Question

Tennis elbow is thought to be aggravated by the impact experienced when hitting the ball. The article "Forces on the Hand in the Tennis One-Handed Backhand" (Int. J. Sport Biomech., 1991: 282-292) reported the force (Newtons on the hand just after impact on a one-handed backhand drive for six advanced players and for eight intermediate players. Advanced: m = 6,11 = 40.3, S1 = 11.3 Intermediate: m = 8,72 = 21.4, 92 = 8.3 In their analysis of the data, the authors assumed that both force distributions were normal and the population variances were not equal. 2.1 Calculate a 99% CI for the difference between true average force for advanced players (11) and true average force for intermediate players (uz). 2.2 Perform the appropriate hypothesis test for these data at the significance level of a=0.01. Be sure to specify the hypotheses, test statistic, the distribution of the test statistic, a p-value or rejection region. Compare the result with the interval you calculated. Does your interval provide compelling evidence for concluding that the two u's are different? 2.3 How would the conclusion change if you did not assume the populations had equal variance? What are the cautions with making this assumption?

Answer 1

The symbols are used as they are defined in the question.

2.1

The 99% confidence interval for is given by

11 - 12

$((\bar{x}_1-\bar{x}_2)\pm s'\sqrt{\frac{1}{m_1}+\frac{1}{m_2}})$

The subscript 1 corresponds to the advanced group and the subscript 2 corresponds to the intermediate group and

(n1 - 1) si + (n2 - 1)s n1 +n2 - 2 V

Hence from the given values,

$s'=\sqrt{\frac{(6-1)11.3^2+(8-1)8.3^2}{6+8-2}}$

$\Rightarrow s'=\sqrt{93.39}=9.663850164$

Hence the 99% confidence interval is given by

$((40.3-21.4)\pm 9.663850164*t_{\;0.005,12}\sqrt{\frac{1}{6}+\frac{1}{8}})$

$i.e.\;\;(18.9\pm 9.663850164*3.055*0.5400617249)$

$i.e.\;\;(18.9\pm 15.94427592)$

2.2

The null and the alternative hypotheses are given by

$H_0:\mu_1=\mu_2\;\;against\;\;H_1:\mu_1\neq\mu_2$

The test statistic is given by

$t=\frac{\bar{x}_1-\bar{x}_2}{s'\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$\Rightarrow t=\frac{40.3-21.4}{9.663850164\sqrt{\frac{1}{6}+\frac{1}{8}}}$

$\Rightarrow t=\frac{18.9}{28.56367464}=0.6616795717$

Under the null hypothesis, the test statistic follows t distribution with df 12

The crtical value is obtained from the Biometrika table as 3.055

As the observed value is less than the critical value, we fail to reject the null hypothesis at 5% level of significance and hence conclude there is no significant difference between the average force of the two groups pf players.

2.3

The conclusion for paired t test may differ than the paired t test . The assumptions include the elimination of outliers.

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