Question

John wishes to study the heights of the women’s basketball team. He completes a simple random sample of women’s basketball team members. 70 71 69.25 68.5 69 70 71 70 70 69.5 74 75.5
John knows that women’s heights are normally distributed. Use the critical value method and a 5% significance level to test the claim that women’s basketball players have heights with a mean greater than 68.6 inches (population mean height of men).
1) What is the significance level ?
2) What is the critical value?
3) What is the test statistic?
4) What is the statistical conclusion? (Reject or Fail to Reject H)

Answer 1

Solution

Answers to the given 4 questions are given below. Detailed working follows at the end.

1) Significance level = 0.05 (i.e., 5%) Answer 1
2) Critical value = 2.2001 Answer 2
3) Test statistic = 3.5844 Answer 3
4) Statistical conclusion: H₀ is rejected. Answer 4

Detailed Working

Let X = Height (inches) of women’s basketball team members.

Then, X ~ N(µ, σ²)

Claim: Women’s basketball players have heights with a mean greater than 68.6 inches

Hypotheses:

Null H₀: µ = µ₀ = 68.5   Vs Alternative H_A: µ > 68.5

Test statistic:

t = (√n)(Xbar - µ₀)/s, where n = sample size; Xbar = sample average; s = sample standard deviation.

Summary of Excel Calculations is given below:

= (√12)(70.6459 – 68.5)/2.0738

= 3.5844

Distribution, Significance Level, α , Critical Value and p-value

Under H₀, t ~ t_{n – 1}

Critical value = upper α% point of t_{n – 1} = t_{11, 0.05} = 2.2001.

Decision:

Since tcal > tcrit, H₀ is rejected.

DONE