You survey a sample of 100 cars drawn from a population with unknown mean price
ID: 3130042 • Letter: Y
Question
You survey a sample of 100 cars drawn from a population with unknown mean price of and standard deviation of = $5,200. Those 100 cars have an average value of $21,345. (a)[6] At the level of significance = 0.02, test H: $22,256 versus H: < $22,256. (b)[1] Justify the test statistic you use in part (a). (c)[2] What is the lowest level of significance to reject H in part (a)? (d)[1] Based on part (c), give one example of that helps reject H in part (a). Hint: Use the sample size to answer part (b). Round off probabilities and zvalues to 5 decimal places.
Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u >= 22256
Ha: u < 22256
As we can see, this is a left tailed test.
Thus, getting the critical z, as alpha = 0.02 ,
alpha = 0.02
zcrit = - 2.053748911
Getting the test statistic, as
X = sample mean = 21345
uo = hypothesized mean = 22256
n = sample size = 100
s = standard deviation = 5200
Thus, z = (X - uo) * sqrt(n) / s = -1.751923077
Also, the p value is
p = 0.039893518
As |z| < 2.054, and P > 0.02, we FAIL TO REJECT THE NULL HYPOTHESIS.
Hence, there is no significant evidence at 0.02 level that the true mean is less than $22256. [CONCLUSION]
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b)
As n > 30, we can approximate the t distribution as a z distribution.
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c)
The lowest level of significant we can reject Ho is the P value,
P = 0.039893518 [ANSWER]
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d)
Hence, we can reject Ho at say, 0.04 level, as 0.04 > 0.03989. [ANSWER]
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