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Q.11

Expert-verified

The Practice of Statistics for AP

Found in: Page 622

Book edition 4th

Author(s) David Moore,Daren Starnes,Dan Yates

Pages 809 pages

ISBN 9781319113339

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Short Answer

Who uses instant messaging? Do younger people use online instant messaging (IM) more often than older people? A random sample of IM users found that $73$ of the $158$ people in the sample aged $18$ to $27$ said they used IM more often than email. In the $28$ to $39$ age group, $26$ of $143$ people used IM more often than email. $9$ Construct and interpret a $90$ % conﬁdence interval for the difference between the proportions of IM users in these age groups who use IM more often than email.

From the given information, it could be interpreted that there is $90 %$ the probability that the difference in the proportion of IM users lies between $0.1959$ and $0.3641$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

It is given in the question that, $x_{1} = 73, n_{1} = 158, x_{2} = 26, n_{2} = 143$

Step 2: Explanation

The formula to compute the confidence interval for the difference in population proportion is :

$C I = ({\hat{p}}_{1} - {\hat{p}}_{2}) \pm z_{α / 2} \times \sqrt{\frac{{\hat{p}}_{1} (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} (1 - {\hat{p}}_{2})}{n_{2}}}$

The $95 %$ confidence interval using Ti- $83$ plus calculator is computed as:

Therefore, the confidence interval is ( $0.1959, 0.3641$ )

Step 3: Final Answer

Therefore, it could be interpreted that there is $90 %$ probability that the difference in the proportion of IM users lies between $0.1959$ and $0.3641$

Most popular questions for Math Textbooks

School $A$ has $400$ students and School $B$ has $2700$ students. A local newspaper wants to compare the distributions of SAT scores for the two schools. Which of he following would be the most useful for making this comparison?

(a) Back-to-back stemplots for A and B

(b) A scatterplot of A versus B

(c) Dotplots for A and B drawn on the same scale

(d) Two relative frequency histograms of A and B drawn on the same scale

(e) Two frequency histograms for A and B drawn on the same scale

Dropping out You have data from interviews with a random sample of students who failed to graduate from a particular college in $7$ years and also from a random sample of students who entered at the same time and did graduate. You will use these data to compare the percentages of students from rural backgrounds among dropouts and graduates.

A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of $100$ students is randomly assigned to two groups, each of size $50$ . One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, $30$ of Instructor A’s students and $22$ of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?

Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one.

State: I want to perform a test of

$H_{0} : p_{1} - p_{2} = 0$

$H_{a} : p_{1} - p_{2} > 0$

where $p_{1} =$ the proportion of Instructor A's students that passed the state exam and $p_{2} =$ the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use $σ = 0.05$

Plan: If conditions are met, I’ll do a two-sample $z$ test for comparing two proportions.

Random The data came from two random samples of $50$ students.

- Normal The counts of successes and failures in the two groups $- 30, 20, 22$ , and $28 -$ are all at least $10$ .

- Independent There are at least 1000 students who take this driving school's class.

Do: From the data, ${\hat{p}}_{1} = \frac{20}{50} = 0.40$ and ${\hat{p}}_{2} = \frac{30}{50} = 0.60$ . So the pooled proportion of successes is

${\hat{p}}_{C} = \frac{22 + 30}{50 + 50} = 0.52$

- Test statistic

localid="1650450621864" $z = \frac{(0.40 - 0.60) - 0}{\sqrt{\frac{0.52 (0.48)}{100} + \frac{0.52 (0.48)}{100}}} = - 2.83$

- $p$ -value From Table A, localid="1650450641188" $P (z \geq - 2.83) = 1 - 0.0023 = 0.9977$ .

Conclude: The $p$ -value, $0.9977$ , is greater than $α = 0.05$ , so we fail to reject the null hypothesis. There is no convincing evidence that Instructor A's pass rate is higher than Instructor B's.

Refer to Exercise $15$ .

(a) Carry out a significance test at the $α = 0.05$ level.

(b) Construct and interpret a $95 %$ confidence interval for the difference between the population proportions. Explain how the confidence interval is consistent with the results of the test in part (a).

Suppose the true proportion of people who use public transportation to get to work in the Washington, D.C., area is $0.45$ In a simple random sample of $250$ people who work in Washington , what is the standard deviation of the sampling distribution of $\hat{p}$ ?

(a) $0.4975$

(b) $0.2475$

(c) $0.0315$

(d) $0.0009$

(e) $0.617$

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