95% ci estimate

Question

The U.S. Dept of Transportation requires tire manufactures to provide tire performance information on the sidewall of the tire so that a prospective customer can be better informed when making a purchase decision. One very important measure of tire performance is the tread wear index, which indicate the tire’s resistance to tread wear compared to a tire with a tire graded with the base of 100. Suppose that a consumer organization wants to estimate the actual tread wear index of a brand name of tires graded 200 that are produced by a certain manufacturer. A random sample of 18 of these tires indicates a sample mean tread wear index of 195.3 and a sample std deviation of 21.4.
a. Assuming that the population of tread wear indices is normally distributed, set up a 95% CI estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name.
b. Do you think that the consumer organization should accuse the manufacturer of producing tires that do not meet the performance information provided on sidewall of the tire? Explain.
c. Tell why an observed tread wear index of 210 for particular tire is not unusual, even though it is outside the CI developed in (a)

Answer 1

H_o: μ = 200

H_A: μ ≠ 200

 

n = 18

x-bar = 195.3

s = 21.4

df = 17

 

a.

Standard Error (SE) = s/√n = 21.4/√18 = 5.044

t_{α/2, df=17} = 2.11 (from a t-table)

Margin of Error (ME) = ±(tα/2, df=17 × SE) = ±(2.11 × 5.044) = ±10.64

Lower Bound of 95% CI = x-bar – ME = 195.3 – 10.64 = 184.66

Upper Bound of 95% CI = x-bar + ME = 195.3 + 10.64 = 205.94

 

 

b.

No, there is no grounds to allegate wrong-doing because the 95% confidence interval contains 200, which means that the sample mean is not statistically significantly different from the claimed tread wear index.

 

c.

Comparing means by z and t distributions is an endeavor to compute probabilities; these methods cannot be legitimately used to determine whether a single observation is within or outside of expected limits.