Question

Loss of hearing can be a significant problem for older adults. Although hearing aids can correct the physical problem, people who have lived a hearing impairment often develop poor communication strategies and social skills. To address this problem, a home education program has been developed to help people who are receiving hearing aids for the first time. The program emphasizes communication skills. To evaluate the program, overall life satisfaction was measured before treatment, again at the end of the training program, and once more at a 6 month follow-up. Data similar to the results obtained in the study are shown below. The higher the number, the greater the life satisfaction reported by the subjects. Do the data indicate a significant improvement in quality of life following the program? Test at the .05 significant level.

Subjects:          Before             After                6 months later

A 3                      7                      8

B 0                      5                      7

C 4                      9                      5

D 1                      7                      4         

1. Calculate f

4. Calculate HDS comparisons (3 of them)
5. Write out your interpretations (6 points)

Answer 1

we use paired t - test :

Given values: (1) calculate sample means on provided data : X̅1 = 7.0 X̅2 = 6.0 therefore mean difference is: μd= X̅1-X̅2 = 7.0-6.0 = 1.0 and thestandard deviation is : s = 2.943920288775949 and the sample size are n = 4 (2) Our test hypothesis is : The following null and alternative hypothesis need to be tested, hypothesis = h0: μd = 0 vs H1: μd < 0 This hypothesis corresponds to a lower tailed, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (3) Test statistics : The calculation of the t-test proceeds as follows, μd 1.0 t = ___________ = ______________ = 0.6794 s/ √(n) 2.943920288775949 / √4 (4) Rejection Criteria : Based on the information provided, the significance level is α = 0.05,and the degrees of freedom are df= 3 and the critical value for a lower tailed test is t tabulated = 2.3534 And the rejection region for this upper-tailed test is R =[t: t < -2.3534] (5) Decision about the null hypothesis : Since it is observed that t calculated means t = 0.6794 > t tabulated =-2.3534 it is then concluded that the null hypothesis is Accepted. Using the P-value approach: The p-value 0.2728, and since p = 0.2728 > α = 0.05 it is then concluded that the null hypothesis is Accepted. (6) Conclusion It is concluded that the null hypothesis Ho is Accepted. Therefore, there is no enough evidence to claim that population mean μd less then than 0 , at the 0.05 significance level.\ 

interpretation:

data dont show significuncy improvement qauality , boths mean are same by statistical test method,(paired test)