Question

Tensile-strength tests were carried out on two different grades of wire rod (Fluidized Bed Patenting of Wire Rods, Wire J., June 1977: 5661), resulting in the accompanying data. Grade Sample Size Sample Mean (kg/mm2 ) Sample SD AISI 1064 129 107.6 1.3 AISI 1078 129 123.6 2.0 Does the data provide compelling evidence for concluding that true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2? Conduct a hypothesis test using a significance of α = 0.05.

Answer 1

Answer:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

Step-by-step explanation:

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2 +10$

Alternative hypothesis: $\mu_1 >\mu_2 +10$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 10$

Alternative hypothesis: $\mu_1 -\mu_2>10$

Our notation on this case :

$n_1 =129$ represent the sample size for group AISI 1078

$n_2 =129$ represent the sample size for group AISI 1064

$\bar X_1 =123.6$ represent the sample mean for the group AISI 1078

$\bar X_2 =107.6$ represent the sample mean for the group AISI 1064

$s_1=2.0$ represent the sample standard deviation for group 1 AISI 1078

$s_2=1.3$ represent the sample standard deviation for group AISI 1064

And now we can calculate the statistic:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

Now we can calculate the degrees of freedom given by:

$df=129+129-2=256$

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.