suppose a sample of 2404 tenth graders is drawn.Of the sampled, 601 read at or below the eighth grade level. using the data, construct the 98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level.
To construct a confidence interval for the population proportion, we can use the formula:
Confidence Interval = Sample Proportion ± Margin of Error
First, let's calculate the sample proportion (p̂), which is the proportion of the sample reading at or below the eighth grade level. The sample proportion (p̂) is calculated by dividing the number of individuals who read at or below the eighth grade level (601) by the total sample size (2404):
p̂ = 601 / 2404 ≈ 0.25
Next, we need to calculate the margin of error. The margin of error depends on the confidence level and the standard deviation.
Since we don't have the population standard deviation, we will use the standard error (SE) instead. The formula to calculate the standard error is:
SE = √ [(p̂ * (1 - p̂)) / n]
where n is the sample size.
SE = √ [(0.25 * (1 - 0.25)) / 2404]
SE ≈ √ (0.1875 / 2404)
SE ≈ √ 0.000078
SE ≈ 0.0088
To calculate the margin of error (ME), we can multiply the standard error (SE) by the critical value ta/2, where "t" is the t-distribution value and a is the significance level.
Since we are looking for a 98% confidence interval, the significance level (alpha) is 1 - confidence level = 1 - 0.98 = 0.02.
Dividing this by 2 (since it will be two-tailed), we have a/2 = 0.02 / 2 = 0.01.
Now, we need to find the t-distribution value for 0.01 in a two-tailed test with degrees of freedom (df) equal to the sample size minus 1. In this case, df = 2404 - 1 = 2403.
Using a t-distribution table or a statistical calculator, the t-value for a two-tailed test at a significance level of 0.01 (df = 2403) is approximately 2.617.
Finally, we can calculate the margin of error using the formula:
ME = t * SE
ME = 2.617 * 0.0088
ME ≈ 0.023
Now we can construct the confidence interval by adding and subtracting the margin of error from the sample proportion:
Confidence Interval = p̂ ± ME
Confidence Interval = 0.25 ± 0.023
Therefore, the 98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is approximately 22.7% to 27.3%.