Part A:
Null hypothesis: The new technique does not improve Natalie's volleyball serve speed.
Alternative hypothesis: The new technique improves Natalie's volleyball serve speed.
Part B:
The appropriate test for this situation is a one-sample t-test. We need to check the following conditions:
1. Random sample: We are told that the coach recorded the speed of 41 random serves, so this condition is met.
2. Normality: We assume that the distribution of serve speeds is approximately normal. We can check this condition by creating a histogram or a normal probability plot of the sample data. If the data is not normal, we can use the central limit theorem since the sample size is large enough (n > 30).
3. Independence: We assume that each serve speed is independent of the others.
Part C:
Using a one-sample t-test with a significance level of 0.05, we need to calculate the test statistic and compare it to the critical value from the t-distribution with 40 degrees of freedom (since n - 1 = 40).
t = (59 - 58) / (2.7 / sqrt(41)) = 3.23
The critical value for a two-tailed test with alpha = 0.05 and 40 degrees of freedom is approximately ±2.021.
Since our test statistic (3.23) is greater than the critical value (2.021), we reject the null hypothesis. We can conclude that there is sufficient evidence at the 0.05 level to suggest that Natalie's new technique has improved her volleyball serve speed.