Question

A recent estimate by a large distributor of gasoline claims that 60% of all cars stopping at their service stations chose unleaded gas and that super unleaded and regular were each selected 20% of the time. In order to check the validity of these proportions, a study was conducted of cars stopping at the distributor's service stations in a large city. The results were as follows:

Gasoline Selected
Regular Unleaded Super Unleaded
51 261 88
Carry out a significance test of the distributors claim

Answer 1

We reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.

Explanation:

There are 3 types of gas listed in the question.

Thus;

n = 3

DF = n - 1

DF = 3 - 1

DF = 2

Let's state the hypotheses;

Null hypothesis; H0: P_regular = P_super unleaded = 20%; P_i leaded = 60%

Alternative hypothesis; Ha: At least 2 proportions differ from the stated value.

Observed values are;

Regular gas; O = 51

Unleaded gas; O = 261

Super Unleaded; O = 88

Total observed values = 51 + 261 + 88 = 400

We are told that super unleaded and regular were each selected 20% of the time and that unleaded gas was chosen 60% of the time.

Thus, expected values are;

Regular gas; E = 20% × 400 = 80

Unleaded gas; E = 60% × 400 = 240

Super Unleaded; E = 20% × 400 = 80

Formula for chi Square goodness of fit is;

X² = Σ[(O - E)²/E]

X² = (51 - 80)²/80) + (261 - 240)²/240) + (88 - 80)²/80)

X² = 13.15

From the chi Square distribution table attached and using; DF = 2 and X² = 13.15, we can trace the p-value to be approximately 0.001

Also from online p-value from chi Square calculator attached, we have p to be approximately 0.001 which is similar to what we got from the table.

Now, if we take the significance level to be 0.05, it means the p-value is less than it and thus we reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.