The mean price for used cars is $10,388. A manager of a Kansas City used car...

The mean price for used cars is $10,388. A manager of a Kansas City used car dealership reviewed a sample of 50 recent used car sales at the dealership in an attempt to determine whether the population mean price for used cars at this particular dealership differed from the national mean. The prices for the sample of 50 cars are contained in the Excel Online file below. Construct a spreadsheet to answer the following questions.

a. Formulate the hypotheses that can be used to determine whether a difference exists in the mean price for used cars at the dealership.

_________<>≤≥=≠
_________<>≤≥=≠

b. What are the test statistic and the -value? Do not round intermediate calculations.

	(to 3 decimals)
-value	(to 4 decimals)

c. At , what is your conclusion?

_________RejectFail to reject the null hypothesis. We have _________insufficientsufficient evidence to conclude that the population mean price at this dealership differs from the national mean price of $10,388.

Sale Price
7578
7906
6681
7582
8961
6341
9960
7651
11516
11561
12434
7403
9519
9233
8371
10637
6517
8497
13000
11777
6340
11165
11495
13010
6708
7387
12805
9800
10479
7422
9384
8591
9989
9603
13181
11862
7767
8958
7402
12143
7115
11482
12807
8663
9603
10203
9628
9776
12420
7246

Sample Size 50

Hypothesized Mean 10388

Level of Significance 0.02

Solutions

Expert Solution

	Values ( X )	Σ ( X_i- X̅ )²
	7578	3893439.312
	7906	2706617.232
	6681	8237933.232
	7582	3877669.872
	8961	348312.4324
	6341	10305255.63
	9960	167133.7924
	7651	3610684.032
	11516	3860517.632
	11561	4039376.432
	12434	8310651.152
	7403	4614677.312
	9519.0	1035.5524
	9233	101238.5124
	8371	1392824.832
	10637	1179005.072
	6517	9206248.272
	8497	1111295.472
	13000	11894359.39
	11777	4954274.672
	6340	10311676.99
	11165	2604414.992
	11495	3778436.192
	13010	11963435.79
	6708	8083672.512
	7387	4683675.072
	12805	10587344.59
	9800	61911.3924
	10479	860849.9524
	7422	4533407.472
	9384	27949.1524
	8591	921945.6324
	9989	191686.3524
	9603	2685.3124
	13181	13175593.23
	11862	5339889.072
	7767	3183298.272
	8958	351862.5124
	7402	4618974.672
	12143	6717530.912
	7115	5934972.992
	11482	3728065.872
	12807	10600363.87
	8663	788863.7124
	9603	2685.3124
	10203	424869.3124
	9628	5901.3124
	9776	50544.0324
	12420	8230128.192
	7246	5313854.832
Total	477559	210893039.4

Mean X̅ = Σ Xi / n
X̅ = 477559 / 50 = 9551.18
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 210893039.38 / 50 -1 ) = 2074.5938

To Test :-
H0 :- µ = 10388
H1 :- µ ≠ 10388

Part b)

Test Statistic :-
t = ( X̅ - µ ) / ( S / √(n))
t = ( 9551.18 - 10388 ) / ( 2074.5938 / √(50) )
t = -2.8522 ≈ - 2.852

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, n-1)
Critical value t(α/2, n-1) = t(0.02 /2, 50-1) = 2.405
| t | > t(α/2, n-1) = 2.8522 > 2.405
Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 2.8522 ) = 0.0063
Reject null hypothesis if P value < α = 0.02 level of significance
P - value = 0.0063 < 0.02 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

Part c)

Reject the null hypothesis. We have sufficient evidence to conclude that the population mean price at this dealership differs from the national mean price of $10,388.

orchestra answered 1 year ago

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