# Confusion about using the t table or z table for confidence intervals

1. Nov 16, 2013

### Bill Nye Tho

Here's the question:

The PCB concentration of a fis hcaught in Lake Michigan was measured by a technique that is known to result in an error of measurement that is normally distributed with a standard deviation of .08 ppm. Suppose the results of 10 independent measurements of this fish are:

11.2, 12.4, 10.8, 11.6, 12.5, 10.1, 11.0, 12.2, 12.4, 10.6

Give a 95 percent interval for the PCB level of this fish.

--------

So, I have a small sample size (10) and a normal distribution. I attempted to use the t-table for this solution but the solutions manual used the Z-distribution table. I can't figure out why.

2. Nov 16, 2013

### lurflurf

We have a known standard deviation. It should probably be .8 ppm typo? So the confidence interval is
$$\bar{x}-Z_{.95/2}\frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+Z_{.95/2}\frac{\sigma}{\sqrt{n}}$$

If the standard deviation is not know, we use the sample standard deviation s to estimate it and the Student's t-distribution

$$\bar{x}-t_{.95/2}\frac{s}{\sqrt{n}}<\mu<\bar{x}+t_{.95/2}\frac{s}{\sqrt{n}}$$

3. Nov 16, 2013

### Bill Nye Tho

That leads me to additional question. I'm given a data set of a sample of 18 students and their IQ scores. Why is the confidence interval for the mean IQ estimated with the Student's t-distribution. I can easily find the variance and the deviation from the data set.

What is the difference between finding the standard deviation myself and having it explicitly stated?

4. Nov 17, 2013

### Hornbein

You can't find the standard deviation yourself. You CAN estimate the standard deviation using your sample.

The distinction between population parameters and sample statistics is very important. You won't get anywhere in statistics until you grasp this.

5. Nov 17, 2013

### Bill Nye Tho

That just cleared up a whoooooole lot of this course for me.

Thanks for that!