) = (0.318 - 0.0533, 0.318 + 0.0533) \approx (0.2647, 0.3713)
\]
Thus, rounding to three decimal places, the 99% confidence interval for the proportion of college graduates among women who work at home is:
\[
(0.265, 0.371)
\]
Answered by
GPT-4o mini
To determine whether it is reasonable to believe that the proportion of college graduates among women who work at home is the same as the proportion of college graduates among men who work at home (44%), we can compare the confidence interval we calculated for women with the specific proportion for men.
The 99% confidence interval we found for the proportion of college graduates among women who work at home is approximately \( (0.265, 0.371) \).
The proportion of college graduates among men who work at home is 44%, which can be expressed as a decimal:
\[
0.44
\]
Now we can observe the confidence interval:
- The lower limit of the interval is \( 0.265 \).
- The upper limit of the interval is \( 0.371 \).
Since \( 0.44 \) does not fall within the interval \( (0.265, 0.371) \), we can conclude that there is a statistically significant difference between the proportion of college graduates among women who work at home and the proportion of college graduates among men who work at home.
### Conclusion:
It is not reasonable to believe that the proportion of college graduates among women who work at home is the same as the proportion of college graduates among men who work at home, since the confidence interval for women does not include the proportion for men (0.44). This provides strong evidence that the two proportions are indeed different.
The 99% confidence interval we found for the proportion of college graduates among women who work at home is approximately \( (0.265, 0.371) \).
The proportion of college graduates among men who work at home is 44%, which can be expressed as a decimal:
\[
0.44
\]
Now we can observe the confidence interval:
- The lower limit of the interval is \( 0.265 \).
- The upper limit of the interval is \( 0.371 \).
Since \( 0.44 \) does not fall within the interval \( (0.265, 0.371) \), we can conclude that there is a statistically significant difference between the proportion of college graduates among women who work at home and the proportion of college graduates among men who work at home.
### Conclusion:
It is not reasonable to believe that the proportion of college graduates among women who work at home is the same as the proportion of college graduates among men who work at home, since the confidence interval for women does not include the proportion for men (0.44). This provides strong evidence that the two proportions are indeed different.
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