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An environmental group conducted a study to determine whether crows in a certain region were ingesting food containing unhealthy levels of lead. A biologist classified lead levels greater than 6.0 parts per million (ppm) as unhealthy. The lead levels of a random sample of 23 crows in the region were measured and recorded. The mean lead level of the 23 crows in the sample was 4.90 ppm and the standard deviation was 1.12 ppm.
a) Construct and interpret a 90 percent confidence interval for the mean lead level of crows in the region.
b) A previous study of crows showed that the population standard deviation was at 2.6 ppm. What minimum sample size would be required to construct a 90 percent confidence interval to have a margin of error within 0.03?

Sagot :

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.

Answer:  (4.50, 5.30)

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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.

[tex]n = \left(\frac{z*\sigma}{E}\right)^2\\\\n \approx \left(\frac{1.645*2.6}{0.03}\right)^2\\\\n \approx 20325.254444 \\\\[/tex]

Round this up to the nearest integer to get 20326. For min sample size problems, always round up.

Answer: 20326

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