The heat conductivity of a particular type of glass: 1.11 1.07 1.11 1.07 1.12 1.
ID: 3157838 • Letter: T
Question
The heat conductivity of a particular type of glass:
1.11 1.07 1.11 1.07 1.12 1.08 1.08 1.15 1.18 1.18 1.12
a.do the data give convincing evidence that the mean heat conductivity of this particular type of glass is less than 1.15 at a 0.05 significance level?
i.State the null and alternative hypotheses.
ii.State the significance level for this problem.
iii.State the test statistic. Use MINITAB to compute the test statistic and P-value. (Copy and paste the results found in the Session window after your answer. Also, use this result to answer part (iv)).
iv.State the P-value.
v.State whether you reject or do not reject the null hypothesis.
vi.State your conclusion in context of the problem.
vii.If the true mean was 1.09, did you make an error? If so, which error? (Answer in complete sentences. You need to explain why or why not you made an error as well as identifying the type of error (Type I or II) if you did make an error. DO NOT answer these questions with a simple "Yes" or "No"; if you do, it will be marked as incorrect.)
Explanation / Answer
1) THE NULL HYPOTHESIS = Ho= u = 1.15
the altenate hypothesis = Ha = u<1.15
2) the significance level of the test = 5% = 0.05
3) test static = (1.11- 1.15)/(0.040/sqrt(11)) = -3.31
4) the degree of freedom = 11-1 = 10
the p value from t table = 0.0039
5) as the p value is less then the significance level therefore we will reject the null hypothesis.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.