Question

Plz quick help, can you show how do this problem on minitab?

10-33. + The overall distance traveled by a golf ball is tested by hitting the ball with Iron Byron, a mechanical golfer with a swing that is said to emulate the distance hit by the legendary champion, Byron Nelson. Ten randomly selected balls of two different brands are tested and the overall distance measured. The data follow: Brand 1: 275, 286, 287, 271, 283, 271, 279, 275, 263, 267 Brand 2: 258, 244, 260, 265, 273, 281, 271, 270, 263, 268 (a) Is there evidence that overall distance is approximately normally distributed? Is an assumption of equal variances justified? (b) Test the hypothesis that both brands of ball have equal mean overall distance. Use o 0.05. What is the P-value? (c) Construct a 95% two-sided CI on the mean difference in overall distance for the two brands of golf balls. (d) What is the power of the statistical test in part (b) to detect a true difference in mean overall distance of 5 yards? (e) What sample size would be required to detect a true dif- ference in mean overall distance of 3 yards with power of approximately 0.75?

Answer 1

Minitab 17

a)

Menu bar-Stat-Basic Statistics-Normality test

MTB > NormTest 'Brand 1';

SUBC>   Title "Brand 1".

Brand 1

3 4 5 6 Brand 1 Normal Mean 275.7 StDev 8.028 10 AD 0.204 P-Value 0.825 70 50 0 40 30 20 10 255 275 Brand 1 260 265 270 280 285 290 295

From above graph, p-value=0.825 > alpha=0.05, accept H0. Brand 1 distance is normally distributed.

MTB > NormTest 'Brand 2';

SUBC>   Title "Brand 2".

Brand 2

From above graph, p-value=0.653 > alpha=0.05, accept H0. Brand 2 distance is normally distributed.

Menu bar-Stat-ANOVA-Vartest

MTB > VarTest 'Brand 1' 'Brand 2';

SUBC> Unstacked;

SUBC> Confidence 95.0;

SUBC> GInterval;

SUBC>   NoDefault;

SUBC> TMethod;

SUBC> TBonferroni;

SUBC> TTest.

Test for Equal Variances: Brand 1, Brand 2

Method

Null hypothesis         All variances are equal

Alternative hypothesis At least one variance is different

Significance level      α = 0.05

95% Bonferroni Confidence Intervals for Standard Deviations

Sample   N    StDev          CI

Brand 1 10   8.0284 (5.40272, 15.3767)

Brand 2 10 10.0449 (5.00388, 25.9897)

Individual confidence level = 97.5%

Tests

                           Test

Method                Statistic P-Value

Multiple comparisons       0.35    0.553

Levene                     0.16    0.698

Test for Equal Variances: Brand 1, Brand 2

p-value= 0.698, accept H0. Variances are equal.

OR

Menu bar-Stat-Basic Statistics-2 variances

MTB > TwoVariances 'Brand 1' 'Brand 2';

SUBC>   Confidence 95.0;

SUBC>   STest 1;

SUBC>   Alternative 0;

SUBC>   GInterval;

SUBC>   NoDefault;

SUBC>   TMethod;

SUBC>   TStatistics;

SUBC>   TConfidence;

SUBC>   TTest.

Test and CI for Two Variances: Brand 1, Brand 2

Method

Null hypothesis         σ(Brand 1) / σ(Brand 2) = 1

Alternative hypothesis σ(Brand 1) / σ(Brand 2) ≠ 1

Significance level      α = 0.05

Statistics

95% CI for

Variable   N   StDev Variance       StDevs

Brand 1   10   8.028    64.456 (5.759, 13.921)

Brand 2   10 10.045   100.900 (5.590, 22.449)

Ratio of standard deviations = 0.799

Ratio of variances = 0.639

95% Confidence Intervals

                            CI for

         CI for StDev      Variance

Method       Ratio           Ratio

Bonett (0.419, 2.264) (0.176, 5.128)

