Try a binomial proportion 2-sample z-test using proportions.
Hypotheses:
Ho: pA = pB
Ha: pA does not equal pB -->this is a two-tailed test (the alternate hypothesis does not show a specific direction).
The formula is:
z = (pA - pB)/√[pq(1/n1 + 1/n2)]
...where n represents the sample sizes, p is (x1 + x2)/(n1 + n2), and q is 1-p.
I'll get you started:
p = (73 + 56)/(5000 + 4500) = ? -->once you have the fraction, convert to a decimal (decimals are easier to use in the formula).
q = 1 - p
pA = 73/5000
pB = 56/4500
Convert all fractions to decimals. Plug those decimal values into the formula and find z. Compare z to the cutoff 0.05 for a two-tailed test (cutoff value is z = + or - 1.96). If the test statistic you calculated exceeds either the positive or negative cutoff z-value, reject the null and conclude a difference. If the test statistic does not exceed either the positive or negative cutoff z-value, do not reject the null (you cannot conclude a difference).
I hope this will help get you started.
Media Disk, Inc. duplicates over a million 3.5” floppy disks each year by copying masters to stacks of blank disks. The company buys its blank stock from two different suppliers, A and B. The manager has decided to check each supplier’s stock by counting the rejected disks in the next run from 5,000 that just arrived from supplier A and 4,500 that just arrived from supplier B. During the run, the disk duplicators rejected 73 of A’s disks and 56 of B’s. State the appropriate hypotheses to test whether the proportions of defective disks from the two suppliers are the same. At a = 0.05, what can Media Disk conclude? This question is due on day 6.
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