Could You Pass This Hardest Inference Procedures Exam?
Which inference procedure would you use in each of these cases? For now, we will assume that the conditions for these procedures are satisfied. Choose the most direct method of solving each problem. Take your time. So, let's try out the quiz. All the best!
- 1.
You are given x(1)and x(2), n(1), n(2) from two independent samples and you believe that there is no difference between the two population proportions.
- A.
1 sample confidence t-interval for the mean
- B.
2 sample confidence z-interval for the difference of proportions
- C.
1 proportion confidence z-interval
- D.
2 sample hypotheses z-test for the difference of proportions
- E.
Paired t-test
Correct Answer
D. 2 sample hypotheses z-test for the difference of proportionsExplanation
You construct the sample proportions from your x and n values and perform a TEST. Tests of proportions are z-tests.Rate this question:
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- 2.
You are given x and n from a random sample and you want an estimate for the population proportion.
- A.
1 sample confidence interval for the mean
- B.
2 sample confidence interval for the difference of proportions
- C.
1 proportion confidence interval
- D.
2 sample hypotheses test for the difference of proportions
- E.
Paired t-test
Correct Answer
C. 1 proportion confidence intervalExplanation
A proportion confidence interval is the appropriate choice when you want to estimate the population proportion based on a random sample. It is used when you have categorical data and want to estimate the proportion of a certain characteristic in the population. This type of confidence interval provides a range of values within which the true population proportion is likely to fall, based on the sample data. It takes into account the sample size and the observed proportion to calculate the confidence interval.Rate this question:
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- 3.
You are given x-bar(1)and x-bar(2),s(1), s(2), n(1), n(2) from two independent samples and you want an estimate of the difference between the two population means.
- A.
1 sample t-confidence interval for the mean
- B.
2 sample t-confidence interval for the difference of means
- C.
1 proportion confidence interval
- D.
2 sample hypotheses t-test for the difference of means
- E.
Paired t-test
Correct Answer
B. 2 sample t-confidence interval for the difference of meansExplanation
This involves a difference of sample means (x-bars) and use of estimators for the population standard deviation. You want an estimate, not a test of whether the sample supports your hypothesis, so we construct a CONFIDENCE INTERVAL for DIFFERENCES and using the T-DISTRIBUTION.Rate this question:
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- 4.
You are given x-bar(1)and x-bar(2),s(1), s(2), n(1), n(2) from two independent samples and you believe that there is no difference between the two population means.
- A.
1 sample confidence t-interval for the mean
- B.
2 sample confidence t-interval for the difference of means
- C.
1 proportion confidence z-interval
- D.
2 sample hypotheses t-test for the difference of means
- E.
Paired t-test
Correct Answer
D. 2 sample hypotheses t-test for the difference of meansExplanation
This involves a difference of sample means (x-bars) and use of estimators for the population standard deviation. This time you want a test of whether the samples support your hypothesis, so we perform a HYPOTHESIS TEST for DIFFERENCES and using the T-DISTRIBUTION.Rate this question:
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- 5.
You are given the difference between pretest and posttest measures for a set of participants. You believe that there is no improvement from pre- to post-test.
- A.
1 sample confidence interval for the mean
- B.
2 sample confidence interval for the difference of proportions
- C.
1 proportion confidence interval
- D.
2 sample hypotheses test for the difference of proportions
- E.
Paired t-test
Correct Answer
E. Paired t-testExplanation
A paired t-test is the appropriate statistical test in this scenario because it is used to compare the means of two related groups. In this case, the pretest and posttest measures are related because they are taken from the same set of participants. The paired t-test allows us to determine if there is a significant difference between the two sets of measures, indicating whether there is improvement or not.Rate this question:
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- 6.
A group of 300 housewives was interviewed to determine if there is a preference for one of two detergents. Detergent A was favoured by 135 housewives; the others favoured Detergent B. Which procedure would you perform to ascertain if the data provide sufficient evidence to indicate a difference in preference for the two detergents?
- A.
Two-sample z-test for means
- B.
One-sample t-test for a mean
- C.
Two-sample t-test for means
- D.
One sample z-test for proportion
- E.
Two sample z-test for proportions
Correct Answer
D. One sample z-test for proportionExplanation
To determine if there is a difference in preference for the two detergents, a one-sample z-test for proportion would be performed. This test is appropriate when comparing proportions from a single sample to a known population proportion. In this case, the proportion of housewives who favored Detergent A (135 out of 300) would be compared to the expected proportion of housewives who prefer Detergent A. The z-test would calculate the test statistic and p-value to determine if the observed difference in proportions is statistically significant.Rate this question:
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- 7.
