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| Section | Objectives |
|---|---|
| Statistical Inference | - Estimation
|
| Descriptive Statistics | - Data summarization
|
| Regression and Correlation | - Relationship analysis
|
| Probability | - Probability rules
|
| Probability Distributions | - Discrete distributions
|
>> Applied-Probability-and-Statistics Lernressourcen <<
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148. Frage
Null hypothesis for a new drug shows no effect. True statement?
Antwort: C
Begründung:
The null hypothesis represents the default claim or baseline assumption in hypothesis testing. For a new drug, a typical null hypothesis states that the drug has no effect, no difference from placebo, or no improvement over the existing standard. The testing process begins by assuming the null hypothesis is true, then evaluates whether the sample evidence is strong enough to reject it. Therefore, the correct statement is that researchers assume no effect until evidence suggests otherwise. This does not mean the drug is proven ineffective; it means the burden of evidence lies with demonstrating an effect. Option B is the alternative hypothesis, not the null. Option C overstates the conclusion because failing to reject the null does not prove the drug has no effect. Option D is false because this type of claim is testable using experimental data and statistical inference.
Study Guide references/topics: null hypothesis, alternative hypothesis, hypothesis testing, statistical evidence.
149. Frage
A commuter has a .4 probability of taking the bus to work, a .4 probability of driving a car, and a .2 probability of cycling. The probability of being late is .15 when taking the bus, .05 when driving a car, and .1 when cycling.
What is the probability of taking the bus and being on time, or driving a car and being on time?
Antwort: A
Begründung:
This item requires multiplying conditional probabilities and then adding mutually exclusive outcomes.
"Taking the bus and being on time" means the commuter takes the bus and is not late. Since the probability of being late by bus is .15, the probability of being on time by bus is 1 # .15 = .85. Therefore, P(bus and on time)
= .4 × .85 = .34. For driving, the probability of being late is .05, so the probability of being on time is 1 # .05
= .95. Therefore, P(car and on time) = .4 × .95 = .38. Because a commuter cannot both take the bus and drive a car on the same trip, the events are mutually exclusive. Add the two joint probabilities: .34 + .38 = .72. The cycling information is not used because the question asks only about bus-on-time or car-on-time outcomes.
References/topics from the Study Guide: probability rules, complements, conditional probability, mutually exclusive events.
150. Frage
Poisson mean = 4. Probability of exactly 2 events?