Try our AP Stats unit 6 test. These questions focus on the ideas of proportion and sample proportion, and using sample proportion as well as a confidence interval (based on a desired degree of confidence) to give some idea of what the population proportion could be. testing and Type I and II errors.
Congratulations - you have completed .
You scored %%SCORE%% out of %%TOTAL%%.
Your performance has been rated as %%RATING%%
Your answers are highlighted below.
Question 1 |
A researcher is conducting quality assurances on a batch of products. To estimate what proportion is defective, the researcher wants to make a one-sample z interval with 99% confidence $(z* = 2.58)$. The margin of error should be no more than 2%. A previous sample suggested 5% of the products are defective. What is the smallest sample size required to fit these parameters?
52 | |
790 | |
791 | |
1024 |
Question 1 Explanation:
The correct answer is (C). To calculate the smallest sample size for the given parameter, we use:
$2.58* \sqrt{\dfrac{(0.05(1-0.05)}{n}} $ $ ≤ 0.02$
$\sqrt{\dfrac{(0.05(1-0.05)}{n}} ≤ \dfrac{0.02}{2.58}$
$\sqrt{\dfrac{0.0475}{n}} ≤ \dfrac{0.02}{2.58}$
$\dfrac{0.0475}{n} ≤ \left(\dfrac{0.02}{2.58}\right)^2$
$0.0475≤n* \left(\dfrac{0.02}{2.58}\right)^2$
$\dfrac{0.0475}{\left(\dfrac{0.02}{2.58}\right)^2}≤n$
$790.4475≤n$
The smallest whole number sample size will be 791.
$2.58* \sqrt{\dfrac{(0.05(1-0.05)}{n}} $ $ ≤ 0.02$
$\sqrt{\dfrac{(0.05(1-0.05)}{n}} ≤ \dfrac{0.02}{2.58}$
$\sqrt{\dfrac{0.0475}{n}} ≤ \dfrac{0.02}{2.58}$
$\dfrac{0.0475}{n} ≤ \left(\dfrac{0.02}{2.58}\right)^2$
$0.0475≤n* \left(\dfrac{0.02}{2.58}\right)^2$
$\dfrac{0.0475}{\left(\dfrac{0.02}{2.58}\right)^2}≤n$
$790.4475≤n$
The smallest whole number sample size will be 791.
Question 2 |
Which of the following statements are true about the relationship between sample size, width of confidence interval, confidence level, and margin of error for a population proportion?
-
The width of the confidence interval increases as the sample size increases
-
The width of the confidence interval increases as the confidence level increases
-
The width of the confidence interval decreases as the margin of error increases
-
The width of the confidence interval is twice the margin of error
I and II | |
II and III | |
I and III | |
II and IV |
Question 2 Explanation:
The correct answer is (D). The width of the confidence interval tends to decrease as the sample size increases, so statement (I) is false. The width of the confidence interval for a population proportion is twice the margin of error, so (III) is false and (IV) is true. The width of the confidence interval increases as the confidence level increases, so (II) is true.
Question 3 |
A teacher is trying to improve student mastery using a new teaching method. The current student mastery is 60%. The teacher does a random sample under the new teaching method and observes a new mean student mastery is 75%. What are appropriate hypotheses for the significance test?
$H_0:p=0.60; H_a: p=0.75$ | |
$H_0:p=0.60; H_a: p>0.60$ | |
$H_0:p>0.60; H_a: p=0.75$ | |
$H_0:p>0.60; H_a: p=0.60$ |
Question 3 Explanation:
The correct answer is (B). The null hypothesis should be a statement of equality using the reported mean. The alternate hypothesis is testing for improvement.
Question 4 |
A student reads a newspaper article that estimates that 60% of students from the local high school attend a four-year college. The student is curious about that number and conducts a separate random sample to survey students. In the random sample of 80 students, 56 reported they were attending a four-year college. Calculate the test statistic for this population proportion.
0.42 | |
0.70 | |
1.30 | |
1.83 |
Question 4 Explanation:
The correct answer is (D). The test statistic is:
$\frac{\text{sample statistic} - \text{null value of parameter}}{\text{standard deviation of statistic}}$
$56÷80 = 0.7$
$\dfrac{0.7-0.6}{\sqrt{\dfrac{0.6(1-0.6)}{80}}} = \dfrac{0.1}{\sqrt{\dfrac{0.24}{80}}} ≈1.83$
$\frac{\text{sample statistic} - \text{null value of parameter}}{\text{standard deviation of statistic}}$
$56÷80 = 0.7$
$\dfrac{0.7-0.6}{\sqrt{\dfrac{0.6(1-0.6)}{80}}} = \dfrac{0.1}{\sqrt{\dfrac{0.24}{80}}} ≈1.83$
Question 5 |
A medical researcher is testing the null hypothesis that a new treatment is no more effective than a previous treatment for a certain disease. What would be a Type I error in this situation?
The researcher concludes the treatment is more effective when it is actually not more effective. | |
The researcher concludes the treatment is not more effective when it is actually more effective. | |
The researcher doesn’t conclude the treatment is more effective when it is not more effective. | |
None of these describe a Type I error. |
Question 5 Explanation:
The correct answer is (A). A Type I error is when the null hypothesis is true, but is rejected. In statement (A), the null hypothesis (the treatment is no more effective) is true, but the researcher does not come to that conclusion.
