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Homework 5 Solutions
Problems 5.7, 5.12, 5.21, 5.28, 5.30, 5.44, 5.52, 5.59, 5.88,5.100*
- Problem should be done with MINITAB. ** Problem should be done by hand and with MINITAB.
5.7 Because the company is collecting samples containing 25 boxes every 10 minutes, there are 720 confidence intervals constructed during every 5-day period.
a. If the level of confidence is 95% for each of the 720 intervals, how many of the intervals would you expect to be in error - that is, fail to contain μ?
We would expect 5% to be in error for a 95% confidence level.
We expect 0. 05 × 720 = 36 of the 720 intervals to be in error.
b. If the sample size is increased from 25 to 50 and the samples are still collected every 10 minutes, how many of the 95% confidence intervals would you expect to be in error during any 5-day period?
Increasing the sample size will make the confidence interval narrower. However, the 95% confidence level means that we would still expect 5% of the intervals to be in error, and that 36 of the 720 intervals won’t contain μ.
c. If the sample size is kept at 25 boxes per sample but the level of confidence is increased from 95% to 99%, how many of the 95% confidence intervals would you expect to be in error during any 5-day period?
I think there is a typo in the book. I think the question meant to ask how many of the 99% confidence intervals would you expect to be in error. I would expect 1%, or 0. 01 × 720 ≈ 7 will be in error.
5.12 A courier company in New York City claims that its mean delivery time to any place in the city is less than 3 hours. To evaluate the claim, the quality control personnel randomly select 50 deliveries and compute the mean delivery time to be ¯y = 2.8 hours with a standard deviation of s = 0.6 hours.
a. Estimate the mean delivery time μ using a 95% confidence interval.
The 95% confidence interval is: y¯ ± zα/ 2 σy¯ = ¯y ± zα/ 2 √σn = 2. 8 ± 1. 96 √^0. 506 = 2. 8 ± 0 .17 = (2. 63 , 2 .97)
b. Based on the 95% confidence interval, does the company’s claim appear reason- able?
Yes, the claim appears reasonable. The confidence interval is completely below 3 hours. c. If a 99% confidence interval was used to estimate μ, would your answer in part (b) change?
The 99% confidence interval is: y¯ ± zα/ 2 σy¯ = ¯y ± zα/ 2 √σn = 2. 8 ± 2. 58 √^0. 506 = 2. 8 ± 0 .22 = (2. 58 , 3 .02)
Our conclusion using the 99% confidence interval is different. This interval is not completely below 3 hours. Thus, there is a chance that the mean is greater than 3 hours.
5.21 A biologist wishes to estimate the effect of an antibiotic on the growth of a particu- lar bacterium by examining the mean amount of bacteria present per plate of culture when a fixed amount of the antibiotic is applied. Previous experimentation with the antibiotic on this type of bacteria indicates that the standard deviation of the amount of bacteria present is approximately 13 cm^2. Use this information to determine the number of observations (cultures that must be developed and then tested) to estimate the mean amount of bacteria present, using a 99% confidence interval with a half-width of 3 cm^2.
We know that ˆσ = 13, E = 3, and zα/ 2 = 2.58. Now, we can use the formula for the sample size to get:
n =
(zα/ 2 )^2 σ^2 E^2
(2. 582 )(13^2 )
We’ll need 125 observations.
5.28 Using a computer software program, simulate 100 samples of size 16 from a normal distribution with μ = 40 and σ = 8. We wish to test the hypotheses H 0 : μ = 40 vs. Ha : μ 6 = 40 with α = 0.10. Assume that σ = 8 and that the population distri- bution is normal when performing the test of hypothesis using each of the 100 samples.
5.30 Refer to Exercises 5.28 and 5.29.
a. Answer the questions posed in these exercises with α = 0.01 in place of α = 0.10. You can use the data sets simulated in Exercises 5.28 and 5.29, but the exact power of the tests will need to be recalculated.
This is a complicated series of questions. I think we need to work all three of 5.28, 5.29, and 5.30 to answer the questions. If we rework 5.28 using α = 0.01 in- stead of 0.10, we find that we incorrectly reject H 0 in 1 out of the 100 simulations.
We’d expect to do this 1% of the time on average, or 1 out 100 tests.
I simulated the data requested for 5.29a, and found that we correctly rejected H 0 18 times.
To answer question 5.29b, we need to make use of the fact that mathematical power for a two-sided hypothesis test can be calculated as:
P W R(μa) = P
z > zα/ 2 −
|μ 0 − μa| σ/
n
For 5.28, with α = 0.10, zα/ 2 = 1.645. Then P W R(41.5) = P
z > 1. 645 − |^408 /−√^4116.^5 |
= P (z > 0 .895) = 1 − 0 .8159 = 0 .1841.
We’d expect 0. 1841 × 100 = 18. 41 ≈ 18 of the tests to correctly reject H 0.
The other situations in Exercise 5.30 are handled similarly, except that we’ll use zα/ 2 = 2.58 corresponding to α = 0.01. I’ve summarized the results in the table below:
Experimental Mathematical Expected α μ σ # Correct Power Correct 0.10 38 8 24 0.2578 25. 0.10 41.5 8 18 0.1841 18. 0.10 43 8 48 0.4404 44. 0.01 38 8 4 0.0571 5. 0.01 41.5 8 2 0.0336 3. 0.01 43 8 12 0.1401 14.
b. Did decreasing α from 0.10 to 0.01 increase or decrease the power of the test? Explain why this change occurred.
Changing the α to 0.01 has dramatically decreased the power. In order to reduce the chance of making a Type I error, the test has increased the critical value from 1.645 to 2.58 which results in a test which is much less likely to reject Ho and hence the power is decreased.
