Question Type 15: Performing and entire statistical test Exercises

A sample of 10 measurements (47,48,49,48,49,50,51,52,53,54) is taken. At the 5% level, test whether the population mean is 50 using a one‐sample t‐test.

A six‐sided die is rolled 60 times with observed face counts: 4, 8, 9, 7, 12, 20. At the 5% significance level, perform a complete chi‐squared goodness‐of‐fit test to determine if the die is fair.

In a study of 100 individuals, 50 smokers and 50 nonsmokers are classified by presence/absence of a respiratory condition: smokers with condition 20, without 30; nonsmokers with 10, without 40. At α=0.05 perform a chi‐squared test of independence between smoking and the condition.

Group A (n=8) and Group B (n=9) have measurements: A: 9,12,8,11,10,9,11,13 B: 7,10,6,9,8,7,11,10,8 Use a pooled‐variance two‐sample t‐test at 5% to determine if their means differ.

A machine claims to fill bottles with mean volume 100 mL. A sample of 15 bottles gives mean 103 mL and s=5 mL. At α=0.05 (one‐tailed), test if the true mean exceeds 100 mL.

A botanist classifies 200 plants into red, blue and green flowers with observed counts 95, 70 and 35. The expected proportions are 0.50, 0.30 and 0.20 respectively. At the 1% level, perform a chi‐squared goodness‐of‐fit test for these proportions.

Two classes’ exam scores are compared. Class 1 (n=20) has \bar x_1=78, s_1=10; Class 2 (n=22) has \bar x_2=72, s_2=12. At α=0.05, perform a pooled two‐sample t‐test for equality of means.

Thirty people are classified by hair color (blonde, brown, black) and eye color (blue, brown, green) with observed counts:

Blond–Blue 5, Blond–Brown 3, Blond–Green 2; Brown–Blue 3, Brown–Brown 7, Brown–Green 0; Black–Blue 2, Black–Brown 4, Black–Green 4.

At 5%, test whether hair and eye color are independent.

A sample of 12 observations yields \bar x=2.5 and s=0.7. Test at α=0.01 whether the population mean is 3.

A quality inspector counts defects per item in 100 items: 0 defects 60, 1 defect 25, 2 defects 10, 3 or more defects 5. Assuming a Poisson distribution, perform a chi‐squared goodness‐of‐fit test at 5%, estimating the parameter from the data.

A survey of 90 individuals records education level (High school, Bachelor, Master) and income bracket (Low, Mid, High). Observed counts:

High school: 12,10,8; Bachelor: 9,12,9; Master: 9,8,13.

At α=0.01, test independence using a chi‐squared test.

Sample 1 (n₁=12): \bar x_1=5, s_1=1; Sample 2 (n₂=15): \bar x_2=6.2, s_2=1.5. Using a Welch two‐sample t‐test at α=0.05 (one‐tailed, H₁: μ₁<μ₂), test the difference.