Question

An ammunition producer claims his best product has an average lifespan of exactly 18 years. A skeptical product evaluator asks for evidence (data) that might be used to evaluate this claim. The product evaluator was provided data collected from a random sample of 50 people who used the product. Using the data, an average product lifespan of 15 years and a standard deviation of 8 years was calculated. Select the 90%, confidence interval for the true mean lifespan of this product.

Answer 1

Answer:

The 90% confidence interval for the true mean lifespan of this product is between 13.1 and 16.9 years.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 50 - 1 = 49

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.6766

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.6766\frac{8}{\sqrt{50}} = 1.9$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 15 - 1.9 = 13.1 years

The upper end of the interval is the sample mean added to M. So it is 15 + 1.9 = 16.9 years

The 90% confidence interval for the true mean lifespan of this product is between 13.1 and 16.9 years.