Problems adapted from Principles of Statistics for Engineers and Scientists by William Navidi.
A certain type of stainless steel powder is supposed to have a mean particle diameter of \(\mu\) = 15 \(\mu\)m. A random sample of 87 particles had a mean diameter of 15.2 \(\mu\)m, with a standard deviation of 1.8 \(\mu\)m. A test is made of \(H_0\): \(\mu\) = 15 versus \(H_a\) : \(\mu \ne\) 15. Find the p-value.
Answer:
p = 0.30002701885400107, cannot conclude that mean particle diameter has deviated.
You purchased a filling machine for candy bags that is supposed to fill each bag with 16 oz of candy. Assume that the weights of filled bags are approximately normally distributed. A random sample of 10 bags yields the following data (in oz):
15.87, 16.02, 15.78, 15.83, 15.69, 15.81, 16.04, 15.81, 15.92, 16.10
a) Find a 95% confidence interval for the upper bound on the fill weight.
b) Can you conclude that the mean fill weight is actually less than 16 oz?
Answer:
upper_bound = 15.962633034102968
p = 0.011446733722292404, appears likely that mean fill weight is actually less than 16 oz.
The article “Utility of Pretest Probability and Exercise Treadmill Test in Korean Women with Suspected Coronary Artery Disease” (Y. Kim, W. Shim, et al., Journal of Women’s Health, 2015:617–622) presents a study of health outcomes in women with symptoms of heart disease. In a sample of 110 women whose test results suggested the presence of coronary artery disease, the mean peak systolic blood pressure was 169.9 mmHg, with a standard deviation of 24.8 mmHg. In a sample of 225 women whose test results suggested an absence of coronary artery disease, the mean peak systolic blood pressure was 163.3 mmHg, with a standard deviation of 25.8 mmHg.
a) Find a 95% confidence interval for the difference in systolic blood pressure between these two groups of women.
b) Can you conclude that the mean systolic blood pressure differs between these two groups of women?
Answer:
conf_interval = [0.8690999212837536, 12.330900078716235]
p = 0.023995856808685323, it does appear as if blood pressure differs
In a study of the relationship of the shape of a tablet to its dissolution time, 6 disk-shaped ibuprofen tablets and 8 oval-shaped ibuprofen tablets were dissolved in water. The dissolve times, in seconds, were as follows:
Disk: 269.0, 249.3, 255.2, 252.7, 247.0, 261.6
Oval: 268.8, 260.0, 273.5, 253.9, 278.5, 289.4, 261.6, 280.2
Can you conclude that the mean dissolve times differ between the two shapes?
Answer:
p = 0.01678438316396713, yes it appears that there is a significant difference in dissolve times.
Breathing rates,in breaths per minute, were measured for a group of 10 subjects at rest, and then during moderate exercise. The results were as follows:
| Subject | Rest | Exercise |
| 1 | 15 | 30 |
| 2 | 16 | 37 |
| 3 | 21 | 39 |
| 4 | 17 | 37 |
| 5 | 18 | 40 |
| 6 | 15 | 39 |
| 7 | 19 | 34 |
| 8 | 21 | 40 |
| 9 | 18 | 38 |
| 10 | 14 | 34 |
Find a 95% confidence interval for the increase in breathing rate due to exercise.
Answer:
conf_interval = [17.371052533037737, 21.42894746696226]
That’s all I’m expecting you to work on, as the problems (minimally) span the cross section of problem types. But if you want more practice here are some additional problems:
An automobile manufacturer wishes to compare the lifetimes of two brands of tire. She obtains samples of six tires of each brand. On each of six cars, she mounts one tire of each brand on each front wheel. The cars are driven until only 20% of the original tread remains. The distances, in miles, for each tire are presented in the following table. Can you conclude that there is a difference between the mean lifetimes of the two brands of tire? State the appropriate null and alternate hypotheses, find the p-value, and state your conclusion.
| Car | Brand 1 | Brand 2 |
| 1 | 36,925 | 34,318 |
| 2 | 45,300 | 42,280 |
| 3 | 36,240 | 35,500 |
| 4 | 32,100 | 31,950 |
| 5 | 37,210 | 38,015 |
| 6 | 48,360 | 47,800 |
| 7 | 38,200 | 33,215 |
Answer:
p = 0.07841963856563927, high enough that we probably shouldn’t reject the null hypothesis, but it seems suspect. would probably want to do further testing.
Many companies have been experimenting with telecommuting, allowing employees to work at home on their computers. Among other things, telecommuting is supposed to reduce the number of sick days taken. Suppose that at one firm, it is known that over the past few years employees have taken a mean of 5.4 sick days. This year, the firm introduces telecommuting. Management chooses a simple random sample of 80 employees to follow in detail, and, at the end of the year, these employees average 4.5 sick days with a standard deviation of 2.7 days. Let \(\mu\) represent the mean number of sick days for all employees of the firm. Find the P-value for testing \(H_0\) : \(\mu \ge\) 5.4 versus \(H_a\): \(\mu <\) 5.4.
Answer:
p = 0.0014345563960383046. quite small. would be unlikely to get such a sample if the null hypothesis were true, so we will reject it.
The article “Wired: Energy Drinks, Jock Identity, Masculine Norms, and Risk Taking” (K. Miller, Journal of American College Health, 2008:481–489) reports that in a sample of 413 male college students, the average number of energy drinks consumed per month was 2.49 with a standard deviation of 4.87, and in a sample of 382 female college students, the average was 1.22 with a standard deviation of 3.24. Can you conclude that the mean number of energy drinks is greater for male students than for female students? <!– 7.1, 12>
Answer:
p = 6.549014480716764e-06. extremely small p value. we are highly confident that the mean number of energy drinks is higher for male students.
In an experiment involving the breaking strength of a certain type of thread used in personal flotation devices, one batch of thread was subjected to a heat treatment for 60 seconds and another batch was treated for 120 seconds. The breaking strengths (in N) of ten threads in each batch were measured. The results were:
60 seconds: 43, 52, 52, 58, 49, 52, 41, 52, 56, 54
120 seconds: 59, 55, 59, 66, 62, 55, 57, 66, 66, 51
Find a 99% confidence interval for the difference in the mean strengths between threads treated for 60 seconds and those treated for 120 seconds.
Answer:
(1.8661029986715194, 15.533897001328487)