Question 6 |
Assuming no other factors are changed, changing which factor can reduce the probability of a Type II error?
Decrease sample size | |
Decrease significance level (α) of test | |
Decrease standard error | |
Decrease the distance from the true parameter to the null |
Question 6 Explanation:
The correct answer is (C). The probability of a Type II error decreases when any of the following occurs:
The standard error is generally the standard deviation over the square root of the sample size, so the smaller the standard deviation the smaller the standard error and the larger the test statistic. A larger test statistic is more likely to lead to rejecting the null, whereas a type II error is when the null is not rejected.
- Increase sample size
- Increase significance level (α) of test
- Decrease standard error
- True parameter value is farther from null
The standard error is generally the standard deviation over the square root of the sample size, so the smaller the standard deviation the smaller the standard error and the larger the test statistic. A larger test statistic is more likely to lead to rejecting the null, whereas a type II error is when the null is not rejected.
Question 7 |
The student council is investigating whether or not students support the new classroom design proposed at their high school. They are curious about the difference of opinion between older (11th and 12th grade) and younger (9th and 10th grade) students. They conduct a random survey of students from each group. Here are the results:
Support new design? |
9th/10th |
11th/12th |
Yes |
72 |
84 |
No |
28 |
36 |
Total |
100 |
120 |
The student council wants to use these results to construct a 95% confidence interval $(z* = 2.58)$ to estimate the difference between the proportions of students in these two groups who support the construction project. Assume all conditions for inference have been met. What is a correct 95% confidence interval based on this data?
$(−0.23, −0.01)$ | |
$(−0.20, −0.04)$ | |
$(−0.08, 0.08)$ | |
$(−0.11, 0.11)$ |
Question 7 Explanation:
The correct answer is (A). The confidence interval can be calculated by:
$(0.72-0.84)± $ $ 1.96 \sqrt{\dfrac{0.72(0.28)}{100}+\dfrac{0.84(0.16)}{100}}$
$=−0.12±0.11361$
$=(−0.23361 , −0.00639)$
$(0.72-0.84)± $ $ 1.96 \sqrt{\dfrac{0.72(0.28)}{100}+\dfrac{0.84(0.16)}{100}}$
$=−0.12±0.11361$
$=(−0.23361 , −0.00639)$
Question 8 |
Which of the following could be appropriate null and alternative hypotheses for a difference of population proportions?
-
$H_0:p_1=p_2$
$H_1:p_1>p_2$ -
$H_0:p_1=p_2$
$H_1:p_1 \neq p_2$ -
$H_0:p_1 > p_2$
$H_1:p_2 < p_1$ -
$H_0:p_1>p_2$
$H_1:p_1=p_2$
I and II | |
III | |
I and IV | |
IV |
Question 8 Explanation:
The correct answer is (A). I is a one-sided test and II is a two-sided test. Both have a clear null hypothesis and an alternative hypothesis which is a set of possibilities that does not overlap with the null. Both hypotheses in III are the same. The alternative hypothesis in IV is a specific equality — this is never supposed to be the case!
Question 9 |
Which of the following needs to be true when getting a confidence interval for a proportion?
$n \hat{p} \geqslant 10$ $\text{ and }$ $n(1-\hat{p}) \leqslant 10$ | |
$n(1-\hat{p}) \geqslant 10$ | |
$n \hat{p} \geqslant 10$ | |
$n \hat{p} \geqslant 10$ $\text{ and }$ $n(1-\hat{p}) \geqslant 10$ |
Question 9 Explanation:
The correct answer is (D). You do need there to be at least 10 successes and 10 failures on average
based on the sample proportion.
$n \hat{p}$ is the expected number of successes and $n(1 − \hat{p})$ is the expected number of failures, just based on the sample proportion $\hat{p}$.
$n \hat{p}$ is the expected number of successes and $n(1 − \hat{p})$ is the expected number of failures, just based on the sample proportion $\hat{p}$.
Question 10 |
A principal wants to gather evidence for the efficacy of a mandated tutoring program. To do this, a random sample of students is taken from those in the voluntary tutoring program, and those not in the program. In the voluntary tutoring program, 75% of students are passing all of their classes. For those not in the program, 65% are passing all classes. The principal performs a significance test on the difference of proportions and the p-value returned is 0.05.
What conclusion should be drawn for a 95% level of confidence?
The p-value does not indicate a statistically significant difference in population proportions, so we should accept the null hypothesis. | |
The p-value does not indicate a statistically significant difference in population proportions, so we should reject the null hypothesis. | |
The p-value indicates a statistically significant difference in the population proportions, so we should accept the null hypothesis. | |
The p-value indicates a statistically significant difference in the population proportions, so we should reject the null hypothesis. |
Question 10 Explanation:
The correct answer is (D). A 95% level of confidence corresponds to a significance level of 1 − 0.95 = 0.05. A p−value of 0.05 is ≤.05, which is enough to indicate a statistically significant difference in the population proportions. This means that we reject the null hypothesis (that the population proportions are equal).
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 10 questions to complete.
|
List |
Next Practice Test:
Quantitative Data: Means >>
AP Statistics Main Menu >>