Question

Question 4 of 10 (1 point) View problem in a pop-up Men spend an average of 29 minutes per day on weekends and holidays exercising and playing sports. They spend an average of 23 minutes per day reading. A random sample of 26 men resulted in a mean of 32.9 minutes exercising with a standard deviation of 6.9 minutes and an average of 21.2 minutes reading withs = 7.2 minutes. At a = 0.10 for both, is there sufficient evidence that these two results differ from the national means? Assume the populations are normally distributed. Source: Time magazine. Part 1 out of 10 Testing the exercise mean. State the hypotheses and identify the claim with the correct hypothesis. Ho: u (select) 0 (select) H1: (select) (select) This hypothesis test is a (select) test. CHECK NEXT

1. state the hypothesis and identify the claim with the correct hypothesis. Also one or two tailed test?

2. Compute the test value

3. Find the P-value

4. Make the decision to reject or not reject the null hypothesis

5. Summarize the results

Answer 1

Testing Excemuise tears Hoi Mzag Hup 29 Test is tuo hailed t-test Test Statictis & X-Mutur s/su :32.9-29 u tas Statistic 609126 p-value 0.008. seme 0.00820.1, hence to is ngented this there is enough evidence to claim that the results different fuani national lend a significa average at loy

Testing Reading Meaw Hoi Me23 H. M#23 Test is two-tailed t-test. Test Statistic: X-M utud Slow Test statistic : 21.2-23 ntas 7.5/26 2 - 1.275 Ip-value: 0.2111, since 0.2144> Ool hence Ho is ust eguted. This there is enough evidence to clain hat the results of reading excercise is at lor. ust different front national average level of significance