Question

A health educator wants to evaluate the effect of a dental film on the frequency with which children brush their teeth. A random selection of 5 children are used for the experiment. First, a baseline of the number of times the children brush their teeth over a month's period is established. Next, the children are shown the dental film and again the number of teeth brushings are recorded for a month. The following data are recorded.   Please analyze this data using a two-tailed test at the α = .05 level.

Participant	Baseline	After Film
1	25	29
2	28	29
3	22	25
4	30	30
5	22	24

What are the degrees of freedom for the test statistic?

Answer 1

degrees of freedom are df= 8  calculation: Given values: (1) The provided sample means are : X̅1 = 25.4 X̅2 = 27.4 and population standard deviations are : s = 3.578 s = 2.702 and the sample size are n1 = 5 and n2 = 5. (2) Our test hypothesis is : The following null and alternative hypothesis need to be tested, hypothesis = h0: μ1 = μ2 vs H1: μ1 ≠ μ1 This hypothesis corresponds to a two-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, z = X̅1 - X̅2 25.4 - 27.4 ---------------------- = -------------------------------- = -0.9974 √(σ1²/n1)+(σ2²/n2) √(3.578²/5) + (2.702²/5) (4) Rejection Criteria : Based on the information provided, the significance level is α = 0.05,and the degrees of freedom are df= 8 and the critical value for a two-tailed test is t tabulated = 1.8595 And the rejection region for this two-tailed test is R =[t: |t| > 1.8595] (5) Decision about the null hypothesis : Since it is observed that t calculated means |t| = 0.9974 < t tabulated = 1.8595 it is then concluded that the null hypothesis is Accepted. Using the P-value approach: The p-value is 0.3478, and since p = 0.3478 > 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. mean μ1 is not different than μ2 ,Therefore, there is no enough evidence to claim that,population mean μ1 is different than μ2 , at the 0.05 significance level.