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AP Statistics with 6 Practice Tests

Martin Sternstein

Chapter 7

Inference for Quantitative Data: Means - all with Video Answers

Educators

Section 26

Quiz 26

01:45

Problem 1

Most recent tests and calculations estimate at the $95 \%$ confidence level that mitochondrial Eve, the maternal ancestor to all living humans, lived $138,000 \pm 18,$ ooo years ago. What is meant by "95\% confidence" in this context?
(A) A confidence interval of the true age of mitochondrial Eve has been calculated using $z$ -scores of ±1.96 .
(B) A confidence interval of the true age of mitochondrial Eve has been calculated using $t$ -scores consistent with $d f$ $=n-1$ and tail probabilities of $\pm 0.025 .$
(C) There is a 0.95 probability that mitochondrial Eve lived between 120,000 and 156,000 years ago.
(D) If 20 random samples of data are obtained by this method and a $95 \%$ confidence interval is calculated from each, the true age of mitochondrial Eve will be in 19 of these intervals.
(E) Of all random samples of data obtained by this method, $95 \%$ will yield intervals that capture the true age of mitochondrial Eve.

Bryan Meares
Bryan Meares
Numerade Educator
01:47

Problem 2

A confidence interval estimate is determined from the GPAs
of a simple random sample of $n$ students. All other things being equal, which of the following will result in a smaller margin of error?
(A) A smaller confidence level
(B) A larger sample standard deviation
(C) A smaller sample size
(D) A larger population size
(E) A smaller sample mean

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator
01:42

Problem 3

One gallon of gasoline is put into each of an SRS of 30 test autos, and the resulting mileage figures are tabulated with $\bar{x}$ $=28.5$ and $s=1.2 .$ Determine a $95 \%$ confidence interval estimate of the mean mileage of all comparable autos.
(A) $28.5 \pm 2.045(1.2)$
(B) $28.5 \pm 2.045\left(\frac{1.2}{\sqrt{29}}\right)$
(C) $28.5 \pm 2.045\left(\frac{1.2}{\sqrt{29}}\right)$
(D) $28.5 \pm 1.96\left(\frac{1.2}{\sqrt{29}}\right)$
(E) $28.5 \pm 1.96\left(\frac{1.2}{\sqrt{29}}\right)$

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator
01:11

Problem 4

What sample size should be chosen to find the mean number of absences per month for school children to within ±0.2 at a
$95 \%$ confidence level if it is known that the standard
deviation is $1.1 ?$
(A) $n \leq \sqrt{\frac{0.2}{1.96 \times 1.1}}$
(B) $n \leq \sqrt{\frac{0.2}{1.96 \times 1.1}}$
(C) $n \leq\left(\frac{1.96 \times 1.1}{0.2}\right)^{2}$
(D) $n \leq \sqrt{\frac{0.2}{1.96 \times 1.1}}$
(E) $n \leq\left(\frac{1.96 \times 1.1}{0.2}\right)^{2}$

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator
01:50

Problem 5

In a study aimed at reducing developmental problems in lowbirth-weight babies (under 2500 grams), 347 infants were exposed to a special educational curriculum while 561 did not receive any special help. After 3 years, the children exposed
to the special curriculum showed a mean IQ of 93.5 with a standard deviation of 19.1; the other children had a mean IQ
of 84.5 with a standard deviation of $19.9 .$ Find a $95 \%$ confidence interval estimate for the difference in mean IQs of all low-birth-weight babies who receive special intervention and those who do not.
(A) $(93.5-84.5) \pm 1.97 \sqrt{\frac{(19.1)^{2}}{347}+\frac{(19.9)^{2}}{561}}$
(B) $(93.5-84.5) \pm 1.97\left(\frac{19.1}{\sqrt{307}}+\frac{19.9}{\sqrt{561}}\right)$
(C) $(93.5-84.5) \pm 1.97 \sqrt{\frac{(19.1)^{2}}{347}+\frac{(19.9)^{2}}{561}}$
(D) $(93.5-84.5) \pm 1.65\left(\frac{19.1}{\sqrt{37}}+\frac{19.9}{\sqrt{561}}\right)$
(E) $(93.5-84.5) \pm 1.65 \sqrt{\frac{(19.1)^{2}+(19.9)^{2}}{347+561}}$

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator
02:08

Problem 6

Nine subjects, 87 to 96 years old, were given 8 weeks of progressive resistance weight training. Strength before and after training for each individual was measured as maximum weight (in kilograms) lifted by left knee extension:
$$
\begin{array}{llllllllllll}
& & & & & & & & & & \text { Mean } & \text { \& } \\
\text { Before: } & 3 & 3.5 & 4 & 6 & 7 & 8 & 8.5 & 12.5 & 15 & 7.5 & 4 \\
\text { After: } & 7 & 17 & 19 & 12 & 19 & 22 & 28 & 20 & 28 & 19.1 & 6 . \\
\text { Difference: } & 4 & 13.5 & 15 & 6 & 12 & 14 & 19.5 & 7.5 & 13 & 11.6 & 4 \\
\hline
\end{array}
$$
Find a 95\% confidence interval estimate for the mean strength gain of all 87 - to 96 -year-olds who are given the training.
(A) $11.6 \pm 3.03$
(B) $11.6 \pm 3.69$
(C) $11.6 \pm 3.76$
(D) $11.6 \pm 5.18$
(E) $11.6 \pm 5.70$

