Question

Robert recorded the number of calls he made at work during the week:

Day	Calls
Monday	20
Tuesday	12
Wednesday	10
Thursday	18

He expected to make 15 calls each day. To determine whether the number of calls follows a uniform distribution, a chi-square test for goodness of fit should be performed (alpha = 0.05).



Using the data above, what is the chi-square test statistic? Answer choices are rounded to the hundredths place.

• a.)

  0.67

• b.)

  0.42

• c.)

  4.54

• d.)

  3.75

SUBMIT MY ANSWER

Answer 1

Answer

The value of Chi-Squared statistic = 4.54

hence, option c) is correct.

Solution:

The following table is obtained:

Days	Calls (Observed)	Expected	(fo-fe)2/fe
Mon	20	60*0.25=15	(20-15)2/15 = 1.67
Tue	12	60*0.25=15	(12-15)2/15 = 0.6
Wed	10	60*0.25=15	(10-15)2/15 = 1.67
Thurs	18	60*0.25=15	(18-15)2/15 = 0.6
$\sum=$	60	60	4.54

where,

fo = observed frrequency = Oi

fe = expected frequency = Ei

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ha​: Some of the population proportions differ from the values stated in the null hypothesis

This corresponds to a Chi-Square test for Goodness of Fit.

Rejection Region

Based on the information provided, the significance level is

α=0.05,

the number of degrees of freedom is df = 4 - 1 = 3

so the rejection region for this test is

R = { χ2 : χ2 > 7.815}

Test Statistics

The Chi-Squared statistic is,

x =Σ (Ο, – Ε)2 Ei = 1.67 + 0.6 + 1.67 + 0.6 = 4.54 1=1

Decision about the null hypothesis

Since it is observed that

$\chi^2 = 4.533 \leq \chi_c^2 = 7.815$

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 some of the population proportions differ from those stated in the null hypothesis, at α=0.05 significance level.