Kim wants to determine a 80 percent confidence interval for the true proportion p of high school students in the area who attend their home basketball games. Out of n randomly selected students she finds that that exactly half attend their home basketball games. About how large would n have to be to get a margin of error less than 0.04 for p?

Q: Kim wants to determine a 99 percent confidence interval for the true proportion of high school…

A: From the given information, Margin error (E) = 0.03 Assume sample proportion = 0.5   The objective…

Q: A researcher found a 95% confidence interval for the proportion of college students who believe in…

A:

Q: Dylan wants to determine a 90 percent confidence interval for the true proportion of high school…

A: The random variable is who attend their home basket ball games.We have to find the sample size.This…

Q: Kim wants to determine a 99% confidence interval for the proportion of high school students in a…

A: We have given that,  Population proportion (p) = 0.53, margin of error (E) = 0.03 and confidence…

Q: Kim wants to determine a 80 percent confidence interval for the true proportion of high school…

A: Given: Margin of error, E=0.04 Confidence level=80%

Q: Dylan wants to determine a 95% confidence interval for the true proportion of high school students…

A:

Q: Beth wants to determine a 90% confidence interval for the proportion of high school students in a…

A: We have given that,  Population proportion (p) = 0.57, margin of error (E) = 0.04 and confidence…

Q: have to be to get a margin of error less than 0.02 for p?

A: confidence level=80% Margin of error(E)=0.02

Q: A health magazine is interested in knowing the population proportion of people who eat fast food…

A:

Q: Kim wants to determine a 80 % confidence interval for the true proportion ? of high school students…

A:

Q: Suppose you want to construct a confidence interval with a margin of error of 0.03 (3%). Find the…

A: The formula for calculating sample size is, n = z2*p^(1-p^)MOE2 Where, z is the z score (default…

Q: A marketing researcher wants to find 95% confidence interval for the mean amount that visitors to a…

A: Given Data: The confidence interval is: CI=95% The standard deviation is: σ=11 The mean of the…

Q: A national survey of 1000 adult citizens of a nation found that 19% dreaded Valentine's Day. The…

A:

Q: In a sample of 300 people, 154 said that they watched news on television. Find the 90% confidence…

A: According to the given information in this question We need to find the 90% confidence interval of…

Q: Kim wants to determine a 95 percent confidence interval for the true proportion p of high school…

A: Margin of error =E =0.03

Q: A recent study indicated that 24% of women over 55 are widows. A researcher wishes to estimate with…

A:

Q: Suppose you calculate a 95% confidence interval for a proportion to to be: 0.39 to 0.78. Determine…

A: Given,   Lower bound = 0.39​ Upper bound = 0.78   Here, we use to determine the margin of error.

Q: Beth wants to determine a 99 percent confidence interval for the true proportion p of high school…

A:

Q: Beth wants to determine a 95 percent confidence interval for the true proportion p of high school…

A: Here we need to find the required sample size inorder to construct a 95% confidence interval for the…

Q: A national survey of 1000 adult citizens of a nation found that 18% dreaded Valentine's Day. The…

A: Solution: Given information: n= 1000 Sample size  p^= 0.18  Sample proportion of adult of citizens…

Q: Dylan wants to determine a 99 percent confidence interval for the true proportion p of high school…

A:

Q: Dylan wants to determine a 99 percent confidence interval for the true proportion p of high school…

A:

Q: Beth wants to determine a 80% confidence interval for the true proportion of high school students in…

A: The formula for the confidence interval for population proportion is given by : C.I =…

Q: 3. In a sample of 300 people, 154 said that they watched news on television. Find the 90% confidence…

A: Given : Sample size of people (n) = 300 Let X : No. of people from sample  who said they watched…

Q: uppose you want to estimate the proportion of employed adults who would continue working if they won…

A: given data, sample proportion p^=0.59 2E=0.3E=0.32=1.5CI=0.99α=1-0.99=0.01we have to find out the…

Q: ement below is the best explanation? ere is 90% confidence that 21% of the adult citizens of the…

A: Given n=2500 Pcap=0.21 ME=0.044 Alpha=1-0.90=0.10 90% confidence level

Q: Kim wants to determine a 90 percent confidence interval for the true proportion p of high school…

A: Margin of error =E =0.02

Q: Kim wants to determine a 95 percent confidence interval for the true proportion p of high school…

A: The question is about sample size estimation Given : Confidence level = 0.95 Prop. of students…

Q: A survey of 95 families showed that 40 of the families owned at least one tv. a. Find the 90%…

A: Number of families in a sample n =95 Let X be the number of families owned at least one tv = 40 p =…

Q: Beth wants to determine a 90 percent confidence interval for the true proportion p of high school…

A: The required margin of error (E) is less than 0.04. The sample proportion p^ is 0.50 (because…

Q: At one large college, 660 out of the 1500 students surveyed said they plan to take another science…

A: Given that,   x=660,n=1500 The sample proportion is, Critical values: By using the standard normal…

Q: Assume that a sample is being used to estimate a population mean . If n = 16, = 38, and s = 3, find…

A: Sample size n=16Sample mean=38Standard deviation=3

Q: C. FIND t a n = 15 for the 98% confidence interval for the mean.

A: Given problem is :

Q: A national survey of 1500 adult citizens of a nation found that 23% dreaded Valentine's Day. The…

A: Given that 23% dreaded Valentine’s Day in a sample of 1500 adult citizens and the margin of error…