Activity 10: Confidence Interval for 1 – Proportion



Activity 4: More Confidence Intervals and Hypothesis Tests

With this lab, we begin Chapter 9. This chapter introduces more confidence intervals and hypothesis tests, but you’ll find a lot of overlap with Chapters 7 and 8. Now that you’ve learned the basic machinery of confidence intervals and hypothesis tests, you’ll find that Chapter 9 merely applies this machinery to new situations; it does not actually present brand new machinery.

We begin in Section 9.3, which is nothing but a rehash of Sections 7.2, 7.3 and 8.2 (but viewed in a new light).

Obtain the data for this activity on the website. When you click the link, the data should open up automatically in Minitab.

1. The derailment of a freight train due to the catastrophic failure of a traction motor armature bearing provided the impetus for the following study. A sample of 17 high-mileage traction motors was selected, and the amount of cone penetration (mm/10) was determined both for the pinion bearing and for the commutator armature bearing (two measurements on each motor). We wish to estimate the population mean difference between penetration for the commutator armature bearing and penetration for the pinion bearing.

a) The key to this problem is that the two-sample data have a natural pairing. Thus, the strategy will be to work only with the differences in each pair. This means the situation reduces to the one-sample situation seen in Chapters 7 and 8. Calculate these differences and put them in column C4. Use Calc > Calculator…. Give this column the title “Diff”.

b) The parameter in this problem is μD. Explain in your own words what μD means. (You may find the material on pp. 382-383 helpful.)

It is the population mean difference between commutator and pinion penetration.

c) Looking at a normal probability plot for the differences computed in part (a), I see a sort of wavy pattern but nothing too serious, particularly because the extremes of the distribution seem close to the line. Obtain and attach this normal probability plot.

[pic]

d) Get descriptive statistics on your differences. Create by hand a 90% confidence interval for μD based on what you obtain. (Note: You should use a t-based confidence interval here; use Table A.5.) There are formulas in Section 9.3, but these formulas are exactly the same as in Section 7.3.

Variable N Mean StDev SE Mean

Diff 17 -4.18 35.85 8.69

The t-multiplier for a 90% CI in this case is t.05,16, which is 1.746 according to Table A.5.

The 90% CI is −4.18 ± 1.746(8.69), which is −4.18 ± 15.17. This can also be expressed as −19.35 to 10.99.

e) Devise a hypothesis test to determine whether the cone penetration of the pinion bearing and the commutator armature bearing are the same or not. Write your hypotheses based on μD below.

H0: μD = 0

Ha: μD ≠ 0

f) Test the hypotheses above using a one-sample t-test (just as you saw in Section 8.2, though there are also formulas in Section 9.3). Calculate the t-statistic by hand, then determine a p-value using Table A.8.

t = −4.18 / 8.69 = −0.481. From Table A.8, the area to the right of .481 with 16 degrees of freedom is .312. This is multiplied by 2 because the test is two-tailed, giving a p-value of .624.

g) Is zero outside the 90% confidence interval in part (d)? Would the null hypothesis be rejected at the 10% level in part (f)? Think about why the answers to these two questions MUST be the same (you don’t have to write that part down).

No and no. If 0 (the null value) is inside the interval, then it’s not a surprising value (at alpha=.10, at least), which means that we wouldn’t reject the null at alpha=.10.

(You didn’t have to answer that last part for credit…)

h) All the work that you did by hand in the preceding activities can of course be done by Minitab automatically (and hopefully after this, you’ll understand how Minitab does the calculations!). Check your work by going to Stat > Basic Statistics > Paired t… and filling in the appropriate boxes. Be sure that you click on Options… and make sure the settings are appropriate for the test and interval you want. Paste the relevant output below and check your previous work.

Paired T for Commutat - Pinion

N Mean StDev SE Mean

Commutat 17 259.88 31.28 7.59

Pinion 17 264.06 27.41 6.65

Difference 17 -4.18 35.85 8.69

90% CI for mean difference: (-19.36, 11.00)

T-Test of mean difference = 0 (vs not = 0): T-Value = -0.48 P-Value = 0.637

2. Suppose you wanted to do Exercise 51 on page 398. Use Minitab to obtain a z-statistic with which you could run a hypothesis test. Paste the relevant output (containing the z-statistic) below. (You don’t actually have to finish the exercise, though you might want to think about how you could easily answer part (a).) To do this, use Stat > Basic Statistics > 2 Proportions… and fill in all the blanks under “Summarized data” appropriately. Also, make sure that you choose a 2-sided alternative hypothesis, type a test difference of 0, and check the “use pooled estimate of p for test” box under Options….

Sample X N Sample p

1 1105 5726 0.192979

2 980 5384 0.182021

Estimate for p(1) - p(2): 0.0109586

95% CI for p(1) - p(2): (-0.00355738, 0.0254746)

Test for p(1) - p(2) = 0 (vs not = 0): Z = 1.48 P-Value = 0.139

The z-statistic is 1.48, as seen in the output above.

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