Pschological tests are often used to determine the hostility levels in people. high scores on the hlt pschological test corresponds to high hostility levels. suppose that 26 college students, all with equally high hlt test scores, were separated into two groups randomly. group 1 (10 students) was treated for high hostility levels by using method a; whereas group 2 (16 students) was treated for high hostility levels by using method b. both groups were treated for a period of 5 months, at which time all 26 students were again given the hlt test. below is the summary data on the hlt test for the two groups. group 1: sample mean is 79 and sample standard deviation is 7group 2: sample mean is 84 and sample standard deviation is 8compute a 95% confidence level for the difference in the means of the two treatment methods and select the best answer below. assume that the independent random samples were taken from normal distributions and that the two population variances are equal.

 the correct answer is we are 95% confident that there is no statistically significant difference in the mean treatment  methods for hostility.  

Step-by-step explanation:

1)  Test for the equality of variances in the two groups to choose the appropriate t-test.  

H0: σ (1)^2 = σ (2)^2  

Ha: σ (1)^2 ≠ σ (2)^2  

Larger variance = 64  

Smaller variance = 49  

F = 1.30612  

Degrees of freedom 15 and 9  

Critical F from the table (with alpha=0.05) = 2.58  

Calculated F is smaller than critical F, so we use the pooled variance t-test.  

Sample 1 size 7  

Sample 2 size 8  

Sample 1 mean 79  

Sample 2 mean 84  

Sample 1 S.D. 7  

Sample 2 S.D. 8  

Pooled S.D = [(n1-1)s1^2+(n2-1)s2^2]/(n1+n2-2)

Pooled variance s = [(6)(49)+(7)(64))] / (13) =  (294+448)/13=742/13=57.076923

Pooled variance s^2 = 57.076923  

Standard error of difference in means = sqrt(1/n1+1/n2) times sqrt(s^2)  

Standard error of difference in means = (0.517549)(7.554927) = 3.910046 (denominator of t)  

Confidence interval = (mean1-mean2) +/- t SE  

t is the critical t with 24 degrees of freedom = 2.056  

(79 - 84) +/- (2.056) (3.910046)  

= (-13.04, 3.04)  

Interval encloses 0  

We are 95% confident that there is no statistically significant difference in the mean treatment  methods for hostility.  

b. 153 students do not participate in sports and seventy-five 12th graders were surveyed

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step-by-step explanation: