Question

The goals scored by two soccer teams are approximately normally distributed and independent random samples of 8 games played by each team this season are given in the following table. 2 3 2 1 2 2 3 2 Peabody Dragons Worchester Tree Frogs 2 1 2 1 2 2 1 At the .01 significance level, does the evidence support the claim that the mean number of goals scored by the Dragons this season is greater than the mean number of goals scored by the Tree Frogs? (3 pts each) 1. State the null and alternate hypothesis: 2. Name the Test: 3. Determine the P-value: (Round to 4 places or 2 nonzero places). 4. Determine whether to Reject Ho or Fail to Reject Ho, and TELL WHY by relating P-value to a: 5. Interpret the result in a complete sentence with wording related to the problem:

Answer 1

Since goals scored by two soccer teams are approximately normally distributed and independent.

Given sample size is small therefore we will use t test for independent means
Given for sample size n=8 Goals Peabody 2 3. 2 1 2 2 3: Dragons Worchester 2 1 2 1 2 Free frogs 2 1 significance level TE010] Calculation of_mean and standard deviation for Group I Peabody Dragons. 1. WIN 1 Mean 77 S.N.Goals Xi Xi -X, (xi-x)² 2 -0.125 0.015625 2 0.8781 0.765625 3 2 -0.125 1.0.015625 4, -1.125 1.265625 5 2 -0.125 0.015625 G. 2 -0.125 0.015625 3 0.875 10.765625 2. 2 -0125 0.015625 fistul 17 2.875 ΣΧι 2 = 2.125 봉 X 7. Sample standard deviation, Elxi-xi-) si sis 2185 7 Pri = 0.641 anet 32 = 0-4T

for Group Tree frogs, S.nl Croals Xi 2 - 3. 4. Xi - GYi-) 0-5 0.25 -0.5 6.25 -0.5 0.25 0.5" 0.28 -0.5 0.25 0.5 0.28 0.5 Or 28 -0.5 0.25 Mean. 5. Exi X2 = 12 n 8 6 2 9 IX = 15 - Total 12 2.00 sample standard deviation Elli-82) Sz n-| 2 Is 7 S = 0.534 and 알 = 0.29 1. Null Hypothesis , How M = M 4 ie there is no difference between meam number of goals scored by both teams. Alternative hypothesis, Hi 4,74 (Right talled test) Le Name f test & t- test for 2 independent means

3. Test statBtR : The test statstz unden He is t XI - X2 st ~ Ant-2 tat m where $? ni+h₂-2 { (n=1):34+ (924)5?? {7x0.41 +7X0:29} こ 8+8-2 14 X 4.9. -- 135 s=0.592 Now t 2.125-15 - 0.592 J7 + t = 2.11 P-value for t-value: 2.11 at df = 14 xx=0.0/ 0.626 4 We fall to reject Ho. Because 0.026 > 0.01 () P-value s. Interpretation & we fail to reject Ho. it means there is not enough evidence te claim that mean number of golls scored by the Dragons is greater than that of Tree frogs: