A leasing firm claims that the mean number of miles driven annually, mu, in its
ID: 3437702 • Letter: A
Question
A leasing firm claims that the mean number of miles driven annually, mu, in its leased cars is less than 13380 miles. A random sample of 23 cars leased from this firm had a mean of 12086 annual miles driven. It is known that the population standard deviation of the number of miles driven in cars from this firm is 3140 miles. Assume that the population is normally distributed, Is there support for the firm?s claim at the 0.05 level of significance? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places, and round your responses as specified in the table. (If necessary, consult a list of formulas.)Explanation / Answer
USING T DISTRIBUTION
Formulating the null and alternative hypotheses,
Ho: u >= 13380
Ha: u < 13380 [ANSWERS, HYPOTHESES]
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As we can see, this is a 1 tailed T test, as n = 23 < 30 is too small to use z. [ANSWER, LEFT TAILED T TEST]
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Thus, getting the critical t,
df = 22
tcrit = -1.717144374
Getting the test statistic, as
X = sample mean = 12086
uo = hypothesized mean = 13380
n = sample size = 23
s = standard deviation = 3140
Thus, t = (X - uo) * sqrt(n) / s = -1.976371335 [ANSWER]
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df = 22
tcrit = -1.717144374 [ANSWER]
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Also, the p value is
p = 0.030391545
Comparing z and zcrit (or, p and significance level), we REJECT THE NULL HYPOTHESIS.
SO: YES, WE CAN SUPOORT THE CLAIM. [ANSWER]
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