Answer to Question #298228 in Psychology for evie

Question

An investigator predicts that dog owners in the country spend more time walking their dogs than do dog owners in the city. The investigator gets a sample of 21 country owners and 23 city owners. The mean number of hours per week that city owners spend walking their dogs is 10.0. The standard deviation of hours spent walking the dog by city owners is 3.0. The mean number of hours country owners spent walking their dogs per week was 15.0. The standard deviation of the number of hours spent walking the dog by owners in the country was 4.0. Do dog owners in the country spend more time walking their dogs than do dog owners in the city?

Answer 1

STEP 1  Identify the test to be done and why: Independent Groups T-Test IV has 2 and ONLY 2 levels DV is quantitative Between-Subjects design (21 country dwellers, 23 city dwellers) STEP 2  Write null and alternative hypotheses: H0: µcity = μcountry H1: μcity ≠ μcountry

STEP 3  Find tcv and write the rejection rule: tCV: α = .01 Directional or non-directional test? Nondirectional df = 42 Because this study has 2 and only two levels in a between subjects design, we use the t-distribution with df = N-2 = 44-2 = 42  use df 40 in Appendix D CV: 2.704

Rejection Rule: Rejection Rule: Reject H0 if tOBS is > 2.704 or ≤ -2.704 STEP 4  Calculate tobs and compare it to the rejection rule: a) ŝx-bar1-xbar2 = ŝpooled √ (1/n1 + 1/n2) = 3.2*√ (1/21 + 1/23) = .9653 b) tOBS = x-bar1 - x-bar2 (10 – 15)/.9653 = -5.18 (rounded) ŝx-bar1-xbar2 -5.18 < -2.704  reject Ho STEP 5  Calculate eta2 and interpret the effect size: eta2 = tobs 2 = -5.182 /(-5.182 + 42) = .3898 = .39 tobs2 + df

Interpret size: large 39% of the variability in time spent walking dogs is accounted for by location.