STATISTICS CASE
Ch11 case Golf
1.
is mean driving distances of current balls is mean driving distances of new balls is mean driving distances of sampled current balls is mean driving distances of sampled new balls
Use the test statistics and normal distribution table to get p-value.
If p-value is smaller than, then we reject H0, which means the mean driving distances of current balls and new balls are different.
2.
From the t distribution table we find that p-value is between 0.05 and 0.1.
With p-value >=0.05, we cannot reject H0, which indicates that there’s no sufficient evidence that the mean driving distances of two balls are different.
There’s no enough proof to show that the coating will make the driving distances of the balls shorter. Since there’s no big difference between the driving forces of two balls, using the more durable new balls to increase the market share of Par, Inc., seems feasible.
3.
From the descriptive statistics we can see that the mean driving distances of two kinds of balls are close. The distances of current balls (270.3) are a little bigger than that of new balls (267.5). With the same maximum, the minimum distance of current balls are 5 yards larger than that of new balls. And the variance of the new balls (97.9) is bigger than that of current balls (76.6), which means that new balls have greater variability of driving distances.
4. =0.05 t0.025=2.0227 d.f=39
The confidence interval of current balls is [267.475, 273.0775]
The confidence interval of current balls is [264.33, 270.67]
Interval estimation of the difference between the means of two populations
=76.85
We take 77 as degree of freedom, t0.025=1.991
Confidence interval [-1.385, 6.935]
5. The management of Par Inc., wants to use the new golf balls to increase market share but are concerned about the effect of coating. The conclusion of the hypothesis testing shows that they don’t need to worry about this problem. However, from the
1.
is mean driving distances of current balls is mean driving distances of new balls is mean driving distances of sampled current balls is mean driving distances of sampled new balls
Use the test statistics and normal distribution table to get p-value.
If p-value is smaller than, then we reject H0, which means the mean driving distances of current balls and new balls are different.
2.
From the t distribution table we find that p-value is between 0.05 and 0.1.
With p-value >=0.05, we cannot reject H0, which indicates that there’s no sufficient evidence that the mean driving distances of two balls are different.
There’s no enough proof to show that the coating will make the driving distances of the balls shorter. Since there’s no big difference between the driving forces of two balls, using the more durable new balls to increase the market share of Par, Inc., seems feasible.
3.
From the descriptive statistics we can see that the mean driving distances of two kinds of balls are close. The distances of current balls (270.3) are a little bigger than that of new balls (267.5). With the same maximum, the minimum distance of current balls are 5 yards larger than that of new balls. And the variance of the new balls (97.9) is bigger than that of current balls (76.6), which means that new balls have greater variability of driving distances.
4. =0.05 t0.025=2.0227 d.f=39
The confidence interval of current balls is [267.475, 273.0775]
The confidence interval of current balls is [264.33, 270.67]
Interval estimation of the difference between the means of two populations
=76.85
We take 77 as degree of freedom, t0.025=1.991
Confidence interval [-1.385, 6.935]
5. The management of Par Inc., wants to use the new golf balls to increase market share but are concerned about the effect of coating. The conclusion of the hypothesis testing shows that they don’t need to worry about this problem. However, from the