91Ó°ÊÓ

Heat exchange happens at a lower rate in materials of low warm conductivity than in materials of high thermal conductivity. For instance, metals typically have high warm conductivity and are extremely efficient at conducting heat.

The measurements of the heat conductivity of $11$randomly selected pieces of a particular type of glass:

$1.111.071.111.071.121.081.081.181.181.181.12$

Calculating the null and alternative hypotheses,

$\begin{array}{r}{H}_0:\mathrm{μ}=1\\ H_a:\mathrm{μ}>1\end{array}$

We know the significance level is $0.05$

sample size n = $11$

standard deviation $s_x=0.043$and sample mean $\stackrel{-}{x}=1.11$

Using

localid="1652949987427" $t=\frac{\stackrel{-}{x}−\mathrm{μ}}{\frac{s}{\sqrt{n}}}=\frac{1.11-1}{{\displaystyle \frac{0.043}{\sqrt{11}}}}=8.527$

The p-value can be obtained using the t-distribution table at $t=8.527$and $df=11-1=10$for one-tailed test is less than 0.00001.

$0.00001<0.05$

Therefore the p-value is less than the significance level. There is sufficient evidence to reject the null hypothesis. Hence, there is convincing evidence that the conductivity of this type of glass is greater than one.