A used-car lot manager wished to su The manager selected a random sample of ten

Question

A used-car lot manager wished to su The manager selected a random sample of ten owners who sold a car that was no more than five years oid and recorded the original asking price, as well as the final selling price. The data is summarized wished to know how successful car owners were in receiving their asking price in the table below ?dependent dada Owner ? Asking Price Selling Price 1350 645 289 11350 200 400 8500 a significance level of a .01 and the partial Excel output provided below, does it appear that ference between car owners asking and selling prices? Do a plete and appropriate hypothesis test using the critical value approach and the 5-step com procedure discussed in class. B. Using only the appropriate statistical table in your textbook, what is the most accurate statement you can make about the p-value of the hypothesis test you conducted in part A? C. Using the partial Excel output provided below, construct a 90% confidence interval estimate of the true mean difference in owners' asking and selling prices. Be certain that your interval estimate is in the correct final format, as discussed in class. Interpret the practical meaning of this interval estimate, in plain English 0.000 hypothesized value 9,385.000 mean Asking Price 9,133.000 mean Selling Price 252.000 mean difference (Asking Price - Selling Price) 420.60274211 std. dev. 133.00626552 std. error 10 n 2. p-value (two-tailed) confidence interval 90% lower confidence interval 90% upper 420-02 r D O.oS

Explanation / Answer

Solution:

Part (a)

Let X = asking price

      Y = selling price

And D = Y – X.

Then, D ~ N(µ, ?2) where ?2 is unknown.

Claim:

Car owners receive substantially less than their original asking price. i.e., µ > 0.

Hypotheses:

Null: H0: µ = µ0= 0 Vs Alternative: HA: µ > 0

Test Statistic:

t = (Dbar - µ0)/SE(Dbar) where

Dbar and s are respectively, sample average and sample standard deviation based on n observations on X and Y.

Calculations

t = (252 - 0)/133.0062

= 1.895

Distribution, Critical Value and p-value:

Under H0, t ~ tn - 1. Hence, for level of significance ?%, Critical Value = upper ?% point of

tn - 1 and p-value = P(tn - 1 > tcal).

From standard statistical tables, upper 1% (given ? 0.01) of t9 (given n = 10) = 2.821

Decision Criterion (Rejection Region):

Reject H0 if tcal > tcrit

Decision:

Since tcal < tcrit, H0 is accepted.

Conclusion:

There is not sufficient evidence to suggest that the claim is valid. i.e., the sample data do not suggest that car owners receive substantially less than their original asking price.

DONE

Part (b)

p-value = P(t9 > 1.895) = 0.0453 ANSWER

Part (c)

90% Confidence Interval for µ is: Dbar ± t9, 0.05.SE(Dbar), where t9, 0.05 = 1.833 (from standard statistical tables) is the upper 5% point of t-distribution with 9 degrees of freedom.

= 252 ± (1.833 x 133.0062)

= 252 ± 243.80

= (8.2, 495.8) ANSWER 1

The above implies that the true difference between the asking price and selling price could be anything from 8 to 496, for 90 percent of times. ANSWER 2

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