One hundred subjects in a psychological study had a mean of 35 on a test instrument designed to measure anger. The study the ó=10. Find a 99% confidence interval for the mean anger score of the population.
99% = mean ± 2.575 SEm
SEm = SD/√n
so 2.575=10/what would n be?
The probability that Luis will pass his statistics test is 0.94. Find the probability that he will fail his statistics test.
To find a 99% confidence interval for the mean anger score of the population, we can use the following formula:
Confidence Interval = sample mean ± (critical value * standard error)
Step 1: Determine the critical value
The critical value is based on the confidence level and the sample size. For a 99% confidence level, we need to find the z-score for a 99% confidence level. The z-score can be obtained from a standard normal distribution table or calculated using statistical software. For a 99% confidence level, the critical value is approximately 2.576.
Step 2: Calculate the standard error
The standard error measures the variability of the mean and is calculated by dividing the standard deviation by the square root of the sample size. In this case, the standard deviation (σ) is given as 10, and the sample size is 100.
Standard Error (SE) = σ / √n
Standard Error = 10 / √100
Standard Error = 10 / 10
Standard Error = 1
Step 3: Calculate the confidence interval
Now we can substitute the values into the confidence interval formula.
Confidence Interval = Sample mean ± (Critical value * Standard error)
Confidence Interval = 35 ± (2.576 * 1)
The lower limit of the 99% confidence interval:
Lower limit = 35 - (2.576 * 1) = 35 - 2.576 = 32.42
The upper limit of the 99% confidence interval:
Upper limit = 35 + (2.576 * 1) = 35 + 2.576 = 37.58
Therefore, the 99% confidence interval for the mean anger score of the population is (32.42, 37.58).