Possible (though not statistically significant) protection against hospitalization given the presence of the risk factor.       Possible (though not statistically significant) risk of hospitalization given the presence of the risk factor.       Statistically significant

The data below are from an independent-measures experiment comparing three different treatment conditions. Use an ANOVA with a = .05 to determine whether these data indicate any significant differences among the treatments    Treatment 1 Treatment 2 Treatment 3   0 1 4 N = 12 0 4 3 G = 24 0 1 6 åX2 = 92 2 0 3           M = 0.5 T = 2 M = 1.5 T = 6 M = 4 T = 16   SS = 3 SS = 9 SS = 6

Test the claim that the mean GPA of night students is significantly different than 2.4 at the 0.2 significance level.The null and alternative hypotheses would be: H0:p=0.6H0:p=0.6H1:p≠0.6H1:p≠0.6 H0:μ=2.4H0:μ=2.4H1:μ≠2.4H1:μ≠2.4 H0:μ=2.4H0:μ=2.4H1:μ>2.4H1:μ>2.4 H0:p=0.6H0:p=0.6H1:p<0.6H1:p<0.6 H0:μ=2.4H0:μ=2.4H1:μ<2.4H1:μ<2.4 H0:p=0.6H0:p=0.6H1:p>0.6H1:p>0.6 The test is: right-tailed left-tailed two-tailed Based on a sample of 31 people, the sample mean GPA was 2.36 with a standard deviation of 0.16The p-value is:  (to 3 decimals)The positive significance level is:  (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis

The following data summarize the results from an independent-measures study comparing three treatment conditions.                                                                              Treatment                                                                     I               II            III                   N = 12                                                                     3              5            6                    G = 60                                                                     5              5           10                 ∑X2 = 392                                                                     3              1           10                                                                     1              5            6                                                                  x̅1 = 3        x̅2 = 4      x̅3 = 8                                                                 T1 = 12      T2 = 16    T3 =32…

Test the claim that the mean GPA of night students is significantly different than 2.3 at the 0.05 significance level.The null and alternative hypothesis would be: H0:μ=2.3H0:μ=2.3H1:μ>2.3H1:μ>2.3 H0:p=0.575H0:p=0.575H1:p>0.575H1:p>0.575 H0:μ=2.3H0:μ=2.3H1:μ<2.3H1:μ<2.3 H0:p=0.575H0:p=0.575H1:p≠0.575H1:p≠0.575 H0:μ=2.3H0:μ=2.3H1:μ≠2.3H1:μ≠2.3 H0:p=0.575H0:p=0.575H1:p<0.575H1:p<0.575 The test is: two-tailed left-tailed right-tailed Based on a sample of 80 people, the sample mean GPA was 2.27 with a standard deviation of 0.02The test statistic is:  (to 2 decimals)The positive critical value is:  (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis

Determine whether the results below appear to have statistical significance, and also determine whether the results have practical significance. In a study of a weight loss program, 6 subjects lost an average of 49 lbs. It is found that there is about a 23% chance of getting such results with a diet that has no effect. (... Does the weight loss program have statistical significance? OA. Yes, the program is statistically significant because the results are unlikely to occur by chance. OB. Yes, the program is statistically significant because the results are likely to occur by chance. OC. No, the program is not statistically significant because the results are likely to occur by chance. O D. No, the program is not statistically significant because the results are unlikely to occur by chance. Does the weight loss program have practical significance? O A. No, the program is not practically significant because the amount of weight lost is trivial. O B. Yes, the program is practically…

Can eating a yogurt a day cause people to lose weight? In a study investigating this: (1) What is the explanatory variable? (2) What is the response variable?

Test the claim that the proportion of men who own cats is significantly different than 90% at the 0.05 significance level.The null and alternative hypothesis would be: H0:μ=0.9H0:μ=0.9H1:μ>0.9H1:μ>0.9 H0:p=0.9H0:p=0.9H1:p>0.9H1:p>0.9 H0:p=0.9H0:p=0.9H1:p≠0.9H1:p≠0.9 H0:μ=0.9H0:μ=0.9H1:μ<0.9H1:μ<0.9 H0:p=0.9H0:p=0.9H1:p<0.9H1:p<0.9 H0:μ=0.9H0:μ=0.9H1:μ≠0.9H1:μ≠0.9 The test is: two-tailed right-tailed left-tailed Based on a sample of 25 people, 93% owned catsThe test statistic is:  (to 2 decimals)The positive critical value is:  (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis Submit QuestionQuestion 12