Calculate the test statistic Calculate the p value What is the conclusion for this hypothesis test? What is the fundamental error with this analysis?
Calculate the test statistic
Calculate the p value
Let us first test whether “the different days of the week have the same frequencies of police calls”, using a Chi-square test of Goodness of fit.
Hypotheses:
The null hypothesis is:
H0 : the different days of the week have the same frequencies of police calls.
The alternative hypothesis is:
H1 : the different days of the week does not have the same frequencies of police calls.
Consider level of significance as 0.01.
Calculation steps:
The calculations have been done in EXCEL.
Denote Oi as the observed frequency (i =1, 2…7) and Ei as the expected frequency(i =1, 2…7). Here, the sample size, n=7.
The expected number of police calls for the day of the week is Ei=∑Oi/n. therefore, the value of the expected number of police calls will be (1,450)/7≈207.143.
Test statistic:
The formula for the test statistic is χ2 = ∑ [(Oi – Ei)2 / Ei], summed over all i .
The Table calculates [(Oi – Ei)2 / Ei]for each (i). So, the value in the first cell will be (157– 207.143)2 /207.143 ≈ 1.6485.
The test statistic value can be calculated by adding all these cell values.
|
Day |
Observed Frequency, Oi |
Expected frequency, Ei |
|
|
Sun |
157 |
207.143 |
12.1381 |
|
Mon |
201 |
207.143 |
0.1822 |
|
Tue |
225 |
207.143 |
1.5394 |
|
Wed |
249 |
207.143 |
8.4580 |
|
Thurs |
174 |
207.143 |
5.3029 |
|
Fri |
205 |
207.143 |
0.0222 |
|
Sat |
239 |
207.143 |
4.8994 |
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