Hypothesis Testing Solved Problems: Z, T, and F Tests

Problem 9: Z-Test for Two Means

A researcher wants to know the intelligence of students in a school. He selected two groups of students. In the first group, there are 150 students having a mean IQ of 75 with a S.D. of 15. In the second group, there are 250 students having a mean IQ of 70 with a S.D. of 20. Is there a significant difference between the means of the two groups?

Solution

Given

For Group I:

• Sample size, n₁ = 150
• Sample mean, x̄₁ = 75
• Standard deviation, s₁ = 15

For Group II:

• Sample size, n₂ = 250
• Sample mean, x̄₂ = 70
• Standard deviation, s₂ = 20

Level of significance: α = 0.05

Since both sample sizes are large (n₁, n₂ > 30), use the Z-test for the difference between two means.

Step 1: State the Hypotheses

• Null Hypothesis (H₀): μ₁ = μ₂ (There is no significant difference between the two means.)
• Alternative Hypothesis (H₁): μ₁ ≠ μ₂ (There is a significant difference.)

Step 2: Test Statistic

Z = (x̄₁ - x̄₂) / √((s₁² / n₁) + (s₂² / n₂))

Step 3: Calculate the Standard Error

SE = √((15² / 150) + (20² / 250))
SE = √((225 / 150) + (400 / 250))
SE = √(1.5 + 1.6) = √3.1 ≈ 1.761

Step 4: Calculate the Z-value

Z = (75 - 70) / 1.761
Z = 5 / 1.761 ≈ 2.84

Step 5: Decision

At the 5% level of significance, the critical value is Z_0.025 = 1.96.

Since 2.84 > 1.96, the calculated value lies in the rejection region. Hence, Reject H₀.

Problem 10: Student's T-Test for Single Mean

Ten individuals are chosen at random from a normal population and their heights are found to be 63, 63, 66, 67, 68, 69, 70, 70, 71, 71 inches. Test if the sample belongs to the population whose mean height is 66 inches.

Ten individuals are chosen at random from a normal population and their heights (in inches) are: 63, 63, 66, 67, 68, 69, 70, 70, 71, 7163,\;63,\;66,\;67,\;68,\;69,\;70,\;70,\;71,\;7163,63,66,67,68,69,70,70,71,71

Solution

Given

• Sample size, n = 10
• Sample observations: 63, 63, 66, 67, 68, 69, 70, 70, 71
• Population mean, μ₀ = 66

Since the sample size is small (n < 30) and the population standard deviation is unknown, use the Student's t-test.

Step 1: State the Hypotheses

• Null Hypothesis (H₀): μ = 66
• Alternative Hypothesis (H₁): μ ≠ 66 (Two-tailed test)

Step 2: Calculate the Sample Mean

∑x = 678
x̄ = 678 / 10 = 67.8

Step 3: Calculate the Sample Standard Deviation

∑(x - x̄)² = 81.60
s = √(81.60 / (10 - 1)) = √9.067 &approx; 3.01

Step 4: Calculate the T-value

The test statistic is: t = (x̄ - μ₀) / (s / √n)
t = (67.8 - 66) / (3.01 / √10)
t = 1.8 / (3.01 / 3.162) = 1.8 / 0.952 &approx; 1.89

Step 5: Decision

Degrees of freedom: n - 1 = 9
At the 5% level of significance, t_{0.025, 9} = 2.262.

Since 1.89 < 2.262, the calculated value does not fall in the rejection region. Hence, Accept H₀. (Accept H₀: 2.56 < 3.35)

Problem 3: F-Test for Equality of Variances

Pumpkins were grown under two experimental conditions. Two random samples of 11 and 9 pumpkins show the sample standard deviations of their weights as 0.8 and 0.5 respectively. Assuming that the weight distributions are normal, test the hypothesis that the true variances are equal.

Given

For Sample 1:

• Sample size, n₁ = 11
• Sample standard deviation, s₁ = 0.8

For Sample 2:

• Sample size, n₂ = 9
• Sample standard deviation, s₂ = 0.5

Level of significance: α = 0.05

Since the population is normal and we are testing the equality of variances, use the F-test.

Step 1: State the Hypotheses

• Null Hypothesis (H₀): σ₁² = σ₂² (The population variances are equal.)
• Alternative Hypothesis (H₁): σ₁² ≠ σ₂² (The population variances are not equal.)

Step 2: Calculate the F-statistic

The F-test statistic is: F = Larger Sample Variance / Smaller Sample Variance

Sample variances:
s₁² = (0.8)² = 0.64
s₂² = (0.5)² = 0.25

Hence, F = 0.64 / 0.25 = 2.56

Step 3: Degrees of Freedom

Numerator degrees of freedom: v₁ = n₁ - 1 = 11 - 1 = 10
Denominator degrees of freedom: v₂ = n₂ - 1 = 9 - 1 = 8

From the F-table at the 5% level of significance: F_0.05(10, 8) &approx; 3.35

Step 4: Decision

Calculated value: F = 2.56
Table value: F_0.05(10, 8) = 3.35

Since 2.56 < 3.35, the calculated F-value is less than the critical value. Therefore, Accept H₀.

Conclusion