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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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    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).2012-05-13

1 Answers 1

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It seems that most of the "given information" is auxiliary to the question. (I mean the details mentioned are not very relevant to the question.)

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. $\mathbb{E} [a X + Y] = a \,\mathbb{E}[X] + \mathbb{E}[Y]$.

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    So to show that the estimator is biased compute the expected value of the estimator and note that it differs from the parameter value. to calculate the bias just take the difference between the two.2012-05-17