The proportion of students in private schools is around 11%. A random sample of 450 students from a wide geographic area indicated that 55 attended private schools. Estimate the true proportion of students attending private schools with 95% confidence. How does it estimate compare to 11%????
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Confidence interval using proportions:
CI95 = p + or - (1.96)(√pq/n)
...where + or - 1.96 represents 95% confidence using a z-table.
Note: p = 55/450 (convert to a decimal); q = 1 - p; and n = 450 (sample size).
I'll let you take it from here.
To estimate the true proportion of students attending private schools with 95% confidence, we can use the formula for the confidence interval for a proportion.
First, let's calculate the point estimate, which is the sample proportion of students attending private schools:
Point Estimate (p̂) = Number of students attending private schools / Total sample size
p̂ = 55 / 450 = 0.1222
Next, we can calculate the margin of error using the following formula:
Margin of Error = Z * sqrt( (p̂ * (1 - p̂)) / n )
Where:
Z is the z-value corresponding to the desired confidence level (95% confidence level corresponds to approximately Z = 1.96).
p̂ is the sample proportion.
n is the sample size.
Margin of Error = 1.96 * sqrt( (0.1222 * (1 - 0.1222)) / 450 )
Now, let's calculate the margin of error:
Margin of Error = 1.96 * sqrt(0.1075 / 450)
Margin of Error ≈ 1.96 * 0.0151 ≈ 0.0296
The 95% confidence interval can be calculated by subtracting and adding the margin of error from the point estimate:
Lower Bound = p̂ - Margin of Error
Lower Bound = 0.1222 - 0.0296 ≈ 0.0926
Upper Bound = p̂ + Margin of Error
Upper Bound = 0.1222 + 0.0296 ≈ 0.1518
Therefore, the 95% confidence interval for the true proportion of students attending private schools is approximately 0.0926 to 0.1518.
To compare the estimated proportion to the given 11%, we can see that the confidence interval (0.0926 to 0.1518) does not include the value of 0.11 (11%). This means that we have evidence to suggest that the true proportion of students attending private schools is significantly different from 11% based on our sample.
To estimate the true proportion of students attending private schools with 95% confidence, we can use the formula for confidence interval for proportions.
The formula is:
\[ \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
where:
- \(\hat{p}\) is the sample proportion (in this case, the proportion of students who attended private schools in the sample)
- \(Z\) is the z-score corresponding to the desired confidence level (in this case, 95% confidence corresponds to a z-score of approximately 1.96)
- \(n\) is the sample size
Given that the sample size is 450 students and 55 attended private schools, we can calculate the sample proportion:
\[ \hat{p} = \frac{55}{450} \approx 0.1222 \]
Now we can plug in the values into the formula:
\[ 0.1222 \pm 1.96 \times \sqrt{\frac{0.1222 \times (1-0.1222)}{450}} \]
Calculating this expression gives us the confidence interval estimate of the true proportion:
\[ 0.1222 \pm 1.96 \times 0.0163 \]
Simplifying, we get:
\[ 0.1222 \pm 0.0319 \]
This means that the estimated true proportion of students attending private schools with 95% confidence is between 0.0903 and 0.1541.
To compare this estimate to the 11% proportion mentioned, we can convert 11% to a decimal:
\[ 11\% = \frac{11}{100} = 0.11 \]
Comparing the estimate (0.0903 to 0.1541) to 0.11, we can see that the estimate does not include the 11% proportion. This suggests that the proportion of students attending private schools, based on the given sample, is not the same as the overall population proportion of 11%.