Answer Happy

2. Create a 95% confidence interval for the mean for this variable in StatCrunch. Provide the Screenshot here and interpret your confidence interval in context. Remember, you only have a sample and, thus, DO NOT KNOW O. Hint: What type of test is this? A z-test or t-test? (20 points). Sigma is unknow we use t-test. Options of One sample confidence interval Mean of variable 95% confidence interval results: Variable Sample Mean Std. Kr. DF Lami U. Lami Bxercise Hours 5.25 0714335543133150116 664 Interpret Confidence Interval: We are 95% confidence that the average hours of exercise that NOVA students do is between 3.8 and 6.7 hours per week.

4. Check the assumptions needed to perform a significance test and do a boxplot in stat crunch to check for outliers. Remember, you only have a sample and, thus, DO NOT KNOW O. Hint: What type of test is this? A z-test or t-test? (10 points): Sigma is unknow we use t-test. Assumptions: Because sigma is unknown, the sample is not random but, in the question, it says even though it's not random we will do this, and n = 52 30, we can use the t-test. The claim is "that the mean hours of exercise that an adult from the age 18-64 years do is 2.5 hours per- week." The null and alternative hypotheses are u = 2.5 Hou = 2.5 hours per week (claim). Hall # 2.5 hours per week This is two tailed tests. Boxplot: There are two outliers in the graph shown below. Options @of 2) Boxplot of Exercise Hours Green marker indicates the sample mean 3 15 Exercise Hours

- 5. Conduct a two-tailed (NOT TWO-SAMPLE - that is entirely different!) significance test using a 5% level of significance and statcrunch and your chosen null hypothesis value (do this even if the assumptions are violated or you have outliers). Use statcrunch and include a screen shot here provide your conclusion interpret your conclusion in context. Remember, you only have a sample and, thus, DO NOT KNOW o. Hint: What type of test is this? A z-test or t-test? (20 points) - Sigma is unknow we use t-test. Options (of 43 One sample Thypothesis test: u: Mean of variable HO:25 HAH*25 Hypothesis test results Variable Sample Mean Std. Err. DF T-Stat P-value Exercise Hours 5.25 0.71433554 51 3.8497315 0.0003 Conclusion (i.e reject or fail to reject H.): The p-value is in region and p-value is smaller than alpha we reject the Ho. P-value = 0.003 and level of significance is 0.05. Interpret Conclusion: There is not enough evidence at the 5% los to support that the mean hours of exercise that an adult from the age 18-64 years can do is 2.5 hours per week.