If a new process for copper mining is to be adopted, it must produce at least 50 tons of ore per day. A 5-day trial gave the following results: 50 47 53 51 52 Do these figures warrant the adoption of the new process? Which procedure would be most appropriate for testing the data?
- A.
One-sample z-test for a mean
- B.
One-sample t-test for a mean
- C.
One sample z-test for proportion
- D.
Chi-Square Test for Goodness of Fit
- E.
Chi-Square test for Homogeneity or Independence
Correct Answer
B. One-sample t-test for a meanExplanation
The figures provided in the 5-day trial are the daily production results of the new process for copper mining. In order to determine if these figures warrant the adoption of the new process, a one-sample t-test for a mean would be the most appropriate procedure to test the data. This test would allow us to compare the average daily production of the new process with the minimum requirement of 50 tons of ore per day. By conducting the t-test, we can determine if the average production is significantly different from the required amount, and make an informed decision on whether to adopt the new process.Rate this question:
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- 8.
Two sets of 60 high school students each were taught algebra by two methods, respectively. The experimental group used programmed learning and no formal lectures; the control group was given formal lectures by a teacher. At the end of the experiment, both groups were given a standardized test, and the number of students scoring above 85% was recorded: 41 out of 60 of the experimental group had scores above 85%; 24 out of 60 in the control group had scores above 85%. Test the hypothesis that the two groups were not different in their performance on the standardized test. Which procedure would be most appropriate for testing the data?
- A.
Wo-sample z-test for means
- B.
Two-sample t-test for means
- C.
One sample z-test for proportion
- D.
Two sample z-test for proportions
- E.
Linear Regression t-test
Correct Answer
D. Two sample z-test for proportionsExplanation
The most appropriate procedure for testing the data in this scenario would be the Two sample z-test for proportions. This is because we are comparing the proportions of students scoring above 85% in the experimental group and the control group. The z-test for proportions allows us to determine if there is a significant difference between the proportions of students in the two groups. By comparing the observed proportions to the expected proportions, we can assess whether the two groups were different in their performance on the standardized test.Rate this question:
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- 9.
Ten sets of identical twins, all wanting to learn French, were divided into two groups, each group containing one of each twin pair. Group 1 was flown to France, where they lived for one month. Group 2 was enrolled in an intensive French course at a local university. At the end of one month, all subjects were given a standard French language exam. Which procedure is appropriate for performing the analysis of the exam scores?
- A.
One-sample z-test for a mean
- B.
Two-sample z-test for means
- C.
One-sample t-test for a mean
- D.
Two-sample t-test for means
- E.
One sample z-test for proportion
Correct Answer
C. One-sample t-test for a meanExplanation
The appropriate procedure for analyzing the exam scores in this scenario is a one-sample t-test for a mean. This is because the study involves comparing the exam scores of the two groups (Group 1 and Group 2) to determine if there is a significant difference in their mean scores. The t-test is used when the sample size is small and the population standard deviation is unknown, which is likely the case in this study.Rate this question:
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- 10.
A vegetable canner claims that the mean fill per 16-ounce can is 16.1 ounces. Several underweight complaints have been lodged against the company, and the canner wants to see if the machine set for the fill mechanism is correct. That is, he wishes to test the hypothesis that µ = 16.1 ounces. Experience with the machine has shown that the variation in fill observed over a number of years is σ =.11 ounces. A random sample of n = 10 cans gave the following measurements in ounces: 16.1, 16.0, 16.2, 15.9, 16.0, 16.1, 16.1, 15.9, 16.1, 16.0. Do these data indicate that µ differs from 16.1 ounces? Which would be the appropriate testing procedure for this scenario?
- A.
One-sample z-test for a mean
- B.
One-sample t-test for a mean
- C.
Two-sample t-test for means
- D.
One sample z-test for proportion
- E.
Hi-Square Test for Goodness of Fit
Correct Answer
A. One-sample z-test for a meanExplanation
The appropriate testing procedure for this scenario would be a one-sample z-test for a mean. This is because we are testing the hypothesis that the mean fill per 16-ounce can is equal to 16.1 ounces. We have the sample mean, sample standard deviation, and the population standard deviation, which allows us to calculate the z-score and compare it to the critical value to determine if there is enough evidence to reject the null hypothesis.Rate this question:
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Mar 18, 2022Quiz Edited by
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Mar 18, 2007Quiz Created by
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