Question

A producer of steel cables wants to know whether the steel cables it produces have an average breaking strength of 5000 pounds. An average breaking strength of less than 5000 pounds would not be adequate, and to produce steel cables with an average breaking strength in excess of 5000 pounds would unnecessarily increase production costs. The producer collects a random sample of 64 steel cable pieces. The breaking strength for each of these cable pieces is recorded in the file SteelCable.xlsx.
1. What is the standard error for this test (three decimal places)?

2. What is the value of the t-statistic for this test (three decimal places)?

3. What is the p-value for this test (four decimal places)?

4. What statistical conclusion can the producer reach regarding the average breaking strength of its steel cables at a 5% significance level?

Cable Breaking Strength
1 4919.00
2 5048.09
3 5482.85
4 5461.08
5 4583.24
6 4926.37
7 5460.99
8 5214.26
9 5286.32
10 4767.98
11 3931.76
12 5191.09
13 5453.64
14 4543.81
15 6060.19
16 6356.22
17 5306.83
18 4515.58
19 4713.03
20 4827.88
21 5873.37
22 4474.94
23 4659.81
24 5255.17
25 5216.45
26 5929.35
27 5258.75
28 4797.17
29 5060.31
30 4434.27
31 5359.47
32 5684.72
33 4959.64
34 4492.24
35 4407.15
36 4936.43
37 5928.43
38 5796.95
39 5470.01
40 5347.04
41 5245.09
42 4723.65
43 5077.84
44 5785.64
45 4390.91
46 5402.91
47 5101.36
48 5471.37
49 5823.75
50 6123.87
51 5168.42
52 4829.37
53 4870.35
54 4884.67
55 4822.61
56 5019.64
57 5560.57
58 5815.23
59 4922.61
60 4393.84
61 5500.39
62 4957.40
63 5258.44
64 5591.10

Answer 1

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:

$S^2= \frac{1}{n-1}*(sumX^2-(\frac{(sumX)^2}{n} ))$

n=64

∑X= 330174.88

∑X²= 1718980202.34

$S^2= \frac{1}{63}*(1718980202.34-(\frac{(330174.88)^2}{64} ))$

S²= 2478376895

S= 497.8329 ≅497.833

2.

For this test the hypotheses are:

H₀: μ = 5000

H₁:  μ ≠ 5000

$t= \frac{Xbar-Mu}{\frac{S}{\sqrt{n} } }$

The sample mean is:

Xbar= ∑X/n)= 330174.88/64=5158.98

‬$t= \frac{5158.98-5000}{\frac{497.833}{\sqrt{64} } }$

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!