a random sample of 20 prices for summer sleeping bags was taken with a sample mean of $83.75 and a sample standard deviation of $29.97
a. Find a 95% confidence interval for the mean price of all summer sleeping bags.
b. what is m i.e., what is the margin of error?
a. 95% = mean ± 1.96 SEm
SEm = SD/√n
b. If you are talking about alpha error, P = .05.
To find the confidence interval and margin of error, we can use the formula:
Confidence Interval = Sample Mean ± (Critical Value × Standard Error)
Margin of Error = Critical Value × Standard Error
Given information:
Sample Mean (x̄) = $83.75
Sample Standard Deviation (s) = $29.97 (Note: If this is the sample standard deviation, we need to use the t-distribution instead of the z-distribution to calculate the confidence interval)
Step 1: Determine the critical value
Since the sample size is less than 30 and the population standard deviation is unknown, we need to use the t-distribution and find the appropriate critical value. We need to use the t-distribution with n-1 degrees of freedom, where n is the sample size. In this case, the sample size is 20, so the degrees of freedom are 19.
Looking up the critical value in a t-distribution table or using a calculator with the appropriate function, we find that for a 95% confidence level and 19 degrees of freedom, the critical value is approximately 2.093.
Step 2: Calculate the standard error (SE)
Standard Error (SE) = Sample Standard Deviation / √(Sample Size)
SE = $29.97 / √(20)
Step 3: Calculate the margin of error (m)
Margin of Error = Critical Value × Standard Error
m = 2.093 × ($29.97 / √(20))
Step 4: Calculate the confidence interval
95% Confidence Interval = Sample Mean ± Margin of Error
CI = $83.75 ± (2.093 × ($29.97 / √(20)))
Now we can calculate the values.
a. 95% confidence interval for the mean price of all summer sleeping bags:
CI = $83.75 ± (2.093 × ($29.97 / √(20)))
b. Margin of Error (m) = 2.093 × ($29.97 / √(20))
Plug in the values to get the final answers.