Question

In: Statistics and Probability

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 ( Xi- 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 SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 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 2 years ago
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