Question

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Single Sample t-Test Coach Brown is training his runners for an upcoming race, but he is concerned about their pace He takes down sample times from his top 7 runners to compare it to his all-time standard race speed of 7.9 minutes per kilometer (). The sample times from his runners were 8.55 minutes per kilometer (M) with a standard deviation estimate of 0.78 (s). Based on this data, complete the six steps of hypothesis testing given that our research question is: "Did the speed of the top 7 runners have a statistical difference compared to the Coach's all-time standard race speed? Step 1: Populations, Distribution, and Assumptions Population 1: Population 2: Distribution: Hypothesis test to be used (Explain): Step 2: Hypotheses Research Hypothesis: Symbolic Research Hypothesis: Null Hypothesis: Symbolic Null Hypothesis: Step 3: Characteristics of the Comparison Distribution Step 4: Critical Values

Answer 1

Step 1: Population 1: All-time standard runners. Population 2: Trained runners by Coach Brown. Distribution: The distribution of the population should be approximately normal. Hypothesis test to be used: t- test for single mean since the population standard deviation is not known. Step 2: Research Hypothesis: The speed of the top 7 runners has a statistical difference compared to Coach's all-time standard race speed. Symbolic Research Hypothesis: H:H€7.9 Null Hypothesis: The speed of the top 7 runners does not have a statistical difference compared to Coach's all-time standard race speed. Symbolic Null Hypothesis: H:=7.91 Step 3: My=8.55 sp=0.78 Step 4:

Consider the level of significance to be 0.05. The degrees of freedom are given as follows: df=n-1 = 7-1 = 6 The critical value of the t distribution at 0.05 level of significance with 6 degrees of freedom is 2.4469 (from t distribution tables). t critical = 2.4469 df =6 Step 5: The test statistic can be computed as follows: _Um-u t= ? 8.55-7.9 0.78/ = 2.2048 Thus, statistic = 2.2048. Step 6: The decision rule is to reject the null hypothesis if the value of the test statistic is greater than the critical value. Since the value of the test statistic (2.2048) is less than the critical value (2.4469), therefore, do not reject the null and hypothesis and conclude that the speed of the top 7 runners does not have a statistical difference compared to Coach's all-time standard race speed.

The effect size is given as follows: U Cohen's d-M SM 8.55-7.9 0.78 =0.8333 Thus, Cohen's d=0.8333). The 90% confidence interval is given as follows: CI = Hly #tas2.dr To The confidence level (c) is given to be 0.90. Therefore, the level of significance is given as follows: a=1-0 =1-0.90 = 0.10 The critical value of the t distribution at 0.10 level of significance with 6 degrees of freedom is 1.9432 (from t distribution tables). Now, the confidence interval can be computed as follows: 0.78 CI=8.55+(1.9432) =8.55+0.5729 =(7.9771,9.1229) Thus, the 90% confidence is (7.9771,9.1229)].