In: Math
Lawyer | Nurse | Teacher | Control | |
8 | 6 | 9 | 8 | |
5 | 7 | 6 | 7 | |
7 | 6 | 8 | 6 | |
7 | 8 | 8 | 7 | |
4 | 9 | 7 | 9 |
A researcher is interested in whether the likeability of a crying woman is affected by the viewer’s knowledge of the woman’s occupation. Twenty participants were shown a video of a woman crying and were asked to rate her likeability on a scale from 1 (not very likable) to 10 (highly likable). Prior to viewing the video, participants were told that the woman was either a lawyer, a nurse, a teacher, or they were not told anything about her occupation (control condition).
(a) State the null and research hypotheses. (b) Calculate the appropriate test statistic. (c) Interpret the test statistics at an alpha level of .05.
**Please show all work and explain!! Thank you!!
(a) H0:Null Hypothesis:
HA: Research Hypothesis:
(b)
From the given data, the following statistics are calculated:
Lawyer | Nurse | Teacher | Control | Total | |
N | 5 | 5 | 5 | 5 | 20 |
31 | 36 | 38 | 37 | 142 | |
Mean | 31/5=6.2 | 36/5=7.2 | 38/5=7.6 | 37/5=7.4 | 142/20=7.1 |
203 | 266 | 294 | 279 | 1042 | |
Std. Dev. | 1.6432 | 1.3038 | 1.1402 | 1.1402 | 1.3338 |
From the above Table, ANOVA Table is formed as follows:
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Sum of Squares | F | P - Value |
Between Treatments | 5.8 | 3 | 5.8/3=1.9333 | 1.9333/1.75=1.10476 | 0.3761 |
Witthin Treatments | 28 | 16 | 28/16=1.75 | ||
Total | 5.8+28=33.8 | 3+16=19 |
(c)
F Ratio = 1.8333/1.75 = 0.371
Degrees of Freedom of Numerator = 3
Degrees of Freedom of Denominator = 16
By Tehnology, P - Value = 0.3761
Since P - Value is greater than = 0.05, the difference is not significant. Fail to reject null hypothesis.
Conclusion:
The data do not support the claim that the lakeability of a crying
woman is affected by the viewer's knowledge of woman's
occupation.