Answer:
Step-by-step explanation:
Hello!
The producer needs his cables to have an average breaking strength of 5000 pounds. A lower average breaking strength means that the cable is not adequate, a higher average breaking strength results in an unnecessary increase in production costs.
The sample is of 64 steel cables and the breaking strength of the pieces was recorded.
I always recommend that the first step to any statistic exercise is to establish the study variable that way you'll have fresh in mind the kind of data set you are working with and the type of distribution to expect from them (for example if it is a discrete variable you'd expect a binomial distribution, a continuous variable leads to a normal distribution (exact or approximate) a categorical variable sets the path to work with non-parametrical statistics such as Chi-Square statistics.)
In this example the study variable is:
X: Breaking strength of a steel cable.
This variable is continuous so first I'll use the sample information to test its distribution. Keep in mind that one of the conditions to use the Student t-test is that the variable has a normal distribution.
The p-value for the normality test is 0.8512, comparing it with the level of significance of the test (α: 0.05) the decision is to not reject the null hypothesis of the normality test, so you can conclude that the breaking strength of the steel cables has a normal distribution:
X~N(μ;σ²)
1.
The standard error of the test is the square root of the variance.
Using the following formula you have to calculate the variance:
[tex]S^2= \frac{1}{n-1}*(sumX^2-(\frac{(sumX)^2}{n} ))[/tex]
n=64
∑X= 330174.88
∑X²= 1718980202.34
[tex]S^2= \frac{1}{63}*(1718980202.34-(\frac{(330174.88)^2}{64} ))[/tex]
S²= 2478376895
S= 497.8329 ≅497.833
2.
For this test the hypotheses are:
H₀: μ = 5000
H₁: μ ≠ 5000
[tex]t= \frac{Xbar-Mu}{\frac{S}{\sqrt{n} } }[/tex]
The sample mean is:
Xbar= ∑X/n)= 330174.88/64=5158.98
[tex]t= \frac{5158.98-5000}{\frac{497.833}{\sqrt{64} } }[/tex]
t= 158.98/62.229= 2.555
3.
This test is two-tailed and so is the p-value. I've used statistics software to calculate it:
p-value 0.0131
4.
Using a significance level of 5%, since the p-value is less than α, the decision is to reject the null hypothesis. You can conclude at this level that the average breaking strength of the steel cables is different than 5000.
I hope this helps!
