Question

In: Statistics and Probability

A manufacturing company produces steel housings for electrical equipment. The main component part of the housing...

A manufacturing company produces steel housings for electrical equipment. The main component part of the housing is a steel trough that is made out of a 14-gauge steel coil. It is produced using a 250-ton progressive punch press with a wipe-down operation that puts two 90-degree forms in the flat steel to make the trough. The distance from one side of the form to the other is critical because of weatherproofing in outdoor applications. The company requires that the width of the trough be between 8.31 inches and 8.61 inches.

The data below represents the widths of the troughs, in inches, for a sample of n = 49:

8.312 8.343   8.317   8.383 8.348   8.410   8.351 8.373   8.481   8.422

8.476 8.382   8.484   8.403 8.414   8.419   8.385 8.465   8.498   8.447

8.436   8.413   8.489 8.414   8.481   8.415 8.479   8.429   8.458 8.462

8.460 8.444   8.429   8.460 8.412   8.420 8.410 8.405   8.323   8.420

8.396 8.447   8.405   8.439 8.411   8.427   8.420 8.498   8.409

(a) At the 0.05 level of significance, is there evidence that the mean width of the troughs is different from 8.46 inches? Be sure to include the following in your answer:

i.                     List the null and alternate hypotheses

ii.                   Find the value of the test statistic

iii.                 List the decision rule.

iv.                 Conclude in a complete sentence and include why you decided as you did – referring to the decision rule.

(b) What assumption about the population distribution is needed in order to conduct the t test in (a)?

Solutions

Expert Solution

(a)

Step 1:

(i)

Ho:   

Ha:   

Null hypothesis states that the mean width of the trough is 8.46 inches.

Step 2: Test statistics

We will use t statistics assuming that the population is normally distributed. Also since the population deviation is not given we will use t statistics

From the sample data we will calculate the sample mean and sample standard deviation:

sample mean = sum of all the terms / no of terms = 412.624 / 49 = 8.421

create below table

data data-mean (data - mean)2
8.312 -0.109 0.011881
8.476 0.055000000000001 0.0030250000000002
8.436 0.015000000000001 0.00022500000000002
8.46 0.039000000000001 0.0015210000000001
8.396 -0.024999999999999 0.00062499999999993
8.343 -0.077999999999999 0.0060839999999999
8.382 -0.039 0.001521
8.413 -0.0079999999999991 6.3999999999986E-5
8.444 0.023000000000001 0.00052900000000007
8.447 0.026 0.00067599999999999
8.317 -0.104 0.010816
8.484 0.063000000000001 0.0039690000000001
8.489 0.068000000000001 0.0046240000000002
8.429 0.0080000000000009 6.4000000000014E-5
8.405 -0.016 0.000256
8.383 -0.038 0.001444
8.403 -0.017999999999999 0.00032399999999996
8.414 -0.0069999999999997 4.8999999999995E-5
8.46 0.039000000000001 0.0015210000000001
8.439 0.018000000000001 0.00032400000000002
8.348 -0.072999999999999 0.0053289999999998
8.414 -0.0069999999999997 4.8999999999995E-5
8.481 0.06 0.0036000000000001
8.412 -0.0089999999999986 8.0999999999974E-5
8.411 -0.0099999999999998 9.9999999999996E-5
8.41 -0.010999999999999 0.00012099999999998
8.419 -0.0019999999999989 3.9999999999956E-6
8.415 -0.0060000000000002 3.6000000000003E-5
8.42 -0.00099999999999945 9.9999999999889E-7
8.427 0.0060000000000002 3.6000000000003E-5
8.351 -0.069999999999999 0.0048999999999998
8.385 -0.036 0.001296
8.479 0.058 0.003364
8.41 -0.010999999999999 0.00012099999999998
8.42 -0.00099999999999945 9.9999999999889E-7
8.373 -0.048 0.002304
8.465 0.044 0.001936
8.429 0.0080000000000009 6.4000000000014E-5
8.41 -0.010999999999999 0.00012099999999998
8.498 0.077 0.005929
8.481 0.06 0.0036000000000001
8.498 0.077 0.005929
8.458 0.037000000000001 0.0013690000000001
8.323 -0.097999999999999 0.0096039999999998
8.409 -0.011999999999999 0.00014399999999997
8.422 0.0010000000000012 1.0000000000024E-6
8.447 0.026 0.00067599999999999
8.462 0.041 0.001681
8.42 -0.00099999999999945 9.9999999999889E-7

(ii) t = -5.922

(iii) Rule : We reject the Null hypothesis is the t statistics is in the critical range.

The t-critical values for a two-tailed test, for a significance level of α=0.05

tc=−2.011 and tc=2.011

Since the t statistics (-5.922)  falls in the rejection area, we reject the Null hypothesis.

(b) assumption is that the data is normally distributed.

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Critical Values T-Distribution -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

orchestra answered 1 year ago
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