In a random sample of 50, the average is 3987 with a population standard deviation of 630. How can one find the 90% confidence interval of the true mean?
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score for .4500 from mean (Z = 1.645)
90% conf. Interval = mean ± 1.645(SD)
z = 50
To find the 90% confidence interval of the true mean, you can use the formula:
Confidence Interval = Sample Mean ± (Z * (Population Standard Deviation / √n))
Where:
- Sample Mean is the average of the random sample (3987 in this case).
- Z is the value from the standard normal distribution that corresponds to the desired level of confidence (90% confidence corresponds to a Z value of approximately 1.645).
- Population Standard Deviation is the known standard deviation of the population (630 in this case).
- n is the sample size (50 in this case).
- √n represents the square root of the sample size.
Now, substituting the given values into the formula, we can calculate the confidence interval as follows:
Confidence Interval = 3987 ± (1.645 * (630 / √50))
Step 1: Calculate the standard error:
Standard Error (SE) = Population Standard Deviation / √n
SE = 630 / √50
Step 2: Multiply the standard error by the Z value:
Margin of Error = Z * SE
Margin of Error = 1.645 * (630 / √50)
Step 3: Calculate the lower and upper limits of the confidence interval:
Lower Limit = Sample Mean - Margin of Error
Upper Limit = Sample Mean + Margin of Error
Calculating the values:
Lower Limit = 3987 - (1.645 * (630 / √50))
Upper Limit = 3987 + (1.645 * (630 / √50))
By substituting the values and performing the calculations, you can find the lower and upper limits of the 90% confidence interval for the true mean.