Question

Two marathon training procedures are tried for comparison purposes. Their efficacy is to be determined in a marathon race. Assume race times are given in minutes. Use α=0.056α=0.056 for all calculations.

The following results were observed:. The data is given in the file below.

	Procedure_A	Procedure_B
1	146	144
2	147	147
3	149	150
4	142	147
5	144	141
6	147	147
7	143	147
8	148	148
9	147	148
10	146	151
11	147	145
12	144	141
13	144	153
14	147	150
15	140	141
16		151
17		144
18		156
19		150
20		153
21		155
22		140
23		138
24		146
25		141
26		151

(a) Do the populations appear to be normal? Hint: use a probability plot to decide
A. Yes, procedure B appears to be normal but procedure A does not appear to be normal
B. Yes, Both procedure A and procedure B appear to be normal
C. Yes, procedure A appears to be normal but procedure B does not appear to be normal
D. No, Both procedure A and procedure B appear to be normal
E. No, neither procedure A appears to be normal nor procedure B appears to be normal
F. Yes, Both procedure A and procedure B appear to be non-normal


(b) Choose the most appropriate statistical hypotheses to see which procedure results in longer marathon times (be very careful in how you formulate your hypothesis, this one is tricky! Use the sample data to make any assumptions necessary.)
A. H0:μ1=μ2HA:μ1>μ2H0:μ1=μ2HA:μ1>μ2
B. H0:μ1=μ2HA:μ1<μ2H0:μ1=μ2HA:μ1<μ2
C. H0:μD=0,HA:μD<0H0:μD=0,HA:μD<0
D. H0:μ1=μ2HA:μ1≠μ2H0:μ1=μ2HA:μ1≠μ2
E. H0:μD=0,HA:μD≠0H0:μD=0,HA:μD≠0
F. H0:μD=0,HA:μD>0H0:μD=0,HA:μD>0


(c) Using technology available to you, test to see if the variances are equal or not?
A. There appears to be more variation in the 2nd procedure then in the 1st.
B. They appear to be equal.
C. There appears to be an unequal amount of variation in the 2nd procedure then in the 1st.
D. There appears to be more variation in the 1st procedure then in the 2nd.


(d) Report the p-value of the test you ran in (c) use at least three decimals in your answer (this is from Levene's test).
P-value =

(e) Test the statistical hypotheses in (b) by carrying out the appropriate statistical test. Find the value of the test statistic for this test, use at least two decimals in your answer (from your test of middle).
Test Statistic =

(f) Determine the PP-value for this test, to at least three decimal places (from your test of middle).
P=

(g) Based on the above calculations, we should reject or not reject  the null hypothesis.

Answer 1

Ans a ) using minitab>graph> boxplot

we have

A. Yes, procedure B appears to be normal but procedure A does not appear to be normal

b ) D. H0:μ1=μ2HA:μ1≠μ2

c ) using minitab>stat>basic stat>2 variance

we have

Test and CI for Two Variances: Procedure_A, Procedure_B

Method

Null hypothesis σ(Procedure_A) / σ(Procedure_B) = 1
Alternative hypothesis σ(Procedure_A) / σ(Procedure_B) ≠ 1
Significance level α = 0.05

Statistics

95% CI for
Variable N StDev Variance StDevs
Procedure_A 15 2.473 6.114 (1.700, 4.138)
Procedure_B 26 4.861 23.626 (3.988, 6.407)

Ratio of standard deviations = 0.509
Ratio of variances = 0.259

95% Confidence Intervals

CI for
CI for StDev Variance
Method Ratio Ratio
Bonett (0.307, 0.802) (0.094, 0.644)
Levene (0.301, 0.863) (0.091, 0.746)

Tests

Test
Method DF1 DF2 Statistic P-Value
Bonett — — — 0.006
Levene 1 39 6.05 0.018

c ) option D is true

D. There appears to be more variation in the 1st procedure then in the 2nd.

d ) P-value = 0.018

e ) using minitab>stat>basic stat>2 sample t

we have

Two-Sample T-Test and CI: Procedure_A, Procedure_B

Two-sample T for Procedure_A vs Procedure_B

N Mean StDev SE Mean
Procedure_A 15 145.40 2.47 0.64
Procedure_B 26 147.12 4.86 0.95

Difference = μ (Procedure_A) - μ (Procedure_B)
Estimate for difference: -1.72
95% CI for difference: (-4.04, 0.61)
T-Test of difference = 0 (vs ≠): T-Value = -1.50 P-Value = 0.143 DF = 38

test statistic = -1.50

f )p value is 0.143

g 0 since p >0.056 ,  we should not reject  the null hypothesis.