Question

2. The null and alternate hypotheses are: HO: Hd = 0 H:Hd0 The following paired observations show the number of traffic citations given for speeding by Officer Dhondt and Officer Meredith of the South Carolina Highway Patrol for the last five months. June September Officer Dhondt Officer Meredith Day July 25 20 May 30 26 August 19 15 26 19 19 Assume the.05 significance level. Do this problem 2 ways: as dependent samples and as independent samples. What is(are) the critical value(s)? What is the test statistic? At the .05 significance level, is there a difference in the mean number of citations given by the two officers?

Answer 1

Case 1 : Pared t test

Hypothesis :

Ho: μD​ = 0

Ha: μD​ ≠ 0

Minitab > Stat > Basic Stat > Paired t test

95% CI for mean difference: (2.717, 6.483) T-Test of mean difference = 0 Pairedt (Test and Confidence Interval) Samples in columns First sample: Dhondt Second sample: Meredith Summarized data (differences) D Worksheet 1 *** Sample size: C1 C2 Mean: Dhondt Meredith Standard deviation: 22 Paired t evaluates the first sample minus the second sample. Select Graphs... Options... Help OK 1 Cancel

95% CI for mean difference: (2.717, 6.483) T-Test of mean difference - 0 Pairedt (Test and Confidence interval) Samples in columns First sample: Dhondt Paired t - Options Confidence level: 35.0 I Worksheet 1 *** C1 C2 Dhondt Meredith C3 Test mean: 0.0 Alternative: not equal 30 19 Help OK Cancel Select Graphs... Options... Help OK Cancel

Output :

- 10-12 22:39:52 Paired T-Test and Cl: Dhondt, Meredith Paired I for Dhondt - Meredith Dhondt Meredith Difference N 5 5 5 Mean 24.40 19.80 4.600 St Dev 4.16 3.96 1.517 SE Mean 1.86 1.77 0.678 95% CI for mean difference: (2.717, 6.483) T-Test of mean difference = 0 (vs not = 0): T-Value - 6.78 P-Value = 0.002

df = 4 (n-1)

t critical = 2.776

P value = 0.002 < 0.05, reject H0

There is enough evidence to claim that population mean difference not equal to 0

Case 2 : Independent samples t test

Hypothesis :

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

Minitab > Stat > Basic Stat > Two sample t test

-5% CI for difference: (-1.32, 10.52) -Test of difference = 0 (vs not =): I-Value = 1.79 P-Value = 0.111 DF = 8 Both use pooled StDev = 4.0620 2-Sample t (Test and Confidence interval) Samples in one column Samples: Dhondt Subscripts: Meredith Samples in different columns Worksheet 1 *** First: Dhondt C1 C2 Second: Meredith Dhondt Meredith 30 Summarized data Sample size: Mean: 22 19 First: Second: Standard deviation: Assume equal variances Select Graphs... Options... Help ок 1 Cancel

1 ISL ULUIlleLeice - U VSTUL - 1 Value - 1.13 Value - U.11. DL-U Both use Pooled StDev = 4.0620 2-Samplet (Test and Confidence interval) 2-Samplet - Options Samples in one colum Samples: Dhond Confidence level: 25.0 85.0 Subscripts: Mered Test difference: 0.0 Samples in different Alternative: not equal First: Dhond C3 -12 Second: Mered Help OK Cancel Worksheet 1 *** + C1 C2 Dhondt Meredith 1 30 19 25 20 26 Mean: deviation: Summarized data Sample size: First: Second: 19 Assume equal variances Select Graphs... 1 Options... Help OK Cancel

Output :

TWO-Sample T-Test and Cl: Dhondt, Meredith Two-sample I for Dhondt vs Meredith Dhondt Meredith N 5 5 Mean 24.40 19.80 StDev 4.16 3.96 SE Mean 1.9 1.8 Difference = mu (Dhondt) - mu (Meredith) Estimate for difference: 4.60 95% CI for difference: (-1.32, 10.52) I-Test of difference = 0 (vs not =) : I-Value = 1.79 P-Value = 0.111 DF = 8 Both use Pooled StDev = 4.0620

df = 8 (n1+n2-2)

t critical = 2.306

P value = 0.111 > 0.05, Do not reject H0

There is not enough evidence to claim that population mean difference not equal to 0

Mean difference = 24.40 - 19.80 = 4.6