BUS 211 Study Guide - Midterm Guide: Standard Deviation, Analysis Of Variance, Statistical Hypothesis Testing
BUS 211 Study Guide
Determining Statistical Processes: T-Tests, Chi-Square, ANOVA
T-Tests
1. Need to deteie hat tpe of data ou’e eig asked to aalze: ualitatie o
quantitative.
2. If you are asked to analyze one qualitative variable (gender) with one quantitative
variable (GPA) (this is a two-category variable), you use t-tests.
a. Example: Does GPA differ by Gender?
3. Solve using t-tests.
a. 1) Determine if variables are equal
i.
b. Reject Ho if p-value is .05, FTR Ho if p-value is .05
c. Look at data and find F & sig under equal variances assumed
i. (F) Levene = 4.086, p-value = .045
d. Since p-value .045 .05, we reject Ho. We have evidence variances are not
equal.
e. 2) Determine if means are equal
i.
ii.
f. Reject Ho if p-value is .05, FTR Ho if p-value is .05
g. Because we found variables , e use ifoatio fo Equal variances not
assumed, t = . ad p-value = .020
h. Since p-value .020 .05, we reject Ho. We have evidence the mean GPA
differs between gender.
T-test summary:
1. Data requirements, one quantitative (ex: major), one qualitative (ex: gender), exactly 2
categories.
2. Determine if variables between the two groups are equal
a. Look at F and sig (p-value)
3. Determine if means between the two groups are equal
a. Look at t and sig (p-value)
Chi-Square Test of Independence
1. Chi-Square is a probability distribution,
2. Is gede ad galig status independent? oe does not rely on the other) – Does
gender determine whether you gamble or not.
3. Data requirements, two qualitative variables (# of categories important)
4. How does work?
a. Of the 134 people who took the survey, 21 claims to be non-gamblers
b. Overall demographic for gender: 41% female, 59% male.
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c. If gender does not determine gambling status, what % of the non-gamblers
would you expect to be female? (Problem 3 in packet)
i. We expect .41(21) = 8 9 female non-gamblers
ii. We expect .59(21) = 12.4 12 male non-gamblers
d. Cross tabulations – allows you to look at 2 variables at the same time :
e. 113 gamblers, we expect .41(113) = 46.3 46 to be female. We expect .59(113) =
66.7 67 to be male
f. Formula to find chi-squared
g. Problem 3 in packet:
i. Ho: gender & gambling status are independent, Ha: gender & gambling
status are dependent
ii. Reject Ho if p-value is .05, FTR Ho if p-value is .05
iii. Test Status:
+
+
+
= 20.5 =
[Pearson Chi Square has 20.5, 3rd column p-value = .000 *]
iv. Since p-value .000 .05, we reject Ho. We have evidence that gender
& gambling are status are dependent.
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Document Summary
We have evidence the mean gpa differs between gender. We expect . 59(113) : problem 3 in packet: status are dependent, ho: gender & gambling status are independent, ha: gender & gambling. + (cid:4666)(cid:2875)(cid:2874) (cid:2874)(cid:2874). (cid:2875)(cid:4667)(cid:3118) (cid:2874)(cid:2874). (cid:2875) (cid:2872)(cid:2874). (cid:2870) [pearson chi square has 20. 5, 3rd column p-value = . 000 *] (cid:3039)(cid:3039) (cid:3030)(cid:3032)(cid:3039)(cid:3039)(cid:3046) (cid:2870)= (cid:4666)(cid:2869)(cid:2876) (cid:2876). (cid:2874)(cid:4667)(cid:3118) (cid:2876). (cid:2874: since p-value . 000 . 05, we reject ho. If we reject ho, we say (cid:862)our results are statistically significant(cid:863). If we fail to reject (ftr) ho, we say (cid:862)our results are statistically non- significant(cid:863). Since my p-value = . 955 . 05, our results are statistically non-significant. We have evidence mean age does not differ by employment status. Since p-value = . 152 . 05, our results are statistically non-significant. We have evidence mean salary does not differ by gender. #3 on white sheet: (chi-square, (cid:2870)= . 337, p-value = . 561) Since p-value = . 561 . 05, our results are statistically non-significant.
