/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q.121 Refer to the information in Exer... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Refer to the information in Exercise 8.120.

a. Construct three 95% confidence intervals.

i. percent of all Asians who would welcome a white person into their families.

ii. percent of all Asians who would welcome a Latino into their families.

iii. percent of all Asians who would welcome a black person into their families.

b. Even though the three-point estimates are different, do any of the confidence intervals overlaps? Which?

c. For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions?

d. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions?

Short Answer

Expert verified

(a) confidence intervals are

(i)CI=(0.7396,0.8404)

(ii)CI=(0.6539,0.7661)

(iii)CI=(0.6014,0.7186)

(b) Overlapping intervals(0.7396,0.8404)and(0.6539,0.7661)(0.6539,0.7661)and(0.6014,0.7186)

(c) This implies that the Asian adult families can welcome a white or Latino person into their families

(d) This implies that the Asian adult families can welcome a white or Black person into their families

Step by step solution

01

part (a) (i) Explanation

i. per cent of all Asians who would welcome a white person into their families.

p=0.79 - sample proportion

n=251- sample size

Using normal approximation, we get

role="math" localid="1650867070853" npandn(1-p)>5

role="math" localid="1650867986307" ⇒n∗p=198.29>5

role="math" localid="1650867993951" ⇒n*(1-p)=52.71>5

Taking binomial random variable as normally distributed, we get

Standard error is p×(1−p)n=0.025709

Margin error for the confidence interval, ME=0.0504

95%confidence interval = Sample mean ±ME

role="math" localid="1650867580599" ⇒CI=0.79±0.0504

⇒CI=(0.7396,0.8404)

02

part (a) (ii) explanation

ii. per cent of all Asians who would welcome a Latino into their families.

p=0.71- sample proportion

n=251- sample size

Using normal approximation, we get

npandn(1−p)>5.

role="math" localid="1650868008893" ⇒n∗p=178.21>5

role="math" localid="1650868013813" ⇒n×(1−p)=72.79>5

Taking binomial random variable as normally distributed, we get

Standard error= 0.71×(1−0.71)251= 0.028641

Margin error for the confidence interval ME=0.0561

95%confidence interval = Sample mean±ME

⇒CI=0.71±0.0561

⇒CI=(0.6539,0.7661)

03

part (a) (iii)

iii. percent of all Asians who would welcome a black person into their families.

p=0.66- sample proportion

n=251- sample size

Using normal approximation, we get

n×(1−p)=85.34>5

Taking binomial random variable as normally distributed, we get

Standard error = 0.66×(1−0.66)251= 0.0299

Margin error for the confidence interval ME = 0.0586

95%confidence interval = Sample mean ±ME

role="math" localid="1650868650350" ⇒CI=0.66±0.0586

⇒CI=(0.6014,0.7186)

04

part (b) explanation

The overlapping confidence intervals are-(0.7396,0.8404)and(0.6539,0.7661),(0.6539,0.7661)and(0.6014,0.7186)

05

part (c) explanation

We can express that there doesn't appear to be a major differentiation between the extent of Asian adults who tell that their families can invite a white individual into their families and furthermore the extent of Asian adults who say that their families can invite a Latino individual into own families.

06

part (d) explanation

We can express that there doesn't appear to be a major differentiation between the extent of Asian adults who tell that their families can invite a white individual into their families and furthermore the extent of Asian adults who say that their families can invite a Black individual into own families.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5

In words, define the random variable X¯ .

Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample

proportion.

Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8;8;10;7;9;9. Assume the underlying distribution is approximately normal.

a. Construct a 90%confidence interval for the population mean grams of fat per serving of chocolate chip cookies

sold in supermarkets.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

b. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done?

c. Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies.

d. Calculate the mean.

e. Is the mean within the interval you calculated in part a? Did you expect it to be? Why or why not?

Suppose that the insurance companies did do a survey. They randomly surveyed 400drivers and found that 320claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.

a. i. x = __________ ii. n = __________ iii. p′ = __________

b. Define the random variables Xand P', in words.

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 95%confidence interval for the population proportion who claim they always buckle up. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.

e. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.