
a.
Find the expected frequencies for testing
a.

Answer to Problem 7CYU
The expected frequencies are,
Categories | Expected frequencies |
1 | 21 |
2 | 17.5 |
3 | 14 |
4 | 10.5 |
5 | 7 |
Explanation of Solution
Calculation:
The observed frequencies of five categories are given.
Expected frequencies:
The expected frequencies are defined as
Let
The total number of trial is obtained as,
Now, the probabilities specified by
Thus, the expected frequencies are,
Categories | Expected frequencies |
1 | |
2 | |
3 | |
4 | |
5 |
b.
Explain whether the chi-square test is appropriate or not.
b.

Answer to Problem 7CYU
The chi-square test is appropriate.
Explanation of Solution
It is given that one of the observed frequencies is less than 5.
It is known that when the null hypothesis
In part (a), it is found that all the expected frequencies corresponding to 5 categories are more than 5.
Hence, the chi-square test is appropriate.
c.
Find the value of
c.

Answer to Problem 7CYU
The value of
Explanation of Solution
Calculation:
Chi-Square statistic:
The chi-square statistic is obtained as
In the given question there are 5 categories.
The observed and expected frequencies are obtained as,
Categories | Observed frequencies | Expected frequencies |
1 | 25 | 21 |
2 | 14 | 17.5 |
3 | 23 | 14 |
4 | 6 | 10.5 |
5 | 2 | 7 |
Now,
Categories | Observed frequencies (O) | Expected frequencies (E) | ||
1 | 25 | 21 | 16 | 0.7619 |
2 | 14 | 17.5 | 12.25 | 0.7 |
3 | 23 | 14 | 81 | 5.7857 |
4 | 6 | 10.5 | 20.25 | 1.9285 |
5 | 2 | 7 | 25 | 3.57143 |
Total | 70 | 70 | 154.5 | 12.748 |
Thus, the value of
d.
Find the degrees of freedom.
d.

Answer to Problem 7CYU
The degrees of freedom is 4.
Explanation of Solution
It is known that under the null hypothesis
Here, there are 5 categories. Thus,
Hence, the degrees of freedom is
Thus, the degrees of freedom is 4.
e.
Find the critical value at
e.

Answer to Problem 7CYU
The critical value at
Explanation of Solution
Calculation:
From parts (c) and (d), it is found that the value of chi-square test statistic is 12.748 with degrees of freedom 4.
Critical value:
In a test of hypotheses, the critical value is the point by which one can reject or accept the null hypothesis.
Critical value:
Software procedure:
Step-by-step software procedure to obtain critical value using MINITAB software is as follows:
- Select Graph>Probability distribution plot > view probability
- Select Chi -Square under distribution.
- In Degrees of freedom, enter 4.
- Choose Probability Value and Right Tail for the region of the curve to shade.
- Enter the Probability value as 0.05 under shaded area.
- Select OK.
- Output using MINITAB software is given below:
Hence, the critical value at
f.
Explain whether the null hypothesis
f.

Answer to Problem 7CYU
The null hypothesis
Explanation of Solution
Interpretation:
The hypotheses are:
Null Hypothesis:
Alternate Hypothesis:
From parts (c) and (e), it is found that the value of chi-square test statistic is 12.748 with degrees of freedom 4 and the critical value is
Rejection rule:
If the
Conclusion:
Here, the
That is,
Thus, the decision is “reject the null hypothesis”.
Therefore, the null hypothesis
g.
Find the critical value at
g.

Answer to Problem 7CYU
The critical value at
Explanation of Solution
Calculation:
From part (c) and (d), it is found that the value of chi-square test statistic is 12.748 with degrees of freedom 4.
Critical value:
Software procedure:
Step-by-step software procedure to obtain critical value using MINITAB software is as follows:
- Select Graph>Probability distribution plot > view probability
- Select Chi -Square under distribution.
- In Degrees of freedom, enter 4.
- Choose Probability Value and Right Tail for the region of the curve to shade.
- Enter the Probability value as 0.01 under shaded area.
- Select OK.
- Output using MINITAB software is given below:
Hence, the critical value at
h.
Explain whether the null hypothesis
h.

Answer to Problem 7CYU
The null hypothesis
Explanation of Solution
Interpretation:
The hypotheses are:
Null Hypothesis:
Alternate Hypothesis:
From part (c) and (g), it is found that the value of chi-square test statistic is 12.748 with degrees of freedom 4 and the critical value is
Rejection rule:
If the
Conclusion:
Here, the
That is,
Thus, the decision is “fail to reject the null hypothesis”.
Therefore, the null hypothesis
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Chapter 10 Solutions
Essential Statistics
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