1 A random sample of 6 steel beams has a mean compressive strength of 58,392 p.s.i (pounds per square inch) with a standard deviation of 648p.s.i.use this information and the level of significance  = 0.05.test weather the true average compressive strength of the steel from which this sample came is 58,000p.s.i.. Assume normality

Answer 1

To test whether the true average compressive strength of the steel is 58,000 p.s.i., we will conduct a one-sample t-test.

Null hypothesis: The true average compressive strength of the steel is 58,000 p.s.i.
Alternative hypothesis: The true average compressive strength of the steel is not 58,000 p.s.i.

Given data:
Sample mean (x̄) = 58,392 p.s.i.
Standard deviation (s) = 648 p.s.i.
Sample size (n) = 6
Population mean (μ) = 58,000 p.s.i.
Level of significance (α) = 0.05

Calculate the t-statistic:
t = (x̄ - μ) / (s / √n)
t = (58392 - 58000) / (648 / √6)
t = 392 / (648 / 2.449)
t = 392 / 265.924
t = 1.473

Degrees of freedom (df) = n - 1 = 6 - 1 = 5

At α = 0.05 significance level, the critical t-value for a two-tailed test with df = 5 is ±2.571.

Since |t| (1.473) < 2.571, we fail to reject the null hypothesis.

Therefore, we do not have enough evidence to conclude that the true average compressive strength of the steel is different from 58,000 p.s.i. at a 0.05 significance level.