Levene (0.387, 2.304) (0.150, 5.308)

Tests

                       Test

Method DF1 DF2 Statistic P-Value

Bonett    1    —       0.35    0.553

Levene    1   18       0.16    0.698

Test and CI for Two Variances: Brand 1, Brand 2

p-value= 0.698, accept H0. Variances are equal.

b)

Menu bar-Stat-Basic Statistics-2 sample t -test

MTB > TwoSample 'Brand 1' 'Brand 2';

SUBC>   Confidence 95.0;

SUBC>   Test 0.0;

SUBC>   Alternative 0;

SUBC>   Pooled.

Two-Sample T-Test and CI: Brand 1, Brand 2

Two-sample T for Brand 1 vs Brand 2

          N    Mean StDev SE Mean

Brand 1 10 275.70   8.03      2.5

Brand 2 10   265.3   10.0      3.2

Difference = μ (Brand 1) - μ (Brand 2)

Estimate for difference: 10.40

95% CI for difference: (1.86, 18.94)

T-Test of difference = 0 (vs ≠): T-Value = 2.56 P-Value = 0.020 DF = 18

Both use Pooled StDev = 9.0927

Comment-p-value=0.020 < alpha=0.05, Reject H0. Thus,Both brands of ball do not have equal mean overall distance.

c)

Menu bar-Quality tools- Tolerance interval

MTB > Name C3 "TBNorm1" C4 "TBNorm2" C5 "TBNPara1" C6 "TBNPara2".

MTB > TolInterval 'Brand 1' 'Brand 2';

SUBC>   GTIPlot;

SUBC>   Confidence 95.0;

SUBC>   PPercent 95.0;

SUBC> SPTLimits 'TBNorm1' - 'TBNorm2';

SUBC> SNPTLimits 'TBNPara1' - 'TBNPara2'.

Tolerance Interval: Brand 1, Brand 2

Method

Confidence level                   95%

Percent of population in interval 95%

Statistics

Variable   N     Mean   StDev

Brand 1   10 275.700   8.028

Brand 2   10 265.300 10.045

95% Tolerance Interval

                                 Nonparametric      Achieved

Variable     Normal Method          Method        Confidence

Brand 1   (248.456, 302.944) (263.000, 287.000)        8.6%

Brand 2   (231.213, 299.387) (244.000, 281.000)        8.6%

Achieved confidence level applies only to nonparametric method

Tolerance Interval Plot for Brand 1

Tolerance Interval Plot for Brand 2

d)

Menu bar-Stat-Power and Sample Size-2-Sample-t

Use pooled standard deviation (See Minitab Help)

MTB > Power;

SUBC>   TTwo;

SUBC>     Sample 10 10;

SUBC>     Difference 5;

SUBC>     Sigma 9.0927;

SUBC>   GPCurve.

Power and Sample Size

2-Sample t Test

Testing mean 1 = mean 2 (versus ≠)

Calculating power for mean 1 = mean 2 + difference

α = 0.05 Assumed standard deviation = 9.0927

            Sample

Difference    Size     Power

         5      10 0.214107

         5      10 0.214107

The sample size is for each group.

Power Curve for 2-Sample t Test

Power is 21.4107 %.

e)

Menu bar-Stat-Power and Sample Size-2-Sample-t

MTB > Power;

SUBC>   TTwo;

SUBC>     Difference 3;

SUBC>     Power 0.75;

SUBC>     Sigma 9.0927;

SUBC>   GPCurve.

Power and Sample Size

2-Sample t Test

Testing mean 1 = mean 2 (versus ≠)

Calculating power for mean 1 = mean 2 + difference

α = 0.05 Assumed standard deviation = 9.0927

            Sample Target

Difference    Size   Power Actual Power

         3     129    0.75      0.751704

The sample size is for each group.

Power Curve for 2-Sample t Test

Sample size is 129.