Question

A random sample of 160 car purchases are selected and categorized by age. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 45-65 group, and 12% for the group over 65. Find the P-value to test the claim that all ages have purchase rates proportional to their driving rates. Use α = 0.05

Age	under 26	26-45	45-65	over 65
purchases	66	39	25	30

please explain how to solve problem on the ti84 calculator. please also explain how you got the answer without the calculator.

ANSWERS

A) 0.01 < P < 0.025

B) P > 0.10

C) P < 0.005

D) 0.005 < P < 0.01

Answer 1

Answer: C) P < 0.005

Explanation:

By hand

The Chi-Square goodness of fit test is used here to test whether the sample are from same population or we can say whether the observed value of frequency are same as the expected frequency.

The Null hypothesis is defined as,

Null hypothesis: There is no significant difference between the observed frequency and the expected frequency.

The observed values are

Observed values
Age	Frequency
under 26	66
26-45	39
45-65	25
over 65	30
Total	160

The expected frequencies are,

Expected values
Age	Frequency
under 26
26-45
45-65
over 65

The Chi-Square statistic is obtained using the formula,

	Observed,	Expected,
under 26	66	28.8	37.2	1383.84	48.05
26-45	39	62.4	-23.4	547.56	8.775
45-65	25	49.6	-24.6	605.16	12.201
over 65	30	19.2	10.8	116.64	6.075
				Sum	75.1008

The P-value for the chi square is obtained from chi square distribution table for chi square = 75.1008 and degree of freedom = k - 1 = 4 - 1 = 3

Since,

at 5% significance level. it can be concluded that the null hypothesis is rejected.

By Ti-84 calculator

Step 1: Go to Stat > Edit then enter observed values in column L1 and expected values in column L3. The screenshot is shown below,

Step 2: Go to Stat > TESTS > GOF-Test

Step 3: Enter df = 3 then enter Calculate.

The result is obtained. The screenshot is shown below,

Result: P < 0.005