Attempt all the Questions (1 mark each):
A research conducted by the University of Michigan claimed that there are more female drivers in the USA than male drivers. A researcher decides to test this claim on his state. In his simple random sample of 900 observations, he noticed that 468 of 900 were women. At = 0.05, is there enough evidence to support the claim?
(Use the P-value method where )
A sample of 100 body temperatures has a mean of 98.8. Assume that is known to be 0.6. Use a 0.05 significance level to test the claim that the mean body temperature of the population is equal to 98.6, as is commonly believed. Is there sufficient evidence to conclude that the common belief is wrong?
(Use the P-value method where )
A simple random sample of 25 filtered 100 mm cigarettes is obtained and the tar content of each cigarette is measured. The sample has a mean of 15 mg and a standard deviation of 4 mg. Use a 0.05 significance level to test the claim that the mean tar content of filtered 100 mm cigarettes is less than 20 mg, which is the mean for unfiltered king size cigarettes. What do the results suggest about the effectiveness of the filters?
(The critical value of ? for (single tail) and 24 d.f is 1.711)
A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2500 occupants not wearing seat belts, 15 were killed. Among 7500 occupants wearing seat belts, 15 were killed. Use this data with 0.05 significance level to test the claim that the fatality rate is higher for those not wearing seat belts. (Write all necessary steps like Hypotheses, Test Statistic, P-value & Conclusion)
Given that .
Listed below are the body lengths (in inches) and weights (in lb) of randomly selected bears:
| Length | 40 64 65 49 47 |
| Weight | 65 356 316 94 86 |
Find the value of the linear correlation coefficient.
Letting y represent weights of bears and letting x represent their lengths, find the regression equation.
Based on the given sample data, what is the best predicted weight of a bear with a length of 72.0 inch ?
Common colds are typically caused by a rhinovirus. In a test of effectiveness of Echinacea, some test subjects were treated with Echinacea extracted with 20% ethanol, some were treated with Echinacea extracted with 60% ethanol, and others were given a placebo. All of the test subjects were then exposed to rhinovirus. Results are summarized in Table given below. Use a 0.05 significance level to test the claim that getting an infection (cold) is independent of the treatment group. What does the result indicate about the effectiveness of Echinacea as a treatment for colds? Given that he critical value of 2 for 1d.f. = 3.841, the critical value of 2 for 2d.f. = 5.991, the critical value of 2 for 3d.f. = 7.815.
| Treatment Group | Total | |||
| Placebo | Echinacea:20%extract | Echinacea:60%extract | ||
| Infected | 85 | 45 | 40 | 170 |
| Not Infected | 15 | 5 | 10 | 30 |
| Total | 100 | 50 | 50 | 200 |
Discussion-2 STAT-490 Pre MBA (Mark-1)
Listed below are randomly selected heights (in inches) of males from year 1880
| Height | 71 62 64 68 68 67 65 65 66 66 |
Find the mean and standard deviation of the sample.
Use a 0.05 significance level to test the claim that heights of men from 1880 have a mean less than 68 inches, which is the mean height given for men today.
(Given that the critical value of for (left tailed) and 9 d.f. is )
Construct a 95% confidence interval estimate of the mean height of males in 1880.
(Given that the critical value of for (two tailed) and 9 d.f. is 2.262)