Question

A GROUP OF 400 CAR BUYERS WAS ANALYZED TO DETERMINE WHETHER THEY INTEND TO PARTICIPATE IN THE TRADE-IN OFFER. THE DEALERS WANT TO FIND OUT WHETHER THE PROPORTION OF PARTICIPANTS IN THE OFFER IS AT LEAST 20%. IT TURNED OUT THAT X = 94 EXPRESSED THEIR WILLINGNESS TO TRADE THEIR CARS IN.

AT THE 5% SIGNIFICANCE LEVEL, DO YOU HAVE SUFFICIENT EVIDENCE THAT THE POPULATION PROPORTION WOULD BE ABOVE 20%?

[A] NULL HYPOTHESIS STATES:

ALTERNATIVE HYPOTHESIS STATES:

[B] SHOW THE TEST STATISTIC VALUE AND THE CRITICAL VALUE(S) NEEDED FOR YOUR DECISION

TEST STATISTIC =

CRITICAL VALUE =

[C] FORMULATE THE REJECTION RULE AND DECISION.

REJECTION RULE STATES: REJECT THE NULL HYPOTHESIS IF <FILL THE BLANK SPACE>

DECISION: WE (REJECT) (DO NOT REJECT) THE NULL HYPOTHESIS  (CIRCLE ONE ANSWER)

Answer 1

(A)

NULL HYPOTHESIS STATES: Here proportion of car buyers they intend to participate in the trade - in offer is less than or equal to 20%. p <= 0.20

ALTERNATIVE HYPOTHESIS STATES: Here proportion of car buyers they intend to participate in the trade - in offer is greater than to 20%. p > 0.20

[B] Sample proportion = p^ = 94/400 = 0.235

standard error of proportion = sqrt(p * (1-p)/n) = sqrt [0.2 * 0.8/400] = 0.02

Test statstic

Z = (p^ - p₀)/se₀ = (0.235 - 0.2)/0.02 = 1.75

Here critical value for 5% significance level

Z_critical = 1.645

Here rejection rule is Z > 1.645 if we reject the null hypothesis.

[C] Here as we see that Z > Z_critical , reject the null hypothesis and our deciston is that we reject the null hypothesis.