Question

According to an avid aquarist, the average number of fish in a $20$-gallon tank is $10$, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. She counts the number of fish in $15$other $20$-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data:$11;10;9;10;10;11;11;10;12;9;7;9;11;10;11$

Short answer

If we take $\alpha =0.05$, we can see that $p<\alpha$. This means that we reject null hypothesis.

There is evidence to conclude that the standard deviation is different than $2$.

Step-by-step solution

Given Information

The given data

$11,10,9,10,10,11,11,10,12,9,7,9,11,10,11$

There are $n=15$tanks. Using $R$, we can easily find the mean and standard deviation of the sample and the results are:

$s=1.222799,\mu =10.06667$

Explanation

We want to test these hypothesis:

$\begin{array}{lr}{H}_0:& \sigma =2\\ H_1:& \sigma \ne 2\end{array}$

Since $n=15$, the number of degrees of freedom is $n-1=14$.

In order to test the hypothesis, we are using ${\chi }^2$distribution with $14$degrees of freedom. The value of the test statistic is given by:

${\chi }^2=\frac{(n-1)s^2}{{\sigma }^2}=\frac{14·1.{22}^2}{2^2}=5.2094$

Final Answer

Using the applet for ${\chi }^2$distribution, we can find the $p$-value and the result is $2·0.01732=0.03464$(because this is two sided test):

Chi-Square Distribution

$X~{\chi }_{\left(df\right)}^2$

If we take $\alpha =0.05$, we can see that $p<\alpha$. This means that we reject the null hypothesis.