# Defining the bias of an estimator T of a population mean µ

I'm trying to revise for an upcoming exam and have come across a past paper question for which I can't quite work out an answer. I can't find any reference in my lecturer's notes and can't quite figure out the best way to approach this problem.

Disclaimer: NOT Homework.

# Given Information

A supermarket sells bags of grapes which are priced according to an approximate weight of 500g. Each bag cannot weigh exactly 500g, and the supermarket admits to a standard deviation in the bag weights of approximatelv 5g around a mean of 500g.

The bag weights can be assumed to be normally distributed. To check that the claimed mean weight of 500g is not misleading, a consumer advocacy group took a sample of ten bags and weighed them.

The weights in grams of the bags were as follows:

500.2, 498.2, 486.3, 494, 502.9, 503.9, 487.9, 496.4, 483.7, 497.4

# Question

Define the bias of an estimator T of a population mean p, and show that the sample mean x̄ is an unbiased estimator for µ.

Any explanation or guidance would be greatly appreciated.

Thanks!

*EDIT: *

I eventually found a formula in the notes which seems like it might apply to this situation. However I am still unsure how to apply it in this situation :(

$$bias(T) = E[T|\theta] − \theta.$$

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So, you are saying that this is a past exam for a course you are taking and that nowhere in the notes the bias of an estimator T of a population mean p is defined? – Did May 13 '12 at 8:09
No, I'm saying that after a good few hours of looking in the notes and online for an appropriate method, I still wasn't clear about how to go about defining it and so I turned to here for help. No need to sound quite so patronising... I did eventually find a formula but am still unsure how to apply it i.e what my T or $\theta$ should be :( – Peter Hamilton May 13 '12 at 8:53
In other words, the answer to Define the bias of an estimator WAS in your notes (unsurprisingly). Now, you need to identify the estimator T in your situation (hint: T depends on the sample) and the parameter theta that this T estimates (hint: theta is a real number). – Did May 13 '12 at 9:16

It seems that your model is as follows: You observe weights $X_1,X_2,\dots,X_n$ i.i.d. from distribution $N(\mu,\sigma^2)$ where $n = 10$, $\sigma = 5g$ and $\mu$ is the unknown parameter (that is, $\mu$ is what statisticians usually call $\theta$). Your estimator is sample mean $T = \frac{1}{n} \sum_{i=1}^n X_i$. What you want to show is $$\mathbb{E}[T \mid \mu] = \mu.$$ Here $\mathbb{E}[T \mid \mu]$ just means that the expectation of $T$ is taken assuming that the underlying sample $X_1,\dots,X_n$ is coming from a probability distribution parametrized by $\mu$ (namely, $N(\mu,\sigma^2)$ in your case.)
Hint: What else you need to know is that the expectation is a linear operator, i.e. $\mathbb{E} [a X + Y] = a \,\mathbb{E}[X] + \mathbb{E}[Y]$.