In: Statistics and Probability
A researcher whether introverts are more likely to prefer online
classes than extraverts. They decide to test this theory by asking
a sample of n = 80 people (a) whether they are an
introvert or an extravert, and (b) whether they prefer online or
InPerson classes.
Below is the data from the study:
Online | InPerson | |
Introvert | ƒo = 8 | ƒo = 19 |
Extravert | ƒo = 23 | ƒo = 30 |
(a) Compute df and determine the critical value for χ2 for a hypothesis test with α = 0.01.
df =
Critical value of χ2 =
(b) Determine the expected frequencies (ƒe) for each of
the cells in the table: (Use 3 decimals)
Online | No Online | |
Introvert | ƒe = | ƒe = |
No Introvert | ƒe = | ƒe = |
(c) Calculate the chi-square statistic
(χ2): (Use 3 decimals)
χ2 =
(d) Write the results as you would write it for the research
literature.
Online |
No Online |
Total |
|
Introvert |
8 |
19 |
27 |
No Introvert |
23 |
30 |
53 |
Total |
31 |
49 |
80 |
The expected values are computed in terms of row and column totals. In fact, the formula is
, where R_i corresponds to the total sum of elements in row i, C_j corresponds to the total sum of elements in column j, and T is the grand total.
The table below shows the calculations to obtain the table with expected values:
Expected Values |
Online |
No Online |
Total |
Introvert |
27 |
||
No Introvert |
53 |
||
Total |
31 |
49 |
80 |
Based on the observed and expected values, the squared distances can be computed according to the following formula: (E - O)^2/E.
The table with squared distances is shown below:
Squared Distances |
Online |
No Online |
Introvert |
||
No Introvert |
Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
H_0: The two variables are independent
H_a: The two variables are dependent
This corresponds to a Chi-Square test of independence.
Rejection Region
Based on the information provided, the significance level is α=0.01 ,
the number of degrees of freedom is df = (2−1)×(2−1)=1,
so then the rejection region for this test is
R={χ2:χ2>6.635}.
Test Statistics
The Chi-Squared statistic is computed as follows:
Decision about the null hypothesis
Since it is observed that χ2=1.428≤χc2=6.635,
it is then concluded that the null hypothesis is not rejected.
Conclusion
It is concluded that the null hypothesis Ho is not rejected.
Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.01 significance level.
The corresponding p-value for the test is
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