Question

You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $\backslash (1,249$ with a very narrow standard deviation of $\backslash )25$. You find a website that has a price comparison for the same computer at a series of stores as follows: $\backslash (1,299;\backslash )1,229.99;\backslash (1,193.08;\backslash )1,279;\backslash (1,224.95;\backslash )1,229.99;\backslash (1,269.95;\backslash )1,249$. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the $5\%$ significance level. As a potential buyer, what would be the practical conclusion from your analysis?

Short answer

Taking $\alpha =0.05$, we can see that $p>\alpha$. This means that we can not reject the null hypothesis.

There is not enough evidence that the variance is greater than ${25}^2=625$.

Step-by-step solution

Given Information

The data from the website is:

$1299,1229.99,1193.08,1279,1224.95,1229.99,1269.95,1249$

Using $R$, we can find the mean and standard deviation of the sample above. The results are:

$\mu =1246.87,s=34.29039$

Explanation

We want to test these hypothesis:

$\begin{array}{ll}{H}_0:& \sigma \le 25\\ H_1:& \sigma >25\end{array}$

Since the sample from the website has $n=8$values, the number of degrees of freedom is $n-1=7$.

We are using ${\chi }^2$distribution with $7$degrees of freedom. Test statistic is given by:

${\chi }^2=\frac{(n-1)s^2}{{\sigma }^2}=\frac{7·34.{29}^2}{{25}^2}=13.169$

Final Answer

Using the applet for ${\chi }^2$distribution, we can find the $p$-value and the result is $0.0681$:

Chi-Square Distribution

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

Taking $\alpha =0.05$, we can see that $p>\alpha$. This means that we can not reject the null hypothesis.

There is not enough evidence that the variance is greater than ${25}^2=625$.