Question

An engineer measured the Brinell hardness of 25 pieces of ductile iron from her company, that were subcritically annealed. She believes that the alloys her company uses in iron, will ultimately make the iron stronger, and thus will have a higher Brinell hardness score. She tests 25 pieces of ductile iron from her company with the resulting data of Brinell hardness scores:

170 167 174 179 179 187 179 183 179
156 163 156 187 156 167 156 174 170
183 179 174 179 170 159 187

The engineer hypothesizes that the mean Brinell hardness score of such ductile iron pieces from her company will greater than known average Brinell score for iron which is 170. At the 5% level of significance, is there enough evidence to conclude this?

State the Claims which means to State the null and alternative hypotheses. Use correct math type. You may want to consider looking back at the symbols assignment from the beginning of class.

a. What is the sample mean?
b. What is the sample standard deviation?
c. Are the normality assumptions met?
d. Conclusion. State your conclusion. Either reject or fail to reject H0 , indicate why, then write the conclusion in terms of the problem.

Answer 1

Yes. The sample size is large enough

H0: μ=170

Ha: μ≠ 170

a) μ= 172.52

b) 10.31

c) yes, the data is distributed around both sides of the mean

d) H0: μ=170 is true

Explanation:

a) mean= sum of all sample values/25

(170+167+174+179+179+187+179+183+179 +156+163+156+187+156+167+156+174+170+ 183+ 179+ 174+ 179+ 170+ 159+ 187)/25 = 172.52

b) standard deviation= √(∑(x-μ)²/25)

Using this formula value of standard deviation was computed to be 10.31

c) Least value is 156, difference between least and mean is 16.52

Max value is 187, difference between max and mean is 14.48

A scatter plot shows that number of values higher than mean is 14 and number of values lower than mean is 11. So we can say that data is distributed around mean

d) Null hypthesis is accepted

Test statistic= (sample mean-expected mean)/standard deviation

=(172.52-170)/10.31

= 0.024

P(Z> 0.024)= (1-0.5080)= 0.492

p-value= 2× 0.492= 0.984

Using 5% or 0.05 significance level,

test statistic is greater than significance level so null hypothesis is accepted