Question

Find the 95% confidence interval for estimating the population mean μ

Answer 1

We first need to determine whether we are dealing with means or proportions in this problem. Since we are given the sample and population mean, we know that we are dealing with means.

Since we have one sample mean, this means we are creating a confidence interval for one sample (1 Samp T Int).

Normally we would check for conditions, but since this is not formulated as a "real-world scenario" type problem, it is hard to check for randomness and independence. Therefore, I will be excluding conditions from this answer.

Confidence Interval Formula

The formula for constructing a confidence interval for means is as follows:

• $\displaystyle \overline{x} \pm t^*\big{(}\frac{\sigma}{\sqrt{n} } \big{)}$

We are given these variables:

• $\overline{x}=50$
• $n=60$
• $\sigma=10$

Plug these values into the formula for the confidence interval:

• $\displaystyle 50\pm t^* \big{(}\frac{10}{\sqrt{60} } \big{)}$

Finding the Critical Value (t*)

In order to find t*, we can use this formula:

• $\displaystyle \frac{1-C}{2}=A$

Calculating the z-score associated with "A" will give us t*.

So, let's plug in the confidence interval 95% (.95) into the formula:

• $\displaystyle \frac{1-.95}{2}=.025$

Use your calculator or a t-table to find the z-score associated with this area under the curve.. you should get:

• $t^*=1.96$

Constructing Confidence Interval

Now, let's finish the confidence interval we created:

• $\displaystyle 50\pm 1.96 \big{(}\frac{10}{\sqrt{60} } \big{)}$

We can calculate the confidence interval, using this formula, to be:

• $\boxed{(47.4697, \ 52.5303)}$

Interpreting the Confidence Interval

We are 95% confident that the true population mean μ lies between 47.4697 and 52.5303.