The mean tread life of a manufacturer's best selling tire is known to be 22,560 miles....

The mean tread life of a manufacturer's best selling tire is known to be 22,560 miles. Because the distribution of individual tire life is positively skewed, it cannot be assumed to be normal. R&D has come up with a new rubber additive that they suspect will increase mean tire life without affecting variability of tire life. A sample of 36 tires with the new additive has a sample mean tread life of 24,470 miles; the sample based estimate of the standard deviation is 1200 miles. At .05 level of significance, test whether the additive is working. Clearly state your hypothesis and conclusion.

Expert Solution

Given that,
population mean(u)=22560
sample mean, x =24470
standard deviation, s =1200
number (n)=36
null, Ho: μ=22560
alternate, H1: μ>22560
level of significance, alpha = 0.05
from standard normal table,right tailed t alpha/2 =1.69
since our test is right-tailed
reject Ho, if to > 1.69
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =24470-22560/(1200/sqrt(36))
to =9.55
| to | =9.55
critical value
the value of |t alpha| with n-1 = 35 d.f is 1.69
we got |to| =9.55 & | t alpha | =1.69
make decision
hence value of | to | > | t alpha| and here we reject Ho
p-value :right tail - Ha : ( p > 9.55 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=22560
alternate, H1: μ>22560
test statistic: 9.55
critical value: 1.69
decision: reject Ho
p-value: 0
we have evidence that new rubber additive will increase mean tire life without affecting variability of tire life

orchestra answered 2 years ago

the tread of life of a particular brand of tire is a random variable best described...

the tread of life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 70000 miles and a standard deviation of 9300 miles. what warranty should the company use if they want 89.25% of the tires to outlast the warranty? round to the nearest integer. (show work please)

A manufacturer claims that the average tread life of a premium tire exceeds the average tread...

A manufacturer claims that the average tread life of a premium tire exceeds the average tread life of economy tire of the same size by at least 5000 miles. To test this claim, 50 pieces of each type of tires were tested under similar conditions. The premium tire had an average tread life of 43000 miles with a sample standard deviation of 2300, while economy tire had an average tread life of 37500 miles with a sample standard deviation of...

The mean life span of a brand name tire is 50,000 miles. Assume that the life...

The mean life span of a brand name tire is 50,000 miles. Assume that the life spans of the tires are normally distributed, and the population standard deviation is 800 miles. a. If you select one tire, what is the probability that its life span is less than 48,500 miles? b. If you select 100 tires, what is the probability that their mean life span is more than 50,200miles? c. What is the distribution of the sample mean in part...

A tire manufacturer produces tires that have a mean life of at least 25000 miles when...

A tire manufacturer produces tires that have a mean life of at least 25000 miles when the production process is working properly. The operations manager stops the production process if there is evidence that the mean tire life is below 25000 miles. The testable hypotheses in this situation are ?0:?=25000H0:μ=25000 vs ??:?<25000HA:μ<25000. 1. Identify the consequences of making a Type I error. A. The manager does not stop production when it is not necessary. B. The manager stops production when...

A tire manufacturer produces tires that have a mean life of at least 32500 miles when...

A tire manufacturer produces tires that have a mean life of at least 32500 miles when the production process is working properly. The operations manager stops the production process if there is evidence that the mean tire life is below 32500 miles. The testable hypotheses in this situation are ?0:?=32500H0:μ=32500 vs ??:?<32500HA:μ<32500. 1. Identify the consequences of making a Type I error. A. The manager stops production when it is necessary. B. The manager stops production when it is not...

A tire manufacturer produces tires that have a mean life of at least 30000 miles when...

A tire manufacturer produces tires that have a mean life of at least 30000 miles when the production process is working properly. The operations manager stops the production process if there is evidence that the mean tire life is below 30000 miles. The testable hypotheses in this situation are ?0:?=30000 H 0 : μ = 30000 vs ??:?<30000 H A : μ < 30000 . 1. Identify the consequences of making a Type I error. A. The manager does not...

A tire manufacturer believes that the tread life of its snow tires can be described by...

A tire manufacturer believes that the tread life of its snow tires can be described by Normal model with a mean of 32,000 miles and a standard deviation of 2500 miles. a). If you buy a set of these tires, would it be reasonable for you to hope that they'll last 40,000 miles? Explain. b). Approximately what fraction of these tires can be expected to last less that 30,000 miles? c). Approximately what fraction of these tires can be expected...

1. A tire company produces a tire that has an average life span of 500 miles...

1. A tire company produces a tire that has an average life span of 500 miles with a standard deviation of 250. The distribution of the life spans of the tires is normal. What is the probability that the tires lasts between 530 and 375 miles? (Round three decimal places) 2. A tire company produces a tire that has an average life span of 480 miles with a standard deviation of 30. The distribution of the life spans of the...

tire manufacturer claims that the life span of its tires is 60,000 miles. Assume the life spans of the tire are normally distributed.

tire manufacturer claims that the life span of its tires is 60,000 miles. Assume the life spans of the tire are normally distributed. You selected 25 tires at random and tested them. The mean life span of the sample is 58,800 miles. The tires had a sample standard deviation, s = 600. Use the .10 level of significance. Which distribution would be indicated? Explain why you chose that distribution. Construct a confidence interval for the mean using the above data....

1) Average tire life is 30,000 miles (μ = 30,000) and σ = 2,300 miles. What is the...

1) Average tire life is 30,000 miles (μ = 30,000) and σ = 2,300 miles. What is the probability that a tire will last 35,000 miles (or more)? 2) If μ = 130 and σ = 20 on an employment test and the company only hires from the top 75% of scores, what is the lowest score you can get and still get hired? 3) The SAT has μ = 500 and σ = 100. What score would put you in the top 15%? 4) For...

Question