1. A company claims that the mean tread life for one of its tyres is 35,000 hour
ID: 2925030 • Letter: 1
Question
1. A company claims that the mean tread life for one of its tyres is 35,000 hours. To test this clai,50 tyres are randomly selected and tested. The sample mean for the tread life was 34,200 hours and the sample standard deviation was 3,100. For denoting the mean tread life, the hypotheses to be tested are: Ho : = 35,000 versus H1 : 35,000. (a) Using the fact that 490 975 2010, calculate a 95% confidence interval for the mean tread life assuming that tread life is normally distributed b) Does your interval provide evidence that the company's claim is false? That is, do you reject Ho? Explain. c) Write a simple statement that summarises your findingsExplanation / Answer
Sample size, n = 50
Sample Mean, x = 34200
Standard Deviation of the sample, stdev = 3100
Since, our sample size is 50, the degrees of freedom, df = n-1 = 49
Also, the confidence interval is 0.95. Subtract it from 1 and this gives the value of alpha = (1-0.95) = 0.05
The t value for alpha = 0.05 with 49 degrees of freedom is 2.0096.
Let's keep this value aside and calculate the Standard error for our sample.
SE = Standard Deviation / Square root(sample size)
= 3100 / SQRT(50)
= 3100 / 7.071
= 438.406
Now, assuming normal distribution, the sample mean should lie between Sample Mean + (T value * SE) and Sample Mean - (T value * SE)
a) For part a, the confidence interval would be:
upper limit = 34200 + (2.0096 *438.406) = 35095.23
lower limit = 34200 - (2.0096 *438.406) = 33304.77
b) No, since the claimed mean value 35000, lies between the 95% confidence interval, we cannot reject null hypothesis(H0)
c) Based on our calculations, within 95% confidence interval we can confirm that the mean tread life for one of the company tyres is 35000 hours and hence, company's claim is true.
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