Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

State the estimated distribution of P.Construct a 92%Confidence Interval for the true proportion of girls in the ages 8to12beginning ice-skating classes at the Ice Chalet.

Short Answer

Expert verified

The result is 92%Confidence Interval for the true proportion of girls in the ages 8to 12beginning iceskating classes at the lce Chalet is 0.7217<p<0.8783.

Step by step solution

01

Given 

p^=x/nis the point estimate of p, where x is the number of favour able events and n is the total number of events.

02

Explanation part 1

Given that there were 64girls and 16boys in that class.

So , x=64andn=64+16=80.

So that, p^=64/80=0.8.

For a sample size of 80, the mean of the sampling distribution of the sample proportion is,

μp^=p=0.8.

The distribution of p^is approximately normal with mean 0.8and standard deviation of 0.044721359.

For the population proportion, pthe 100(1-α)%confidence interval is:

p^±zα2p^(1-p^)n

the sample size is n, and for level of significance αthe standard normal distribution critical value is zα2.

Considerp, the proportion of students accepted into law school from the general population, and n, the sample size.

The confidence level is92%.

The level of significance is α=1-0.92=0.08. Hence, α/2=0.04.

03

Explanation part 2

If of the observations must lie within an interval, the remaining must lie outside the interval.

Due to symmetry, of the population will be above the top limit, while the remaining will be below the lower limit of the interval.

Thus, the upper limit of the interval is such that, lie below it.

As a result,

The confidence interval is,

Cl=p^±za2p^(1p^)n=0.8±(1.751)(0.8)(10.8)80=(0.8±(1.751)(0.044721359))=(0.8±0.0783071)(0.7217,0.8783)

the result of confidence Interval for the true proportion of girls in the ages to beginning iceskating classes at the lce Chalet is.

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 Vaia!

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

The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let X=the age of a Winter Foothill College student.

role="math" localid="1648388299408" x¯=_____

Out of a random sample of 65freshmen at State University, 31students have declared a major. Use the “plus-four” method to find a 96% confidence interval for the true proportion of freshmen at State University who have declared a major.

Of 1,050randomly selected adults, 360identified themselves as manual laborers, 280identified themselves as non-manual wage earners, 250identified themselves as midlevel managers, and 160identified themselves as executives. In the survey, 82%of manual laborers preferred trucks, 62%of non-manual wage earners preferred trucks, 54%of mid-level managers preferred trucks, and 26%of executives preferred trucks

The sampling error given in the survey is ±2%. Explain what the ±2% means.

The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal.

Identify the following:

a. x¯ = _____

b. σ = _____

c. n = _____

A sample of 20heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2pounds with a standard deviation of 0.1pounds. The population standard deviation is known to be 0.2pounds.

What would happen if 40 heads of lettuce were sampled instead of 20, and the confidence level remained the same?

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.

Sign-up for free