A study measuring fitness level in teens randomly sampled 100 female teens and 102 male teens with a higher score indicating more fitness. Researchers suspected that the fitness level for female teens is lower than for male teens, and wanted to check whether the data would supported this hypothesis.If μ1 and μ2 represent the mean fitness level for female teens and male teens respectively, which of the following is an appropriate pair of hypotheses in this case?H0: µ1 − µ2 = 0Ha: µ1 − µ2 < 0H0: µ1 − µ2 < 0Ha: µ1 − µ2 = 0H0: µ1 = µ2Ha: µ1 > µ2H0: µ1 − µ2 = 0Ha: µ1 - µ2 > 0

A study measuring fitness level in teens randomly sampled 112 male teens and 101 female teens with a higher score indicating more fitness. Researchers suspected that the fitness level for male teens (μ1) is higher than for female teens (μ2), and wanted to check whether the data would supported this hypothesis.The following hypotheses were tested:H0: µ1 = µ2Ha: µ1 > µ2Sample N Mean SD1(M) 112 7.38 6.952(F) 102 7.15 6.31t-test of difference: df = 210 t-value = 0.25 p-value = 0.400Which of the following is an appropriate conclusion based on the output? The data provide sufficient evidence to conclude that male and female teens do not differ in mean fitness score. The data do not provide sufficient evidence to reject H0, so we accept it, and conclude that female and male teens do not differ in mean fitness score. The data do not provide sufficient evidence to conclude that the mean fitness score of male teens is higher than that of female teens. The data provide sufficient evidence to reject H0 and to conclude that the mean fitness score for male teens is higher than that of female teens.

A Canadian study measuring depression level in teens (as reported in the Journal of Adolescence, vol. 25, 2002) randomly sampled 112 male teens and 101 female teens, and scored them on a common depression scale (higher score representing more depression). The researchers suspected that the mean depression score for male teens is higher than for female teens, and wanted to check whether data would support this hypothesis.The following hypotheses were tested:H0: μ1 = μ2Ha: μ1 > μ2The following is the (edited) output for the test:Which of the following is an appropriate conclusion based on the output? The data do not provide sufficient evidence to conclude that the mean depression score of male teens is larger than that of female teens. The data provide sufficient evidence to conclude that male and female teens do not differ in mean depression score. The data do not provide sufficient evidence to reject H0, so we accept it, and conclude that male and female teens do not differ in mean depression score. The data provide sufficient evidence to reject H0 and to conclude that the mean depression score for male teens is larger than that of female teens.

The statement of the alternative hypothesis Ha: μ1 - μ2 < 0, states:Question 5Select one:a.that population mean 1 is smaller than population mean 2.b.that population mean 2 is smaller than population mean 1.c.that population mean 1 is larger than population mean 2.d.that the two population means are the same.

Given the alternative hypothesis Ha: µ1 ≠ µ2 ≠ µ3 ≠ …….. µk. Which are the correct interpretations?*1 pointThe population mean is not same in all the groupsAll groups should be different from each otherPopulation means and hypothesized valueare equalThe population mean is same in all the groups

Watch your weight: A study published in 2008 reported that the mean weight of men aged 40-59 was 89.3 kg. Another study, published in 2016, reported that a sample of 192 men aged 40-59 had an average weight of 87 kg. Assume the population standard deviation is =σ14.2 kg. Can you conclude that the mean weight of men aged 40-59 is lower in 2016 than in 2008? Use the =α0.05 level of significance.Part 1 of 5(a) State the appropriate null and alternate hypotheses.H0: =μ89.3H1: <μ89.3The hypothesis test is a ▼left-tailed test.Part: 1 / 51 of 5 Parts CompletePart 2 of 5(b) Compute the value of the test statistic. Round your answer to two decimal places.z=