A leasing firm claims that the mean number of miles driven annually, , in its leased cars is less than miles. A random sample of
1 answer:
Complete question is;
A leasing firm claims that the mean number of miles driven annually, u , in its leased cars is less than 12800 miles. A random sample of 50 cars leased from this firm had a mean of 12499 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. Is there support for the firm's claim at the 0.05 level of significance?
Perform a one-tailed test.
null hypothesis?
alternative hypothesis?
type of test statistic?
value of the test statistic?
the p-value?
Can we support the leasing firm’s claim that the mean number of miles driven annually is less than 12800 miles?
Answer:
A) Null hypothesis; μ ≥ 12800
Alternative hypothesis; μ < 12800
B) type of test statistic: z-score test
C) test statistic; z = -0.678
D) p - value = 0.2489
E) No, we can't support the leasing firms claim
Step-by-step explanation:
We are given;
Population mean; μ = 12800 miles
Sample mean; x¯ = 12499 miles
Sample size; n = 50
Population standard deviation; σ = 3140
Let's define the hypothesis;
Null hypothesis; μ ≥ 12800
Alternative hypothesis; μ < 12800
The alternative hypothesis is the claim.
Since sample size is greater than 30, we will use z-score test.
Formula for z-score;
z = (x¯ - μ)/(σ/√n)
z = (12499 - 12800)/(3140/√50)
z = -0.678
From online p-value from z-score calculator attached, using z = -0.678 with one tailed hypothesis and significance value of 0.05,we have;
p - value = P(z < -0.678) = 0.2489
The p-value is greater than the significance value, so we will fail to reject the null hypothesis and conclude that there is no sufficient information to support the leasing firm's claim
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