A test of the 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. The manufacturer claims a true mean breaking strength of 8000 lbs. You believe the manufacturer might be wrong, but you are uncertain whether the true mean is larger or smaller.

Can we support the manufacturer’s claim at a level of significance equal to 0.05?

Question

Cybro Cybro

15-11-2024
Mathematics

High School

A test of the 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. The manufacturer claims a true mean breaking strength of 8000 lbs. You believe the manufacturer might be wrong, but you are uncertain whether the true mean is larger or smaller.

Can we support the manufacturer’s claim at a level of significance equal to 0.05?

Answer :

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mboluwape mboluwape · Answer 1 · 2024-01-26T23:44:29-05:00

We have sufficient evidence to support the claim that the true mean breaking strength is not equal to 8000 lbs at a significance level of 0.05.

To determine if we can support the manufacturer's claim of a true mean breaking strength of 8000 lbs, perform a hypothesis test.

Hypotheses:

Null hypothesis (H0): The true mean breaking strength is 8000 lbs.

Alternative hypothesis (Ha): The true mean breaking strength is not equal to 8000 lbs.

Use a two-tailed test with a significance level of 0.05.

This means that if the calculated p-value is less than 0.05, we'll reject the null hypothesis in favor of the alternative hypothesis.

Given:

Sample mean (x) = 7750 lbs

Sample standard deviation (s) = 145 lbs

Sample size (n) = 6

Claimed true mean (μ) = 8000 lbs

Significance level (α) = 0.05

To perform the hypothesis test, calculate the test statistic using the t-distribution:

t = (x - μ) / (s / √n)

t = (7750 - 8000) / (145 / √6)

t ≈ -2.585.

Next, determine the critical value(s) for a two-tailed test with a significance level of 0.05 and degrees of freedom (df) equal to n - 1.

df = 6 - 1 = 5

Using a t-table or a statistical software, the critical values for a two-tailed test with a significance level of 0.05 and 5 degrees of freedom are approximately -2.571 (lower critical value) and 2.571 (upper critical value).

Since the absolute value of our calculated t-value (-2.585) is greater than the critical value of 2.571, we can reject the null hypothesis.

Therefore, we have sufficient evidence to support the claim that the true mean breaking strength is not equal to 8000 lbs at a significance level of 0.05.