STATS Florida State University Chi Square Hypothesis Testing Questions

I’m working on a statistics test / quiz prep and need an explanation to help me learn.

Homework 6: Chapter 10, 11, 12 (Applications)

Worth: 20 points

Name of Student: ____________________

Re upload with answers.

Confidence Intervals

1. The administrators of a hospital are concerned, and number of work shifts missed by nursing staff. The state-wide average for nursing absences for all hospitals in the state is 11.30 shifts per year. The attendance records 250 nurses were drawn as a sample to develop a baseline for performance tracking. The average shift absence rate in the sample was 11.90 shifts. With a standard deviation of 3.90. Create a 95 percent and 99 percent confidence interval to estimate the absence rate for this particular hospital. Show all calculations.
2. The number of vehicle accidents are sampled from a city of 20 major traffic intersections several times a day to assess traffic congestion and problems. The police department uses the information to allocate resources to traffic control. A performance standard of an average of 2.0 accidents per hour per area is established. A traffic emergency control system is mobilized, and resources directed to traffic control once the traffic condition is equal to or worse than the standard (2.0 accidents per hour per area)
3. What is the critical value for a chi-square test with 13 degrees of freedom at 1 percent (.01) level of significance?
4. In an electric appliance factory 400 refrigerators are produced with an assumed failure rate of 6 percent. The norm for refrigerators failing is 8 percent. The null hypothesis is that the two distributions are similar, and the alternate hypothesis is that they are dissimilar.
5. Explain the importance of the four test assumptions of the independent samples t-test
6. The following table shows the result of the Kolmogorov-Smirnov test for a variable and three of its transformations
7. Consider the data in the following table. It is group of 12 welfare recipients that participate in a training course. Before and After abilities are measured through a standardized test. Each individual took a test before the start of training and the same test after the training was completed. The paired results for each individual participant are reported. Enter the data into a SPSS a data set.
8. Please conduct one way ANOVA analysis on the data in the following table.

1. 95% confidence interval
2. 99% confidence interval

Using the data below. Should the police department mobilize the traffic emergency control system? Why or why not?

W10.1 Data Table: Number of Accidents

Areas	Number of Vehicle Accidents
1	0
2	3
3	1
4	2
5	0
6	0
7	2
8	3
9	1
10	3
11	2
12	2
13	5
14	1
15	2
16	0
17	2
18	1
19	0
20	2

Chi-Square Hypothesis Testing.

Critical value______________________

(Use the following table to find critical values…

https://people.smp.uq.edu.au/YoniNazarathy/stat_models_B_course_spring_07/distributions/chisqtab.pdf Can also use Appendix B in textbook.

1. If the chi-square test statistic were 16.98, what would you conclude regarding the null hypothesis?
2. If the chi-square value were 68.03 what would conclude regarding the null hypothesis?

Goodness of Fit Test

Apply the Chi-square formula to test the hypothesis at the 5 % level of significance

calculate the chi-square, calculate degree of freedom (r-1), calculate critical value at 5 percent, is null hypothesis accepted or rejected.

	Actual (Observed)	Norm (Expected)
Passed	376	368
Failed	24	32
Total	400	400

T-Tests

a.

b.

c.

d.

Tests for Normality

	Kolmogorov-Smirnov
Variable	Statistic	Degree of Freedom	Significance
Index	.095	102	.025
SQRT Index	.304	102	.2
Log Index	.098	102	.006
Squared Log Index	.057	102	.200

1. Explain what is Kolmogorov-Smirnov test, what does it test for, how do you interpret it?
2. Which variable(s) should the analyst use?

See this YouTube tutorial for an explanation of how the test is conducted in SPSS and interpreted,

https://www.youtube.com/watch?v=NEC3ZPl8o2c

Paired T-Test

1. Run Normality Test– Kolmogorov-Smirnov (SPSS Hint; Analyze-Descriptives-Explore_ select normality tests and histograms in plots dialogue box). Are the before and after test scores a normal distribution?
2. Use the paired samples t-test to determine if there was evidence of improvement. (95% level of confidence)
3. Use a nonparametric alternative (Wilcoxon signed rank test) (Hint: SPSS Analyze-Nonparametric-Related Samples)

What is your overall conclusion?

Table W12.4

Welfare Recipient	Before Test Score	After Test Score
1	4.5	6.7
2	3.2	4.2
3	5.8	5.2
4	3.9	4.3
5	4.2	4.1
6	3.9	4.8
7	2.6	3.2
8	5.2	4.8
9	4.5	4.5
10	3.9	4.1
11	3.8	3.6
12	4.2	5.9

Report and interpret the SPSS for both tests. Where there any differences between the two tests?

Bottom line conclusion, was there statistically significant improvements in mean test scores? What this training program effective?

Clip and Paste SPSS output into homework.

ANOVA

The daily number of accidents per 1,000 students is used to assess an educational program. In school A, safety workshops were conducted for teachers and students (treatment group). In school B and C, no safety workshops were conducted (control group). Assume normal distributions of the date in each group. Determine whether the program in School A has a lower number of accidents.

Number of Accidents per 1,000 students reported
SCHOOL A	SCHOOL B	SCHOOL C
3.82	6.58	9.47
1.01	.59	4.86
5.96	7.00	8.65
8.99	7.17	5.54
8.85	7.83	6.45
5.54	8.68	2.71
.14	1.17	8.63
4.07	5.50	9.30
7.65	6.84	7.37
1.39	4.09	7.09

Look at these two Tutorial videos of how to conduct one -way Anova test in SPSS. The second one is must helpful to show you have to manipulate and set up your data in SPSS. (Hint: you need a categorial variable i.e. group which identifies the school and another column with all the accident data.

Excellent exposition

https://www.youtube.com/watch?v=jYn5Jv7Gh4s

Must helpful in rearranging your data (Hint the table about will have to be rearranged in SPSS data view.

https://www.youtube.com/watch?v=OEOeXpxSjf8

Simple Linear Regression (one IV, one DV)

• 9.Definitions and Concepts
• Using the Productivity dataset that came with SPSS installation (SPSS Student or User file), run a simple linear regression to examine the bivariate relationship between employees perception of productivity of work center (Productivity) and the perceptions of having adequate authority to do one’s job (Jobauthr). Interpret regression results. What is the R squared, is the regression coefficient for Jobauthr significant at 5% level.

1. For what purpose is simple regression used
2. What is the interpretation of the coefficient of determination, R square and the values it can assume?
3. What is Pearson’s correlation coefficient, r? What values can it assume, how is it interpreted.
4. What is the regression equation written in algebraic form?
5. Hos is simple regression on used for hypothesis testing.

In SPSS, make sure to click on Statistics after you have dragged and downed independent variable and dependent variable. and select Descriptives so Pearson’s correlation coefficient will be produced

Run a multiple regression, where you regress,

Variables: Jobknowl (job knowledge), jobauthr(job authority, wktrtmt (employees perception of how workers are treated on productivity

What is the R-square, which regression coefficients are significant at 5 percent level?

(Hint: Open the Dataset (Productivity.sav (SPSS format) — Commands: Analyze-Regression-Linear