Respuesta :
Answer:
We conclude that the new technique is not an improvement over the old technique at 5% level of significance.
Step-by-step explanation:
We are given that the average speed of a baseball line drive is 83 miles per hour.
His coach recorded the speed of 42 random hits during practice and found that his average speed using the new technique was 84.2 miles per hour, with a standard deviation of 4.7 miles per hour.
Let [tex]\mu[/tex] = true average speed using the new technique.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 83 miles per hour {means that the new technique is not an improvement over the old technique}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 83 miles per hour {means that the new technique is an improvement over the old technique}
The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average speed using the new technique = 84.2 mph
s = sample standard deviation = 4.7 mph
n = sample of hits = 42
So, the test statistics = [tex]\frac{84.2-83}{\frac{4.7}{\sqrt{42} } }[/tex] ~ [tex]t_4_1[/tex]
= 1.655
The value of t test statistics is 1.655.
Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of 1.6832 at 41 degree of freedom for right-tailed test.
Since our test statistic is less than the critical value of t as 1.655 < 1.6832, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.
Therefore, we conclude that the new technique is not an improvement over the old technique.
