Vehicles heading south on Hosea Kutâko Avenue may go straight, turn right or turn left at the Sam Nujoma Avenue intersection. The city traffic engineer believes that equal proportions of cars normally head in the three different directions: straight, turn right and turn left. One hundred-and-eighty vehicles were observed on one day, with the following results: Can we conclude that the traffic engineer is correct? Use the 0.10 level of significance.

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Solution::

Here, n 210 and k = 4
=

The hypothesis are given by,

The null hypothesis is given by,

Ho: Equal proportion of car in three direction

The alternative hypothesis is given by,

HA: Not equal proportion of car in three direction

Here we apply chi square goodness of fit test.

E = np = 60 Where n = 180 and P
3
OE(O_i-E)
Straight7260121442.4
Right Turn4560-152253.75
Left Turn6360390.15
Total18018003786.3

The value of test statistic is given by,

\chi^2=\sum_{i=1}^{4}\frac{(O_i-E_i)^2}{E_i}=\sum _{i=1}^{4}\frac{(O_i-np)^2}{np}

O: for observe frequency

E: for expected frequency

² = 6.3

So the value of test statistic is 6.3

The critical value is given by,

2
Xan-1 X6.10,3-1=X0.10.2= 4.61
2
=

[Critical value we obtain from using sowtware]

Decision about the null hypothesis

\chi^2>\chi^2_{\alpha,n-1}
i.e, 6.3 4.61
V

Here the value of test statistic is greater than critical value.

i.e, The value of test statistic fall under critical region.

So we reject the null hypothesis.

Conclusion::

There is no equal proportion of car in three direction at 10% level of significance.