## Mean Claim

#### Problem

The contents of twelve boxes of matches were recorded as:

34, 31, 29, 35, 33, 30, 31, 28, 27, 35, 32, 31

On the box it stated, "Average contents 32 matches", but the sample mean can be shown to be about 31.3.

When the company received a complaint that the average contents is less than 32, they claimed that approximately one in five samples of size twelve would have a mean less than 31.3.

By using the sample data, test this claim and determine the minimum sample mean before the claim can be rejected at a 5% significance level.

#### Solution

Sample mean, $\bar{x} = \dfrac{\sum x}{n} = \dfrac{376}{12}$.
Sample variance, $s^2 = \dfrac{\sum x^2}{n} - \bar{x}^2 = \dfrac{11856}{12} - \left( \dfrac{376}{12} \right)^2 = \dfrac{56}{9}$.

We are told that the population mean, $\mu = 32$, but because we do not know the population variance we must use the sample variance. As $n \lt 30$, we will use the best estimator, $\hat{\sigma}^2 = \dfrac{n}{n-1} s^2 = \dfrac{12}{11} \dfrac{56}{9} = \dfrac{224}{33}$.

By the central limit theorem, the distribution of sample means will be approximately normal: $\bar{X} \sim N \left (\mu, \dfrac{\sigma^2}{n} \right)$. As $\dfrac{\hat{\sigma^2}}{n} = \dfrac{\frac{224}{33}}{12} = \dfrac{56}{99}$, we shall assume that $\bar{X} \approx \left (32,\dfrac{56}{99} \right )$.

If the sample mean is $\dfrac{376}{16} \approx 31.3$, then the standardised normal score, $z = \dfrac{\frac{376}{12} - 32}{\sqrt{\frac{56}{99}}} \approx -0.886$. Using normal probability distribution tables, $P \left (\bar{X} \lt \dfrac{376}{12} \right) = P(z \lt -0.89) = 0.18783 \approx 19$%, which agrees with the claim that approximately one in five such samples would have a mean less than 31.3.

Given the claim that the population mean is 32, let us establish the lower bound for the sample mean if 5% is considered to be statistically significant; from the tables we obtain a critical $z$-score of -1.6449.

As $z = \dfrac{\bar{X} - \mu}{\sqrt{\frac{\sigma^2}{n}}}$, $-1.6449 = \dfrac{m - 32}{\sqrt{\frac{56}{99}}} \implies m \approx 30.8$. In other words, the sample mean would have to be less than 30.8 before the result is considered statistically significant at a 5% level and the claim that the population mean is 32 could be rejected. Hence we would accept a sample mean of 31.3.

Technically the central limit theorem states that the distribution of sample means is approximately normal and that approximation improves as $n$ increases. For small samples, with $n \lt 30$, we should use the Student's $t$-distribution. With $n = 12$, the number of degrees of freedom are 11 and so the critical $t$-score for a 5% significance level is 1.796.

Solving $\dfrac{m - 32}{\sqrt{\frac{56}{99}}} = -1.796$ we get $m \approx 30.6$. So we would still draw the same conclusion that a sample mean of 31.3 is not statistically significant at a 5% level and we can accept the manufacturer's claim that the average contents is 32 matches.

Problem ID: 306 (20 Jan 2007)     Difficulty: 3 Star

Only Show Problem