Question

Tennis elbow is thought to be aggravated by the impact experienced when hitting the ball. A recent article reported the force (N) on the hand just after impact on a one-handed backhand drive for eight advanced players and for ten intermediate players. The advanced players have a mean force of 38.1 and standard deviation 10.3. And the intermediate players have a mean force of 27.8 and standard deviation 7.5. Use a 99% confidence interval to determine if there is a significant difference in mean force between the two groups of players.

Answer 1

Since the P-value (1.9646) is greater than the significance level (0.10), we cannot accept the null hypothesis. Thus, we have enough statistical evidence to suggest that there is a significant difference between the two groups of players.

Explanation:

Null hypothesis: μ1 - μ2 = 0

Alternative hypothesis: μ1 - μ2 ≠ 0

These hypotheses gives a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Thus, to solve this: the significance level is 0.10. A two-sample t-test of the null hypothesis will be conducted.

Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = √[(s₁²/n₁) + (s₂²/n₂)]

thus we have SE = √[(10.3²/8) + (7.5²/10]

= √(13.261 + 5.625)

= √(18.886)

SE = 4.3458.

Degree of freedom is

DF = (s₁²/n₁ + s₂²/n₂)² / { [ (s₁² / n₁)² / (n₁ - 1) ] + [ (s₂² / n₂)² / (n₂ - 1) ] }

DF = (10.3²/8 + 7.5²/10)² / { [ (10.3²/8)² / (8-1) ] + [ (7.5²/10)² / (10-1) ] }

DF = (13.261 + 5.625)² / [(13.261)² / 7] + [ (5.625)² / 9]

DF = (18.886)² / {(175.854/7) + (31.6406/9)

(356.68) / (25.122 + 3.5156)

356.68 / 28.6376

DF = 12.45

The test statistic t = [ (x₁ - x₂) - d ] / SE

where x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means which is zero, and SE is the standard error.

Thus we have [ (38.1 - 27.8) - 0] / 4.3458

t = [ (10.3) - 0] /4.3458

t = 10.3 / 4.3458

t = 2.3701.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 12 degrees of freedom is more extreme than 2.3701; that is, less than 2.3701 or greater than 2.3701.

Use the t Distribution Calculator to find P(t < 2.3701) = 0.9823, and P(t > -2.3701) = 0.9823. Thus, the P-value = 0.9823 + 0.9823 = 1.9646.

Since the P-value (1.9646) is greater than the significance level (0.10), we cannot accept the null hypothesis. Thus, we have enough statistical evidence to suggest that there is a significant difference between the two groups of players.