Question

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. Of the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly consistent with the claim? Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) A.) State the distribution to use for the test. (Round your answers to two decimal places.) B.) What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.) C.) What is the p-value? (Round your answer to four decimal places.) D.) Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion. (i) Alpha (Enter an exact number as an integer, fraction, or decimal.) E.) Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to the nearest whole number.)

Answer 1

Given that,
population mean(u)=50000
sample mean, x =46500
standard deviation, s =9800
number (n)=28
null, Ho: μ=50000
alternate, H1: μ!=50000
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.052
since our test is two-tailed
reject Ho, if to < -2.052 OR if to > 2.052
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =46500-50000/(9800/sqrt(28))
to =-1.8898
| to | =1.8898
critical value
the value of |t α| with n-1 = 27 d.f is 2.052
we got |to| =1.8898 & | t α | =2.052
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.8898 ) = 0.0696
hence value of p0.05 < 0.0696,here we do not reject Ho
ANSWERS
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A.
null, Ho: μ=50000
alternate, H1: μ!=50000
B.
test statistic: -1.8898
critical value: -2.052 , 2.052
decision: do not reject Ho
C.
p-value: 0.0696
D.
we do not have enough evidence to support the claim that its deluxe tire averages at least 50,000 miles before it needs to be replaced.
E.
TRADITIONAL METHOD
given that,
sample mean, x =46500
standard deviation, s =9800
sample size, n =28
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 9800/ sqrt ( 28) )
= 1852.026
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 27 d.f is 2.052
margin of error = 2.052 * 1852.026
= 3800.357
III.
CI = x ± margin of error
confidence interval = [ 46500 ± 3800.357 ]
= [ 42699.643 , 50300.357 ]
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DIRECT METHOD
given that,
sample mean, x =46500
standard deviation, s =9800
sample size, n =28
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 27 d.f is 2.052
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 46500 ± t a/2 ( 9800/ Sqrt ( 28) ]
= [ 46500-(2.052 * 1852.026) , 46500+(2.052 * 1852.026) ]
= [ 42699.643 , 50300.357 ]
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interpretations:
1) we are 95% sure that the interval [ 42699.643 , 50300.357 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
95% sure that the interval [ 42700,50300]