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Answer :
At a 0.05 significance level, the t-test with a t-statistic of approximately -4.22 and 5 degrees of freedom rejects the manufacturer's claim, indicating a significant difference in breaking strength.
Hypotheses:
- Null hypothesis [tex](\(H_0\)): \(\mu = 8000\)[/tex] lbs (The true mean breaking strength is equal to the claimed mean.)
- Alternative hypothesis [tex](\(H_1\)): \(\mu \neq 8000\)[/tex] lbs (The true mean breaking strength is not equal to the claimed mean.)
Given:
- Sample mean [tex](\(\bar{X}\))[/tex] = 7750 lbs
- Sample standard deviation (s) = 145 lbs
- Sample size (n) = 6
- Claimed true mean [tex](\(\mu\))[/tex] = 8000 lbs
Calculate T-Statistic:
[tex]\[ t = \frac{(\bar{X} - \mu)}{(\frac{s}{\sqrt{n}})} \]\[t = \frac{(7750 - 8000)}{\left(\frac{145}{\sqrt{6}}\right)} \approx -4.22 \][/tex]
Degrees of Freedom (df):
df = n - 1 = 6 - 1 = 5
Critical Value:
- At [tex]\(\alpha = 0.05\)[/tex] with (df = 5), the critical values are approximately [tex]\(\pm2.571\)[/tex].
Decision Rule:
- If [tex]\(|t| > 2.571\)[/tex], reject [tex]\(H_0\)[/tex].
Decision:
- Since [tex]\(|-4.22| > 2.571\)[/tex], we reject [tex]\(H_0\)[/tex].
Conclusion:
- At a significance level of 0.05, there is enough evidence to reject the manufacturer's claim that the true mean breaking strength is 8000 lbs. The data suggests a significant difference.
The question is:
A test of breaking strengths of 6 ropes manufactured by a company showed a (sample) mean breaking strength of 7750 lbs and a (sample) standard deviation of 145 lbs, whereas the manufacturer claimed a true mean breaking strength of 8000 lbs. You think that the manufacturer is wrong and that the true mean is not 8000 lbs, but you are agnostic as to whether it is larger or smaller. Can we support the manufacturer’s claim at a level of significance equal to 0.05? (Be sure to write down your null and alternative hypothesis. If you do not, you will lose many points
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