identify what is the 90 percent confidence interval for occupational prestige for respondents with only a bachelor's degree (N=750), given you have known their mean=52.74, standard deviation=12.67.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion of 5% in the smaller area. (90% leaves 5% in each direction.)
90% confidence interval = mean ± 1.645(SEm)
SEm = SD/√(n-1)
This gives confidence interval for means rather than individual scores.
To calculate the 90 percent confidence interval for occupational prestige for respondents with only a bachelor's degree, we can use the following formula:
Confidence Interval = Mean ± (Z * (Standard Deviation / √N))
Where:
- Mean represents the sample mean.
- Z represents the Z-score corresponding to the desired confidence level.
- Standard Deviation represents the sample standard deviation.
- N represents the sample size.
Since we want a 90 percent confidence interval, we need to find the corresponding Z-score. We can use a Z-score table or a statistical calculator to determine the value.
Using a Z-score table, the Z-score for a 90 percent confidence level is approximately 1.645.
Now, let's substitute the values into the formula:
Confidence Interval = 52.74 ± (1.645 * (12.67 / √750))
First, calculate the value inside the parentheses:
(12.67 / √750) ≈ 0.4617
Now substitute this value into the equation:
Confidence Interval = 52.74 ± (1.645 * 0.4617)
Next, calculate the value inside the second parentheses:
(1.645 * 0.4617) ≈ 0.7592
Finally, substitute this value into the equation:
Confidence Interval = 52.74 ± 0.7592
To calculate the upper and lower limits of the confidence interval, add and subtract the value from the mean:
Upper Limit = 52.74 + 0.7592 ≈ 53.50
Lower Limit = 52.74 - 0.7592 ≈ 51.98
Therefore, the 90 percent confidence interval for occupational prestige for respondents with only a bachelor's degree is approximately 51.98 to 53.50.