Each student in a class was classified according to the scale: 17 - 20 years Cat

Question

Each student in a class was classified according to the scale: 17 - 20 years Category 1; 21
- 24 years Category 2; 25 years & above Category 3.
The distribution of their final grade in Statistics according to the Age classification was
developed and processed using MINITAB and the following output was obtained:
MTB > ChiSquare ‘Grade A’ ‘Grade B’ ‘Grade C’ ‘Grade D’ ‘Grade E’ .
Chi-Square Test for Association: Grade, Age
Rows: Age Columns: Grade
A B C D E All
1 7 13 17 16 18 71
8.61 13.98 * 13.98 14.52
0.300 0.069 0.423 0.290 0.833

2 6 11 19 9 9 **
6.55 10.64 15.14 10.64 11.05
0.045 0.012 *** 0.252 0.379

3 3 2 1 1 0 7
0.85 1.38 1.96 1.38 1.43
5.456 0.280 0.472 0.104 1.432
All 16 26 37 26 27 132
Cell Contents: Count
Expected count
Contribution to Chi-square
Chi-Square = 11.333, DF = 8, P-Value = 0.1835
WARNING * 5 cells with expected counts less than 5.0

(a) State the null and alternative hypotheses for this investigation>
(b) Calculate the missing values marked *, ** and ***.
(c) State clearly any assumptions made in calculating the values in Part (b).
(d) Comment on the warning statement to the end of the output.
(e) Explain how you would adjust the output so as to eliminate the warning.
(f) State a conclusion from this investigation.

Explanation / Answer

(a) State the null and alternative hypotheses for this investigation.

Answer:

Null hypothesis: H0: The two categorical variables age category and grade are not associated.

Alternative hypothesis: Ha: The two categorical variables age category and grade are associated.

(b) Calculate the missing values marked *, ** and ***.

Answer:

First Missing value is the expected frequency. The formula for expected frequencies is given as below:

Expected Frequency = Row total * Column total / Grand total

First missing expected frequency = 37*71/132 = 19.9015

Second missing value is the row total for second row.

Second missing value = 6+11+19+9+9 = 54

Third missing value = (O – E)^2/E = (19 – 15.14)^2/15.14 = 0.98412

(c) State clearly any assumptions made in calculating the values in Part (b).

Answer:

We assume that all observations are independently distributed over both categorical variables.

(d) Comment on the warning statement to the end of the output.

Answer:

The warning statement says that there are five cells of expected frequencies for which value is less than 5.0. If there are more cells are with value less than five then there would be a significant effect on the validity of the chi square test.

(e) Explain how you would adjust the output so as to eliminate the warning.

Answer:

We would adjust the output so as to eliminate the warning by removing or ignoring the significant cells for which expected counts is less than 5.0.

(f) State a conclusion from this investigation.

Answer:

For the given chi square test, the p-value is given as 0.1835 which is greater than the given level of significance or alpha value 0.05, so we cannot reject the null hypothesis. So, we conclude that there is sufficient evidence that the two categorical variables age category and grade are not associated. (Or there is no any statistically significant association exists between the two categorical variables age category and grade.)

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