Question

A small lawnmower company produced 1,500 lawnmowers in 2008. In an effort to determine how maintenance-free these units were, the company decided to conduct a multiyear study of the 2008 lawnmowers. A sample of 200 owners of these lawnmowers was drawn randomly from company records and contacted. The owners were given an 800 number and asked to call the company when the first major repair was required for the lawnmowers. Owners who no longer used the lawnmower to cut their grass were disqualified. After many years, 183 of the owners had reported. The other 17 disqualified themselves. The average number of years until the first major repair was 3.3 for the 183 owners reporting. It is believed that the population standard deviation was 1.47 years. If the company wants to advertise an average number of years of repair-free lawn mowing for this lawnmower, what is the point estimate? Construct a 95% confidence interval for the average number of years until the first major repair.

Answer 1

Answer:

The 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

Step-by-step explanation:

The (1 - α)% confidence interval for the average using the finite correction factor is:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

The information provided is:

$N=1500\\n=183\\\sigma=1.47\\\bar x=3.3$

The critical value of z for 95% confidence level is,

z = 1.96

Compute the 95% confidence interval for the average number of years until the first major repair as follows:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

     $=3.3\pm 1.96\times\frac{1.47}{\sqrt{183}}\times\sqrt{\frac{1500-183}{1500-1}}\\\\=3.3\pm 0.19964\\\\=(3.10036, 3.49964)\\\\\approx (3.1, 3.5)$

Thus, the 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).