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AP Stats - Chapter 10 Test

Authored by Jason Bratten

Mathematics

9th - 12th Grade

CCSS covered

Used 63+ times

AP Stats - Chapter 10 Test
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10 questions

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1.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Choosing to run a test with a lower risk of α,that is, a smaller level of significance, results in

a lower risk of Type II error and lower power.

a lower risk of Type II error and higher power.

a higher risk of Type II error and lower power.

a higher risk of Type II error and higher power.

no change in risk of Type II error or in power.

2.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

A factory manager would like to compare two different assembly techniques by looking at the average number of units assembled by an employee during a certain time period. The manager uses random selection to choose 200 employees from the factory, and randomly assigns each employee to one of the two techniques. After training the employees on their respective techniques, the manager records the number of units each of the employees

assembles within the allotted time. Which of the following would be the most appropriate inferential statistical test

in this situation?

One-sample z-test

Two-sample t-test

Paired t-test

One-sample t-test

None of these tests are appropriate

Tags

CCSS.HSS.ID.A.4

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Which of the following statements is (are) true about the t-distribution with k degrees of freedom?

I. The t-distribution is symmetric.

II. The t-distribution with k degrees of freedom has a smaller variance than the t- distribution with k + 1 degrees of freedom.

III. The t-distribution has a larger variance than the standard normal (z) distribution.

I only

II only

III only

I and II

I and III

Tags

CCSS.HSS.ID.A.3

4.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

In an attempt to estimate the mean carbon content of the day's steel output, an engineer chooses a random sample of 15 rods from that day. The process is newly

developed, and there are no historical data on the variability of the process, although graphic displays of the engineer's sample results indicate no heavy skew or outliers. In estimating this day's mean

carbon content, the primary reason the engineer should use a t-confidence interval rather than a z-

confidence interval is because the engineer

is estimating the population mean using the sample mean.

is using the sample variance as an estimate of the population variance.

is using data, rather than theory, to judge that the carbon content is normal.

is using data from a specific day only.

has a small sample, and a z-confidence interval should never be used with a small sample.

Tags

CCSS.HSS.IC.B.4

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

It is appropriate to use a t-distribution with 20 degrees of freedom for which of the following?
I. Constructing a confidence interval from an SRS with n = 19.

II. Doing a hypothesis test where Ho: μ1μ2=0 H_o:\ \mu_1-\mu_2=0\   and Ha: μ1μ2>0H_a:\ \mu_1-\mu_2>0  with n1=8n_1=8  and n2=12n_2=12  .
III. Testing for differences between two proportions where n1=21n_1=21  and n2=29n_2=29  

I only

II only

III only

I, II, and III

None are appropriate

6.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

What is the critical t-value for finding a 96 percent confidence interval estimate from a sample of 18

observations?

2.054

2.205

2.214

2.224

2.235

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A relief fund is set up to collect donations for the families affected by recent storms. A random sample of 400 people shows that 28% of the 200 who were contacted by telephone actually made contributions, compared to only 18% of the 200 who received first class mail requests. Which formula calculates the 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail?

(0.280.18)±1.96(0.23)(0.77)200\left(0.28-0.18\right)\pm1.96\sqrt{\frac{\left(0.23\right)\left(0.77\right)}{200}}

(0.280.18)±1.96(0.23)(0.77)200+(0.23)(0.77)200\left(0.28-0.18\right)\pm1.96\sqrt{\frac{\left(0.23\right)\left(0.77\right)}{200}+\frac{\left(0.23\right)\left(0.77\right)}{200}}

(0.280.18)±1.96(0.23)(0.77)400\left(0.28-0.18\right)\pm1.96\sqrt{\frac{\left(0.23\right)\left(0.77\right)}{400}}

(0.280.18)±1.96(0.28)(0.72)200+(0.18)(0.82)200\left(0.28-0.18\right)\pm1.96\sqrt{\frac{\left(0.28\right)\left(0.72\right)}{200}+\frac{\left(0.18\right)\left(0.82\right)}{200}}

(0.280.18)±1.96(0.28)(0.72)400+(0.18)(0.82)400\left(0.28-0.18\right)\pm1.96\sqrt{\frac{\left(0.28\right)\left(0.72\right)}{400}+\frac{\left(0.18\right)\left(0.82\right)}{400}}

Tags

CCSS.HSS.IC.B.4

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