# Solved-6.31 A manufacturer claims that the average tensile strength of thread A exceeds the average tensile strength of thread

#### ByDr. Raju Chaudhari

Oct 6, 2020

6.13 A manufacturer claims that the average tensile strength of thread A exceeds the average tensile strength of thread B by at least 12 kilograms. To test this claim, 50 pieces of each type of thread were tested under similar conditions. Type A thread had an aver- age tensile strength of 86.7 kilograms with a standard deviation of 6.28 kilograms, while type B thread had an average tensile strength of 77.8 kilograms with a standard deviation of 5.61 kilograms. Test the manufacturer's claim using a 0.05 level of significance.

## Solution

Given that the sample size $n_1 = 50$, $n_2 = 50$, sample mean $\overline{x}_1= 86.7$,
$\overline{x}_2= 77.8$, standard deviation $\sigma_1 = 6.28$ and $\sigma_2 = 5.61$.

State the hypothesis testing problem

The hypothesis testing problem is

$H_0 : \mu_1 - \mu_2=12$ against $H_1 : \mu_1 -\mu_2 < 12$ ($\textit{left-tailed}$)

Define test statistic The test statistic is

 \begin{aligned} Z&=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 -\mu_2)-d0}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}. \end{aligned}

The test statistic $Z$ follows standard normal distribution $N(0,1)$.

Specify the level of significance The significance level is $\alpha = 0.05$.

Determine the critical value

As the alternative hypothesis is $\textit{left-tailed}$, the critical value of $Z$ $\text{is}$ $\text{-1.64}$.

The rejection region (i.e. critical region) is $\text{Z < -1.64}$.

Computation

The test statistic for testing above hypothesis under the null hypothesis is

 \begin{aligned} Z_{obs}&=\frac{(\overline{x}_1 -\overline{x}_1)-d0}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\\ &= \frac{86.7-77.8-12}{\sqrt{\frac{6.28^2}{50}+\frac{5.61^2}{50}}}\\ &= -2.603 \end{aligned}

Decision

The rejection region (i.e. critical region) is $\text{Z < -1.64}$. The test statistic is $Z_{obs} =-2.603$ which falls $inside$ the critical region, we $\textit{reject}$ the null hypothesis.