Question

Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean of 1085 and a standard deviation of 52. Grade 2 has 15 observations yielding a sample mean of 1034 and a standard deviation of 61. Assume the populations to be approximately normal with equal variances. State and conclude your hypothesis at the 0.05 level of significance if grade 1 and grade 2 true means are equal to each other.

Answer 1

Answer:

$t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167$

$p_v =2*P(t_{23}$

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.

Step-by-step explanation:

Data given

$n_1 =10$ represent the sample size for group 1

$n_2 =15$ represent the sample size for group 2

$\bar X_1 =1085$ represent the sample mean for the group 1

$\bar X_2 =1034$ represent the sample mean for the group 2

$s_1=52$ represent the sample standard deviation for group 1

$s_2=61$ represent the sample standard deviation for group 2

We are assuming that the two samples are normally distributed with equal variances and that means:

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

System of hypothesis

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

The statistic is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The degrees of freedom are given by:

$n_1+n_2 -2$

And the pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

Replacing we got:

$S^2_p =\frac{(10-1)(52)^2 +(15 -1)(61)^2}{10 +15 -2}=3323.043$

And the deviation would be:

$S_p=57.646$

The degrees of freedom are:

$df=10+15-2=23$

The statistic would be:

$t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167$

The p value would be

$p_v =2*P(t_{23}$

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.