g The average midterm exam score for students from a large undergraduate course is 69. A random sample of 32 students is selecte
1 answer:
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Part A
Given info:
- xbar = sample mean = 4.90 ppm
- s = sample standard deviation = 1.12 ppm
- n = 23 = sample size
Because n > 30 is not true, and we don't know the population standard deviation (sigma), this means we must use a T distribution.
The degrees of freedom here are n-1 = 23-1 = 22.
At 90% confidence and the degrees of freedom mentioned, the t critical value is roughly t = 1.717
Use a T distribution table or calculator to determine this. If you don't have a calculator for the task, then you can search out "inverse T calculator" and there are tons of free options to pick from.
The margin of error E is
E = t*s/sqrt(n)
E = 1.717*1.12/sqrt(23)
E = 0.400982
This is approximate and accurate to 6 decimal places.
The confidence interval is going to be xbar plus or minus that E value
L = lower bound = xbar - E = 4.90 - 0.400982 = 4.499018 = 4.50
U = upper bound = xbar + E = 4.90 + 0.400982 = 5.300982 = 5.30
The confidence interval in the format of (L, U) is (4.50, 5.30)
You could also express it as the format L < mu < U and it would be 4.50 < mu < 5.30; however, I'll stick to the first method.
<h3>Answer: (4.50, 5.30)</h3>=====================================================
Part B
Since we know sigma = 2.6 is the population standard deviation, we can use a Z distribution now.
At 90% confidence, the z critical value is roughly 1.645; use a table or calculator to determine this.
Round this up to the nearest integer to get 20326. For min sample size problems, <u>always</u> round up.
<h3>Answer: 20326</h3>