Question

In: Math

A farmer randomly samples 600 eggs and finds that 28 are bad. Find a 95% confidence...

A farmer randomly samples 600 eggs and finds that 28 are bad. Find a 95% confidence interval for the actual proportion of bad eggs.

Expert Solution

Solution :

Given that,

n = 600

x = 28

Point estimate = sample proportion = = x / n = 28 / 600 = 0.047

1 - = 1 - 0.047 = 0.953

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_{/ 2} * $\sqrt$ (( * (1 - )) / n)

= 1.96 * ( $\sqrt$ ((0.047*0.953) /600 )

= 0.017

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.047 - 0.017< p <0.047 + 0.017

0.030< p <0.064

The 95% confidence interval for the population proportion p is : (0.030 , 0.064)

milcah answered 8 months ago

A farmer focusing on the production of eco-friendly chicken eggs collects the following data about his output. In a sample of 50 eggs, the farmer finds the average

1. Chicken Eggs- from module 5 slides A farmer focusing on the production of eco-friendly chicken eggs collects the following data about his output. In a sample of 50 eggs, the farmer finds the average egg to weigh 47 grams. The standard deviation of the egg weight is 2 grams and the distribution of weights resembles a normal distribution reasonably closely. The farmer can sell the eggs to a local distributor. However, they have to be in the interval between 44 grams and...

Construct a 95% confidence interval for mu 1minusmu 2. Two samples are randomly selected from each...

Construct a 95% confidence interval for mu 1minusmu 2. Two samples are randomly selected from each population. The sample statistics are given below. n 1 equals 50, n 2 equals 60, x overbar 1 equals25, x overbar 2 equals23, sigma 1 equals1.5, sigma 2 equals1.9

Find a 95% confidence interval for the difference of two means when you have independent samples,...

Find a 95% confidence interval for the difference of two means when you have independent samples, both of size 36, from two populations, in which the variances are not known to be equal. In this case, n1=36, s12=8.2, the first sample mean is equal to 7.9; n2 =36, s22=10.4, the second sample mean is equal to 9.1 Interpret the interval once you find it. Namely, can you conclude one mean is greater than the other mean and if so, which...

1. If n=28, x¯ (x-bar)=50, and s=6, find the margin of error at a 95% confidence...

1. If n=28, x¯ (x-bar)=50, and s=6, find the margin of error at a 95% confidence level (use at least two decimal places) 2. What is the margin of error for a poll with a sample size of 2400 people? Round your answer to the nearest hundredth of a percent. % 3. If you want a poll to have a margin of error of 3.24%, how large will your sample have to be? Round your answer to the nearest whole...

A store randomly samples 603 shoppers over the course of a year and finds that 142...

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail. A 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694) The store management can be 95% confident that any sample of shoppers would have a proportion of visitors whose visit is because...

Given the follow data for two pooled samples, calculate %95 confidence interval for each mean, 95%...

Given the follow data for two pooled samples, calculate %95 confidence interval for each mean, 95% confidence interval for difference of means, and one-tailed p-value for the null hypothesis, “Y not greater than X”. a. sample X: n=12, mean = 20.0, std. deviation = 3.1 sample Y: n = 12, mean = 22.0, std. deviation = 3.1 b. Same as part a, except the mean of Y is 23.0 c. Same as part a, except the standard deviation of Y...

An ecologist randomly samples 19 plants of a specific species and measures their heights. He finds...

An ecologist randomly samples 19 plants of a specific species and measures their heights. He finds that this sample has a mean of 15 cm and a standard deviation of 4 cm. If we assume that the height measurements are normally distributed, find a 90% confidence interval for the mean height of all plants of this species. Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary,...

Confidence Intervals (Proportions) 1 Find the margin of error and 95% confidence interval for the following...

Confidence Intervals (Proportions) 1 Find the margin of error and 95% confidence interval for the following surveys. Round all answers to 2 decimal places. (a) A survey of 500 people finds that 56% plan to vote for Smith for governor. Margin of Error (as a percentage): Confidence Interval: % to % (b) A survey of 1500 people finds that 47% support stricter penalties for child abuse. Margin of Error (as a percentage): Confidence Interval: % to % 2 Assume that...

A 95% confidence interval for a difference in proportions p1-p2 if the samples have n1=60 with...

A 95% confidence interval for a difference in proportions p1-p2 if the samples have n1=60 with p^1=0.69 and n2=60 with p^2=0.56, and the standard error is SE=0.09.

If n=600 and ˆp=0.1, construct a 95% confidence interval about the population proportion. ROUND ANSWER TO...

If n=600 and ˆp=0.1, construct a 95% confidence interval about the population proportion. ROUND ANSWER TO THREE DECIMAL! Preliminary: Is it safe to assume that n≤0.05 of all subjects in the population? Yes No Verify nˆp(1−ˆp)≥10 = ? ROUND ANSWER TO ONE DECIMAL. 3. nˆp(1−ˆp)= ? 4. Confidence Interval: What is the 95% confidence interval to estimate the population proportion? ROUND ANSWER TO 3 DECIMAL PLACES!!!!! 5. ??????? < p < ???