SOLUTION: You measure 21 backpacks' weights, and find they have a mean weight of 45 ounces. Assume the population standard deviation is 8.3 ounces. Based on this, what is the maximal margin
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-> SOLUTION: You measure 21 backpacks' weights, and find they have a mean weight of 45 ounces. Assume the population standard deviation is 8.3 ounces. Based on this, what is the maximal margin
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Question 1196386: You measure 21 backpacks' weights, and find they have a mean weight of 45 ounces. Assume the population standard deviation is 8.3 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean backpack weight.
Give your answer as a decimal, to two places
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You can put this solution on YOUR website! sample size = 21
sample mean = 45
population standard deviation = 8.3
sample standard error = 8.3 / sqrt(21) = 1.811208489.
confidence level = .99
critical alpha = .01 / 2 = .005.
critical z-score = plus or minus 2.575829303.
z-score formula = z = (x - m) / s
z is the critical z-score.
x is the critical raw score
m is the sample mean
s is the standard error.
solving for x, you get:
x1 = z1 * se + 45 = -2.57582903 * 1.811208489 + 45 = 40.3346361.
x2 = z2 * se + 45 = 2.57582903 * 1.811208489 + 45 = 49.6653639.
maximum margin of error = (x2 - x1) / 2 = 4.6653639.
your solution is 4.67 rounded to two decimal places.
confirmation that the confidence level = 99% is shown below.
