The shares of the U. S. automobile market held in 1990 by General Motors, Japanese manufacturers, Ford, Chrysler, and other manufacturers were, respectively, 36%, 26%, 21%, 9%, and 8%. Suppose that a new survey of 1,000 new-car buyers shows the following purchase frequencies:

GM Japanese Ford Chrysler Other
193 384 170 90 163

(a) Show that it is appropriate to carry out a chi-square test using these data.

(b) Determine whether the current market shares differ from those of 1990. Use α = .05.

Step-by-step explanation:

Hello!

The objective is to test if the purchase frequencies of new-car buyers follow the distribution of the shares of the U.S. automobile market for 1990.

You have one variable of interest:

X: Brand a new-car buyer prefers, categorized: GM, Japanese, Ford, Chrysler and Other

n= 1000

Observed frequencies

GM 193

Japanese 384

Ford 170

Chrysler 90

Other 163

a) The test to use to analyze if the observed purchase frequencies follow the market distribution you have to conduct a Goodness to Fit Chi-Square test.

The conditions for this test are:

- Independent observations

In this case, we will assume that each buyer surveyed is independent of the others.

- For 3+ categories: each expected frequency (Ei) must be at least 1 and at most 20% of the Ei are allowed to be less than 5.

In our case we have a total of 5 categories, 20% of 5 is 1, only one expected frequency is allowed to have a value less than 5.

I'll check this by calculating all expected frequencies using the formula: Ei= n*Pi (Pi= theoretical proportion that corresponds to the i-category)

E(GM)= n*P(GM)= 1000*0.36= 360

E(Jap)= n*P(Jap)= 1000*0.26= 260

E(Ford)= n*P(Ford)= 1000*0.21= 210

E(Chrys)= n*P(Chrys)= 1000*0.09= 90

E(Other)= n*P(Other)= 1000*0.08= 80

Note: If all calculations are done correctly then ∑Ei=n.

This is a quick way to check if the calculations are done correctly.

As you can see all conditions for the test are met.

b) The hypotheses for this test are:

H₀: P(GM)= 0.36; P(Jap)= 0.26; P(Ford)= 0.21; P(Chrys)= 0.09; P(Other)= 0.08

H₁: At least one of the expected frequencies is different from the observed ones.

α: 0.05

k= number of categories of the variable.

This test is one-tailed right this mean you'll reject the null hypothesis to high values of X²

Decision rule using the critical value approach:

If ≥ 9.488, reject the null hypothesis

If < 9.488, don't reject the null hypothesis

The value of the statistic under the null hypothesis is greater than the critical value, so the decision is to reject the null hypothesis.

Using a 5% level of significance, there is significant evidence to conclude that the current market greatly differs from the preference distribution of 1990.

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation: