How Do You Calculate a 95% Confidence Interval for Potato Weights?

[Gregg]

Homework Statement

3 (a) A sample of 50 washed baking potatoes was selected at random from a large batch.
The weights of the 50 potatoes were found to have a mean of 234 grams and a standard
deviation of 25.1 grams.
Construct a 95% confidence interval for the mean weight of potatoes in the batch.
(4 marks)

Homework Equations

$\bar{x} = \mu$

$s = \frac{\sigma}{\sqrt{n}}$


$z = \frac{x-\mu}{s}$

The Attempt at a Solution

The confidence interval 95% means p = 0.975

$z = \frac{x-\mu}{s}$


$\pm 1.96 = \frac{x-234}{\frac{25.1}{\sqrt{50}}}$


$\Rightarrow 227-241$

Why is the standard deviation

$s = \frac{25.1}{\sqrt{50}}$

and not just 25.1. The question states that the s.d. is 25.1!

 

[Random Variable]

$\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$ is approximately N(0,1) for large n (assuming that the original distribtuion is not skewed)

 

[Mark44]

Why is the standard deviation

$s = \frac{25.1}{\sqrt{50}}$

and not just 25.1. The question states that the s.d. is 25.1!

There are two things going on here. On the one hand there are the population standard deviation ($\sigma$ and sample standard deviation s. On the other is the standard deviation of the mean, which is defined as:
$$\sigma_{mean} = \frac{\sigma}{\sqrt{n}}$$

The wikipedia article here--http://en.wikipedia.org/wiki/Standard_deviation--talks about the st. dev. of the mean in the section titled Relationship between standard deviation and mean.

 

[Gregg]

There are two things going on here. On the one hand there are the population standard deviation ($\sigma$ and sample standard deviation s. On the other is the standard deviation of the mean, which is defined as:
$$\sigma_{mean} = \frac{\sigma}{\sqrt{n}}$$

The wikipedia article here--http://en.wikipedia.org/wiki/Standard_deviation--talks about the st. dev. of the mean in the section titled Relationship between standard deviation and mean.

That's what's confusing, the standard deviation of the mean is the same as the standard deviation of the sample divided by root n?

 

[Random Variable]

That's what's confusing, the standard deviation of the mean is the same as the standard deviation of the sample divided by root n?

It's approximately the same.

 

[rclakmal]

s =SIGMA/SQRT(N)

STANDS FOR STANDARD ERROR NOT FOR STANDARD DEVAITION!

 

[Gregg]

s =SIGMA/SQRT(N)

STANDS FOR STANDARD ERROR NOT FOR STANDARD DEVAITION!

The standard deviation of the sample mean is sometimes called the standard error.