Question

The minimal Brinell hardness for a specific grade of ductile iron is 130. An engineer has a sample of 25 pieces of this type of iron that was subcritically annealed with the Brinell hardness values as given below. The engineer would like to know whether this annealing process results in the proper Brinell hardness on average.
135 149 132 142 124 130 122 128 120 128 127 123 136 141 130 139 134 135 130 141 149 137 137 140 148

State the appropriate hypotheses; conduct a hypothesis test using α = 0.05 utilizing the classical approach, confidence interval approach, or p-value approach; state the decision regarding the hypotheses; and make a conclusion.

Answer 1

Answer:

$t=\frac{134.28-130}{\frac{8.229}{\sqrt{25}}}=2.60$  

Critical value

$df=n-1=25-1=24$  

We need to find a critical value in the t distribution with df=29 who accumulates 0.025 of the area on each tail and we got:

$t_{\alpha/2} = \pm 2.06$

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance

P-value  

Since is a two-sided tailed test the p value would given by:  

$p_v =2*P(t_{(24)}>2.60)=0.0157$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis and we can say that the true mean is not equal to 130 the specification.  

Step-by-step explanation:

Data given and notation  

135 149 132 142 124 130 122 128 120 128 127 123 136 141 130 139 134 135 130 141 149 137 137 140 148

We can calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And the deviation with:

$s= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=134.28$ represent the sample mean  

$s=8.229$ represent the sample standard deviation  

$n=25$ sample size  

$\mu_o =130$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is equal to 130 :  

Null hypothesis:$\mu = 130$  

Alternative hypothesis:$\mu \neq 130$  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

$t=\frac{134.28-130}{\frac{8.229}{\sqrt{25}}}=2.60$  

Critical value

We need to calculate the degrees of freedom first given by:  

$df=n-1=25-1=24$  

We need to find a critical value in the t distribution with df=29 who accumulates 0.025 of the area on each tail and we got:

$t_{\alpha/2} = \pm 2.06$

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance

P-value  

Since is a two-sided tailed test the p value would given by:  

$p_v =2*P(t_{(24)}>2.60)=0.0157$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis and we can say that the true mean is not equal to 130 the specification.