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator
01:31

Problem 7

A company manufactures a synthetic rubber bungee cord with a braided covering of natural rubber and a minimum breaking strength of $450 \mathrm{~kg}$. If the mean breaking strength of a sample drops below a specified level, the production process is halted and the machinery inspected. Which of the following would result from a Type I error?
(A) Halting the production process when too many cords break
(B) Halting the production process when the breaking strength is below the specified level
(C) Halting the production process when the breaking strength is within specifications
(D) Allowing the production process to continue when the breaking strength is below specifications
(E) Allowing the production process to continue when the breaking strength is within specifications

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator
01:22

Problem 8

A coffee-dispensing machine is supposed to deliver 12 ounces of liquid into a large paper cup, but a consumer believes that the actual amount is less. As a test, he plans to obtain a
random sample of 5 cups of the dispensed liquid and, if the mean content is less than 11.5 ounces, to reject the 12 -ounce
claim. If the machine operates with a known standard deviation of 0.9 ounces, what is the probability that the
consumer will mistakenly reject the 12-ounce claim even though the claim is true? (Assume that all conditions for
inference are met.)
(A) $P\left(t>\frac{11.5-12}{0.9 / \sqrt{5}}\right)$
(B) $P\left(t<\frac{11.5-12}{0.9 / \sqrt{5}}\right)$
(C) $P\left(z>\frac{11.5-12}{0.0 / \sqrt{5}}\right)$
(D) $P\left(z>\frac{11.5-12}{0.9 / \sqrt{5}}\right)$
(E) $\left(\begin{array}{l}5 \\ 2\end{array}\right)(0.4)^{2}(0.6)^{3}$

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator
01:34

Problem 9

A fast-food chain advertises that its large bag of french fries has a weight of 150 grams. Some high school students, who enjoy french fries at every lunch, suspect that they are
getting less than the advertised amount. With a scale borrowed from their physics teacher, they weigh a random sample of 15 bags. What is the conclusion if the sample mean
is $145.8 \mathrm{~g}$ and standard deviation is $12.81 \mathrm{~g}$ ? (Assume that all conditions for inference are met.)
(A) There is sufficient evidence to prove the fast-food chain advertisement is true.
(B) There is sufficient evidence to prove the fast-food chain advertisement is false.
(C) The students have sufficient evidence to reject the fastfood chain's claim.
(D) The students do not have sufficient evidence to reject the fast-food chain's claim.
(E) There is not sufficient data to reach any conclusion.

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator
01:53

Problem 10

Given an experiment with $H_{0}: \mu=35, H_{\mathrm{a}}: \mu<35,$ and a possible correct value of $32,$ you obtain a sample statistic of $\sigma_{\bar{x}_{1}-\bar{x}_{2}}$. After doing analysis, you realize that the sample size $n$ is actually larger than you first thought. Which of the following results from reworking with the increase in sample size?
(A) Decrease in probability of a Type I error; decrease in probability of a Type II error, decrease in power
(B) Increase in probability of a Type I error, increase in probability of a Type II error, decrease in power
(C) Decrease in probability of a Type I error; decrease in probability of a Type II error; increase in power
(D) Increase in probability of a Type I error, decrease in probability of a Type II error, decrease in power
(E) Decrease in probability of a Type I error; increase in probability of a Type II error; increase in power

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator
01:35

Problem 11

Thirty students volunteer to test which of two strategies for taking multiple-choice exams leads to higher average results. Each student flips a coin, and if heads, uses Strategy A on the
first exam and Strategy $\mathrm{B}$ on the second, while if tails, uses Strategy B on the first exam and Strategy A on the second. The average of all 30 Strategy A results is then compared to the average of all 30 Strategy B results. What is the conclusion at the $5 \%$ significance level if a two-sample hypothesis test, $H_{0}: \mu_{1}=\mu_{2}, H_{\mathrm{a}}: \mu_{1} \neq \mu_{2},$ results in a $P$ -value of
$0.18 ?$
(A) The observed difference in average scores is significant.
(B) The observed difference in average scores is not significant.
(C) A conclusion is not possible without knowing the average scores resulting from using each strategy.
(D) A conclusion is not possible without knowing the average scores and the standard deviations resulting from using each strategy.
(E) A two-sample hypothesis test should not be used here.

Jerrah Biggerstaff
Jerrah Biggerstaff
Numerade Educator