The standard deviation of test scores on a certain achievement test
Is 11.3. If a random sample of 81 students had a sample score of 74.6 , find a 90% confidence interval
estimate for the average score of all students.
Since it is two-tailed, find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.05 rather than .10) and insert its Z score in the equation below.
90% = mean ± 1.645 SEm
SEm = SD/√n
To find the 90% confidence interval estimate for the average score of all students, we can use the formula:
Confidence Interval = sample mean ± (critical value) * (standard deviation / sqrt(sample size))
First, let's find the critical value for a 90% confidence level.
Since we have a large sample size (81 students), we can use the Z distribution.
Using a confidence level of 90%, we need to find the critical value that leaves 5% of the data in the tails. Since it is a two-tailed test, each tail will have 2.5% of the data.
Consulting the Z-table or using a statistical calculator, we find that the critical value for a 90% confidence level is approximately 1.645.
Now, let's plug in the values we have into the formula:
Confidence Interval = 74.6 ± (1.645) * (11.3 / sqrt(81))
First, we need to calculate the standard error of the mean using the formula: standard deviation / sqrt(sample size):
Standard error = 11.3 / sqrt(81) = 11.3 / 9 = 1.2556 (rounded to 4 decimal places)
Now, we can plug this value into the formula:
Confidence Interval = 74.6 ± (1.645) * (1.2556)
Calculating the upper and lower limits of the confidence interval:
Upper limit = 74.6 + (1.645) * (1.2556) = 74.6 + 2.0659 ≈ 76.66
Lower limit = 74.6 - (1.645) * (1.2556) = 74.6 - 2.0659 ≈ 72.54
Therefore, the 90% confidence interval estimate for the average score of all students is approximately 72.54 to 76.66.
To find the 90% confidence interval estimate for the average score of all students, we will need to calculate the margin of error and then use it to construct the interval.
Step 1: Calculate the standard error of the mean (SEM).
The standard error of the mean is the standard deviation divided by the square root of the sample size.
SEM = standard deviation / √sample size
SEM = 11.3 / √81
SEM = 11.3 / 9
SEM = 1.2556 (rounded to four decimal places)
Step 2: Calculate the margin of error (ME).
The margin of error is the critical value multiplied by the standard error of the mean.
For a 90% confidence level, the critical value is 1.645 (from the standard normal distribution table).
ME = critical value * SEM
ME = 1.645 * 1.2556
ME = 2.0645 (rounded to four decimal places)
Step 3: Calculate the confidence interval.
The confidence interval is calculated by subtracting and adding the margin of error to the sample mean.
Lower Limit = sample mean - ME
Lower Limit = 74.6 - 2.0645
Lower Limit = 72.5355 (rounded to four decimal places)
Upper Limit = sample mean + ME
Upper Limit = 74.6 + 2.0645
Upper Limit = 76.6645 (rounded to four decimal places)
Therefore, the 90% confidence interval estimate for the average score of all students is (72.5355, 76.6645).