Question

1. Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a population standard deviation of 47 pounds. Suppose we take a sample of 30 new players and we find that the average weight from that sample is 237 pounds. We are interested in seeing if the weight of NFL players is decreasing.
   1. What is the standard error?
   2. What is the margin of error at 90% confidence?
   3. Using my sample of 30, what would be the 90% confidence interval for the population mean?
   4. If I wanted to control my margin of error and set it to 5 at 90% confidence, what sample size would I need to take instead of the 30?
   5. What are the null and alternative hypotheses?
   6. What is the critical value at 90% confidence?
   7. Calculate the test statistic (using the sample of 30 and NOT the answer from part d).
   8. Find the p-value.
   1. What conclusion would be made here at the 90% confidence level?

Answer 1

Part a

Standard error = σ/sqrt(n)

We are given

σ = 47

n = 30

Standard error = 47/sqrt(30) = 8.580986734

Standard error = 8.5810

Part b

We are given

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

Margin of error = E = Z*Standard error = 1.6449*8.5810 = 14.11489

Margin of error = 14.11489

Part c

Confidence interval = Xbar -/+ E

Xbar = 237

Lower limit = 237 - 14.11489 = 222.8851

Upper limit = 237 + 14.11489 = 251.1149

Part d

The sample size formula is given as below:

n = (Z*σ/E)^2

We are given

Confidence level = 90%

So, critical value Z = 1.6449

(by using z-table)

Margin of error = E = 5

Population standard deviation = σ = 47

Sample size is given as below:

n = (1.6449*47/5)^2

n = 239.0753

n = 240

Required sample size = 240

Part e

Null hypothesis: H0: the weight of NFL players is not decreasing.

Alternative hypothesis: Ha: the weight of NFL players is decreasing.

Part f

Test is lower tailed test.

Confidence level = 90%

So, critical value Z = -1.2816

(by using z-table)

Part g

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

Z = (237 – 247)/[47/sqrt(30)]

Z = -1.1654

Test statistic = -1.1654

Part h

P-value = 0.1219

(by using z-table)

Part i

Confidence level = 90% or c = 0.90

So, α = 1 – 0.90 = 0.10

P-value > α = 0.10

So, we do not reject the null hypothesis

So, there is not sufficient evidence to conclude that the weight of NFL players is decreasing.