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
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.