You company manufactures pedicle screws of different lengths for vertebral fusio

Question

You company manufactures pedicle screws of different lengths for vertebral fusion surgeries. Over the course of one year, you find 24 out of 80 screws 30 mm in length came loose, whereas only 15 out of 65 screws 60 mm in length came loose.

a. (4 pts) Use Minitab to test the hypothesis that a higher proportion of 30 mm pedicle screws came loose when compared to the proportion of 60 mm pedicle screws that came loose for a level = 0.05.

b. (2 pts) Explain the reason for choosing your alternative hypothesis for the test in part A.

c. (2 pts) Use the P-value from part A to determine whether or not you accept or reject the null hypothesis.

d. (2 pts) Use the 95% confidence interval from part A to determine whether or not you accept or reject the null hypothesis.

e. (4 pts) Use the Calc Probability Distributions menu to calculate the P-value obtained from the Minitab test in part A. You must show the Minitab output to receive full credit.

please show minitab output window!!

Explanation / Answer

(a)

Test and CI for Two Proportions

Sample X N Sample p
1 24 80 0.300000
2 15 65 0.230769

Difference = p (1) - p (2)
Estimate for difference: 0.0692308
95% lower bound for difference: -0.0511473
Test for difference = 0 (vs > 0): Z = 0.93 P-Value = 0.175

(b) Since we want to test whether p1 > p2, the alternative hypothesis is Ha: p1 > p2

(c) 0.175 > 0.05, so we fail to reject ho

There is no sufficient evidence that p1 > p2

(d) The 95% confidence interval is [-0.0511, 0.1896]

Since the above interval includes 0, the null hypothesis can't be rejected.

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