A butter machine makes sticks of butter that average 4.0 ounces in weight with a

Question

A butter machine makes sticks of butter that average 4.0 ounces in weight with a standard deviation of 0.16 ounces. There are 4 sticks to a package. (a) A package weighs ___ ounces, give or take ___ ounces or so. (b) A store purchases 100 packages, Estimate the chance that it gets 100 pounds of butter (1600 ounces), to within 8 ounces. According to one investigator's model, his data are like 400 draws made at random from a large box. The null hypothesis is that the average of the box equals 80; the alternative is theta the average of the box is more than 80. In fact, his data averaged out to 85.4, with a SD of about 45. Compute z and P. What should he conclude?

Explanation / Answer

3(a)

Let's denote the stick by X, and package by Y.

So, Y = 4X, because the package contains 4 sticks

Mean weight of a stick = 4 ounces

So, mean weight of the package = 4*4 = 16 ounces

SD of stick weight = 0.16 ounces, so SD of package = 40.5*0.16 = 0.32 ounces

So, the package weighs 16 ounces, give or take 0.32 ounces or so.

(b)

Expected total weight of 100 packages = 100*4*4 = 1600 ounces

SD of total weight = 4000.5*0.16 = 3.2 ounces

Calculating the z-scores at the two extremes:

z1 = ((1600-8)-1600)/3.2 = -2.5

z2 = ((1600+8)-1600)/3.2 = 2.5

At z1, p-value is: p1 = 0.00621

At z2, p-value is: p2 = 0.99379

Required probability = p2-p1 = 0.98758 = 98.758%

(4)

Sample mean, X' = 85.4

Sample size, n = 400

Sample SD, S = 45

Standard error, S' = S/n0.5 = 45/4000.5 = 2.25

z = (X'-80)/S' = (85.4-80)/2.25 = 2.4

At this z score, looking at the z-table for right tailed test, p = 0.0082

So, at the significance level of 0.05, result is significant.

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