Question

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(5p) A manufacturer of baseball bats selects every 10th bat that comes off the production line and weighs it. The weights (in ounces) from one's day production, in order of time, are shown in below: 31.1 31.7 31.5 31.8 31.6 32.3 32.8 33.1 32.1 32.3 33.1 33.3 33.5 33.6 33.5 33.5 Test for randomness above and below the mean (of these values) at a 5% significance level

Answer 1

Here we are to test,

$H_0:$ The weights are random.

versus

  is false

Mean of the given weights(in ounces) = 32.55

Now we form a sequence by assigning '1' to the points which are greater than the mean and '0' to the points which are less than or equal to the mean. To make the above dialect clear we can construct the following table:

X	X-mean(X)	Number Assigned
31.1 31.7 31.5 31.8 31.6 32.3 32.8 33.1 32.1 32.3 33.1 33.3 33.5 33.6 33.5 33.5	-1.45 -0.85 -1.05 -0.75 -0.95 -0.25 0.25 0.55 -0.45 -0.25 0.55 0.75 0.95 1.05 0.95 0.95	0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1
Total= 520.8

Hence, the sequence formed is: { 0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 }

Number of '0's =m=8

Number of '1' s=n=8

Number of runs of '0'= n1= 2

Number of runs of '1' = n2 = 2

To test $H_0$ , the test statistic used is 'U' , the total number of runs.

The tabulated value of U = U(tab) = n1+n2=4

The critical value of U at 5% level of significance in this case is 4. (Using Critical values Table of a Run Test ) .Since the tabulated value of U does not exceed the critical value, we reject the Null hypothesis and conclude that the data given is not random.