Statistics Homework 4: T-tests, ANOVA, and Data Analysis

This statistics assignment provides a comprehensive analysis of various statistical concepts. It begins by defining and differentiating between ratio, ordinal, and nominal data types. It then explains the suitability of t-tests for ratio data and provides examples of hypotheses that can be examined using t-tests and ANOVAs. The assignment also clarifies the difference between independent and dependent measures and highlights the importance of descriptive statistics like mean and standard deviation. Furthermore, it includes the calculation of ANOVAs for two specific problems, demonstrating the process of hypothesis testing and the interpretation of results, including p-values and levels of significance. The solution concludes with a discussion on the value of t-tests and ANOVAs compared to descriptive statistics.

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Homework 4
An example of ratio data is height of students. An example of ordinal data is grades of
students classified in A, B,C & D based on performance. An example of nominal data is
gender. The data type appropriate for a t test would be ratio since it has numerical values.
For t test, we need mean and standard deviation which are sensible only for a ratio data.
A hypothesis which can be examined using t test is whether the performance of boys and girls
of a given class tend to differ significantly. A suitable hypothesis to be tested using ANOVA
is to compare the performance of three sections of Grade 8 in a science exam.
Independent measures are those where the values of the respective datasets are independent
of each other. On the contrary, dependent measures are those where the respective values
tend to be dependent on one another.
The descriptive statistics that would be helpful are mean and standard deviation since these
are used to determine the test statistic and hence useful for hypothesis testing. Descriptive
statistics normally indicate the key characteristics of the sample data. The t test and ANOVA
tend to present information about the population characteristics which descriptive statistics do
not provide.
Calculating ANOVAs
Problem 1
Null Hypothesis: μSingle = μTwin = μTriple
Alternative Hypothesis: Atleast one of the above means is different from the other two
Level of significance = 0.05
The output obtained from Excel is pasted below.

Since p value (0.018) < Level of significance, hence null hypothesis rejected. This implies
that the difference between children is statistically significant.
Problem 2
Null Hypothesis: μ5months = μ20months = μ35months
Alternative Hypothesis: Atleast one of the above means is different from the other two
Level of significance = 0.05
The output obtained from Excel is pasted below.

Since p value (0.000) < Level of significance, hence null hypothesis rejected. This implies
that the difference between time to get home for children of different ages is statistically
significant.