5.44 The researcher was interested in determining whether μ was less than 14. The sample data yielded n = 40, ¯y = 13.5, s = 3.62. Determine the level of significance for testing H 0 : μ ≥ 14 vs. Ha : μ < 14. Is there significant evidence in the data to support the claim that μ is less than 14 using α = 0.05?
The test statistic is z = (^) σ/¯y−√μ^0 n = (^313). 62.^5 /−√^1440 = − 0. 874
P-value = P (z < − 0 .87) = 0.1922.
Since this is greater than α = 0.05, we conclude that we do not have significant evi- dence that μ is less than 14.
5.52* A new reading program was being evaluated in the fourth grade at an elementary school. A random sample of 20 students was thoroughly tested to determine reading speed and reading comprehension. Based on a fixed-length standardized test reading passage, the following speeds (in minutes) and comprehension scores (based on a 100- point scale) were obtained.
a. Use the reading speed data to place a 95% confidence interval on μ, the average reading speed, for all fourth-grade students in the large school from which the sample was selected.
Using MINITAB gave the following results:
One-Sample T: Speed
Variable N Mean StDev SE Mean 95% CI Speed 20 9.10000 2.57314 0.57537 (7.89573, 10.30427)
b. Plot the reading speed data using a normal probability plot or boxplot to eval- uate whether the data appear to be a random sample from a normal population distribution.
Both plots are given below. Based on the plots, the data do appear to be a random sample from a normal distribution. The boxplot doesn’t show extreme skewness or outliers, and the normal probability plot is quite linear.
We calculate a 100(1 − α)% confidence interval using: ¯y ± tα/ 2 √sn. In this case, we have n − 1 = 9 degrees of freedom. Looking on Table 2, we see that tα/ 2 = 1. 833 in this case.
Untreated: ¯y ± tα/ 2 √sn = 43. 6 ± (1.833) √^5. 107 = 43. 6 ± 3 .304 = (40. 3 , 46 .9)
Treated: y¯ ± tα/ 2 √sn = 36. 1 ± (1.833) √^4. 109 = 36. 1 ± 2 .840 = (33. 3 , 38 .9)
We are 90% confident that the average height of untreated shrubs is between 40. cm and 46.9 cm. We are 90% confident that the average height of treated shrubs is between 33.3 cm and 38.9 cm.
b. Do the two confidence intervals overlap? What conclusions can you make about the effectiveness of dikegulac as a growth retardant?
The two intervals do not overlap. This would indicate that the average heights of the treated and untreated shrubs are significantly different.
5.88* In a standard dissolution test for tablets of a particular drug product, the manu- facturer must obtain the dissolution rate for a batch of tablets prior to release of the batch. Suppose that the dissolution test consists of assays for 24 randomly selected individual 25-mg tablets. For each test, the tablet is suspended in an acid bath and then assayed after 30 minutes. The results of the 24 assays are given here.
a. Using a graphical display, determine whether the data appear to be a random sample from a normal distribution.
The normal probability plot of the data is shown below. The plot shows a little curvature, but is still pretty linear. It does seem reasonable to assume these data are a random sample from a normal distribution.
b. Estimate the mean dissolution rate for the batch of tablets, for both a point es- timate and a 99% confidence interval.
We can use ¯y as our point estimate, and use a t-interval. Using MINITAB, we get:
Descriptive Statistics: Results Variable Mean StDev Results 19.829 0.
One-Sample T: Results Variable N Mean StDev SE Mean 99% CI Results 24 19.8292 0.4319 0.0882 (19.5817, 20.0766)
c. Is there significant evidence that the batch of pills has a mean dissolution rate less than 20 mg (80% of the labeled amount in the tablets)? Use α = 0.01.
We are testing the hypotheses H 0 : μ = 20 vs. Ha : μ < 20. I used MINITAB, and found the test statistic of t = − 1 .94 with a corresponding p-value of 0.033. Since this is greater than α = 0.01, we conclude that we do not have significant evidence that the mean dissolution rate is less than 20 mg.
One-Sample T: Results Test of mu = 20 vs < 20 99% Upper Variable N Mean StDev SE Mean Bound T P Results 24 19.8292 0.4319 0.0882 20.0495 -1.94 0.
d. Calculate the probability of a Type II error if the true dissolution rate is 19.6 mg.
The probability of a Type II error for a 1-sided hypothesis test can be calculated as:
One-Sample T: 6am, 2pm, 10pm
Variable N Mean StDev SE Mean 95% CI
6am 15 0.128133 0.035516 0.009170 (0.108465, 0.147802)
2pm 15 0.116000 0.040601 0.010483 (0.093516, 0.138484)
10pm 15 0.141600 0.042798 0.011050 (0.117899, 0.165301)
b. Does there appear to be a significant difference in average SO 2 emissions over the three time periods?
No, there doesn’t appear to be a significant difference in the average SO 2 emis- sions over the three time periods. The three C.I.’s have a considerable overlap.
c. Combining the data over the entire day, is the average SO 2 emissions using the new scrubber less than 0.145, the average daily value for the old scrubber?
We are testing the hypotheses H 0 : μ = 0.145 vs. Ha : μ < 0 .145. I used MINITAB to do this test for the combined data.
MINITAB calculated a test statistic of t = − 2 .74 with a corresponding p-value of 0.004. This is very small. Thus there is significant evidence that the average SO 2 level using the new scrubber is less than 0.145.
Note: A Z-test would also work here, as the sample size is pretty large when we combine all of the data through the day.
One-Sample T: Emissions Test of mu = 0.145 vs < 0. 95% Upper Variable N Mean StDev SE Mean Bound T P Emissions 45 0.128578 0.040255 0.006001 0.138661 -2.74 0.
