In: Math
Are phone calls equally likely to occur any day of the week? The day of the week for each of 392 randomly selected phone calls was observed. The results are displayed in the table below. Use an αα = 0.10 significance level.
Outcome | Frequency | Expected Frequency |
---|---|---|
Sunday | 55 | |
Monday | 54 | |
Tuesday | 64 | |
Wednesday | 50 | |
Thursday | 58 | |
Friday | 57 | |
Saturday | 54 |
Hint: Helpful Video [+]
Help
Solution:
Here, we have to use chi square test for goodness of fit.
Null hypothesis: H0: Phone calls are equally likely to occur any day of the week.
Alternative hypothesis: Ha: Phone calls are not equally likely to occur any day of the week.
We are given level of significance = α = 0.10
Test statistic formula is given as below:
Chi square = ∑[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
We are given
N = 7
Degrees of freedom = df = N – 1 = 7 – 1 = 6
α = 0.10
Critical value = 10.645
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
Day |
O |
E |
(O - E)^2/E |
Sunday |
55 |
56 |
0.017857143 |
Monday |
54 |
56 |
0.071428571 |
Tuesday |
64 |
56 |
1.142857143 |
Wednesday |
50 |
56 |
0.642857143 |
Thursday |
58 |
56 |
0.071428571 |
Friday |
57 |
56 |
0.017857143 |
Saturday |
54 |
56 |
0.071428571 |
Total |
392 |
392 |
2.035714286 |
Chi square = ∑[(O – E)^2/E] = 2.035714286
Chi square test statistic = 2.036
P-value = 0.9164
(By using Chi square table or excel)
P-value > α = 0.10
So, we do not reject the null hypothesis
There is sufficient evidence to conclude that Phone calls are equally likely to occur any day of the week.
What is the correct statistical test to use?
Answer: Goodness-of-Fit
What are the null and alternative hypotheses?
.
The degrees of freedom = df = N – 1 = 7 – 1 = 6
The test-statistic for this data = 2.036
The p-value for this sample = 0.9164
The p-value is greater than α.
Based on this, we should fail to reject the null
Thus, the final conclusion is