More SkittlesWhat sample size would be required to reduce the standard deviation of the sampling distribution to one-half the value you found in Exercise 35(b)? Justify your answer.

Short Answer

Expert verified

Sample size is120

Step by step solution

01

Given Information

It is given that sample sizen=30

02

Calculation

The standard deviation of sampling distribution for sample proportion is

p^σp^=p(1-p)n

We need to reduce it to half of the value.

σp^2=12p(1-p)n=12·p(1-p)n=p(1-p)4n

When it is reduced to half, nis replaced by 4n.

It means sample size need to be multiplied by 4.

Hence, 4n=4(30)=120

Required sample size is120

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

Tall girls? To see if the claim made in Exercise 12is true at their high school, an Ap Statistics class chooses an SRS of twenty 16-year-old females at the school and measures their heights. In their sample, the mean height is 64.7inches. Does this provide convincing evidence that 16-year-old females at this school are taller than 64inches, on average?

a. What is the evidence that the average height of all 16-year-old females at this school is greater than 64inches, on average?

b. Provide two explanations for the evidence described in part (a).

We used technology to simulate choosing 250SRSs of size n=20from a population of three hundred 16-year-old females whose heights follow a Normal distribution with mean localid="1654113150676" μ=64inches and standard deviation μ=2.5inches. The dotplot shows x=the sample mean height for each of the 250simulated samples.

c. There is one dot on the graph at 62.5. Explain what this value represents.

d. Would it be surprising to get a sample mean of x=64.7or larger in an SRS of size 20when μ=64inches and σ=2.5inches? Justify your answer.

e. Based on your previous answers, is there convincing evidence that the average height of all 16-year-old females at this school is greater than 64inches? Explain your reasoning.

Suppose we select an SRS of size n=100n=100from a large population having proportion p of successes. Let pp^be the proportion of successes in the sample. For which value of p would it be safe to use the Normal approximation to the sampling distribution of pp^?

a. 0.01

b. 0.09

c. 0.85

d. 0.975

e. 0.999

What does the CLT say? Asked what the central limit theorem says, a student replies, "As you take larger and larger samples from a population, the histogram of the sample values looks more and more Normal." Is the student right? Explain your answer.

Sample proportions List all 6possible SRSS of size n=2, calculate the proportion of red cars in the sample, and display the sampling distribution of the sample proportion on a dotplot. Is the sample proportion an unbiased estimator of the population proportion? Explain your answer.

COLORAGE
RED1
WHITE5
SILVER8
RED20

More sample minimums List all 4possible SRSs of size n=3, calculate the minimum age for each sample, and display the sampling distribution of the sample minimum on a dot plot with the same scale as the dot plot in Exercise 20. How does the variability of this sampling distribution compare with the variability of the sampling distribution from Exercise 20? What does this indicate about increasing the sample size?

From exercise20:

Car NumberColorAge
1
Red1
2
White5
3
Silver8
4
Red20
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