Question

1. A track coach is trying to determine if a new training program will lead to better standing long jump results. She selects 20 of her athletes, tests their standing long jump distances, then has them take part in a 12-week training program. After they complete the program, she tests their standing long jump distances. (The distances below are recorded in meters.) Before After 2.35 2.34 2.42 2.48 2.26 2.29 2.58 2.62 2.62 2.64 2.16 2.18 2.40 2.44 2.62 2.67 2.35 2.39 2.47 2.44 (a) Construct a 90% confidence interval for the mean difference in jump distance (after-before). [5 pts] (6) The creators of the program claim that their program significantly increases jump distance. Test this claim at the 5% significance level. i. Define the parameter(s) and state the hypotheses: [3pts] ii. Find the observed sample statistic (using your calculator) and the test statistic (using a formula). Use proper notation for the sample statistic ( X or p ), and clearly identify your test statistic as either a z-score or a t-score. [4pt]

We do not use critical values at all in our course so please avoid using them when giving me explanations, as they can cause confusion on my end between your explanation and what im learning on my course

Answer 1

Let us denote the difference

d = after - before

1. a)

Let à denotes the sample mean of difference, s denotes the sample standard deviation of difference and n denotes the sample size. Here, 0.0276 d = 0.026, sź = 0.0008, sd = 0.0276, n = 10, standard error of mean Sd vn = 0.008718 V10 The margin of error is Sd 0.0276 0.0276 ME *t 1-0.9 *t0.05,9 * 1.833113 = 0.015981 0.016 n 2 V10 V10 a 90 percent confidence interval for the population mean of difference ud is Sd = = *t 1-0.9 ,9 0.028 = 0.026 * 10.05,9 710 = 0.026 + 0.028 V10 * 1.833113 = 0.026 +0.015981 = (0.010019, 0.041981) ~ (0.01,0.04)
b) parameter : population mean of difference = $\mu_d$

Let ud denotes the population mean of differnce. Hypotheses : To test null hypothesis Hofd = 0 against alternative hypothesis H1 :fed > 0 This is a one sided right tailed test as alternative hypothesis H contains only > sign. Let à denotes the sample mean of difference, sa denotes the sample standard deviation of difference and n denotes the sample size. Here, d = 0.0260, så = 0.0008, sd = 0.0276, n = 10 The test statistic can be written as : vncă – 0.0) t= sd which under H, follows at distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > t0.05,n–1 = 1.833113 0.4 - 0.3 0.2 0.1 0.0 Now, Value of test statistic: The value of the test statistic tobs 10(0.0260 - 0) 0.0276 = 2.982405 2.98 associated degrees of freedom = 9 The critical value =tcritical = t0.05,9 = 1.833113 ~ 1.833 P - value = P(ty > 2.982405) = 1- P(tg < 2.982405) = 1-0.992305 = 0.007695 0.0077 Conclusion : Since p-value < 0.05 and tobs > tcritical = 1.833113, so we reject H, at 0.05 level of significance Hence, we can conclude that mean difference is significantly greater than 0