**Estimating the Mean Amount Spent on Child's Birthday Gift: Confidence Interval Calculation**### Survey InformationIn a recent survey, 33 individuals were asked about the amount of money they spent on their child's last birthday gift. The data collected exhibited a roughly bell-shaped distribution, characterized by the following statistics:- **Mean:** $40- **Standard Deviation:** $3### Confidence Interval CalculationTo provide a reliable estimate of the average amount spent, we will calculate a 95% confidence interval for the mean amount parents spend on their child's birthday gift. We will round our answer to the nearest hundredth where necessary.The formula for a confidence interval for a mean is given by:\[ \text{CI} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]where:- $ \bar{x} $ is the sample mean- $ z $ is the z-score corresponding to the desired confidence level (for 95%, $ z \approx 1.96 $)- $ \sigma $ is the sample standard deviation- $ n $ is the sample size- $ \sqrt{n} $ is the square root of the sample sizeLet’s plug in the values:- $ \bar{x} = 40 $- $ \sigma = 3 $- $ n = 33 $First, calculate the standard error (SE):\[ \text{SE} = \frac{3}{\sqrt{33}} \approx 0.52 \]Next, determine the margin of error (ME):\[ \text{ME} = 1.96 \times 0.52 \approx 1.02 \]Finally, construct the confidence interval:\[ \text{CI} = 40 \pm 1.02 \]\[ \text{CI} = (38.98, 41.02) \]### InterpretationWe can be 95% confident that the true mean amount of money spent on a child's birthday gift by the surveyed population falls between $38.98 and $41.02.This calculation provides valuable insight for parents, retailers, and researchers into typical spending behaviors for children's birthday gifts.

na survey, 33 people were asked how much they spent on their child's last birthday gift. The rest vere roughly bell-shaped with a mean of $40 and standard deviation of $3. alculate, state, and interpret a 95% confidence interval to estimate the mean amount of money arents spend on their child's birthday gift. Round to the nearest 100th where necessary.

na survey, 33 people were asked how much they spent on their child's last birthday gift. The rest vere roughly bell-shaped with a mean of $40 and standard deviation of $3. alculate, state, and interpret a 95% confidence interval to estimate the mean amount of money arents spend on their child's birthday gift. Round to the nearest 100th where necessary.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A survey was conducted that asked 976 people how many books they had read in the past year. Results indicated that x= 14.2 books and s= 17.7 books. Construct a 90% confidence interval for the mean number of books people read Interpret the Construct a 90% confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to two decimal places as needed.) O A. There is 90% confidence that the population mean number of books read is between and O B. If repeated samples are taken, 90% of them will have a sample mean between and OC. There is a 90% probability that the true mean number of books read is between and O Time Remaining: 00:58:23 Next Statcrunch 350 PM O 67°F Sunny ヘ回 12/1/2021 P Type here to search DELL F9 F10 F11 F12 PrtScr Insert Delete F3 F4 F5 F6 F7 F8 Esc F1 F2 @ # $ % & Backspace 2 3 4 6 7 1 W E R Y U Tab Enter A S D J K Caps Lock
It has been reported that 84% of federal government employees use e- mail. If a sample of 235 federal government employees is selected, find the mean, variance, and standard deviation of the number who use e-mail. Round your answers to three decimal places. a. Find mean
A mechanic sells a brand of automobile tire that has a life expectancy that is normally distributed, with a mean life of 35,000 miles and a standard deviation of 2100 miles. He wants to give a guarantee for free replacement of tires that don't wear well. How should he word his guarantee if he is willing to replace approximately 10% of the tires? Click to view page 1 of the table. Click to view page 2 of the table. Tires that wear out by miles will be replaced free of charge. (Round to the nearest mile as needed.)
A particular fruit's weights are normally distributed, with a mean of 334 grams and a standard deviation of 6 grams. The heaviest 11% of fruits weigh more than how many grams? Give your answer to the nearest gram. Check Answer
A survey was conducted that asked 1013 people how many books they had read in the past year. Results indicated that x=10.3 books and s=16.6books. Construct a 90% confidence interval for the mean number of books people read. Interpret the interval. Construct a 90% confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to two decimal places as needed.) A. There is a 90% probability that the true mean number of books read is between ___ and ____ B. There is 90% confidence that the population mean number of books read is between ___ and ____. C. If repeated samples are taken, 90% of them will have a sample mean between ___ and ___.
A city took a survey of 1,550 households to find out how many televisions each owned. The mean was 2.1 televisions, with a standard deviation of 1.7. What should the researchers do to decrease the margin of error? Answer in complete sentences. UPLOAD WRITER C2
In a survey, 19 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $44.7 and standard deviation of $16.8. Estimate how much a typical parent would spend on their child's birthday gift (use a 98% confidence level). Give your answers to 3 decimal places. Express your answer in the format of E. $ + $
Recently, a random sample of 13–18 year olds was asked, "How much do you currently have in savings?" The data in the table represent the responses to the survey. Approximate the mean and standard deviation amount of savings. LOADING... Click the icon to view the frequency distribution for the amount of savings. Savings Frequency $0– $199 343 $200– $399 94 $400– $599 62 $600– $799 21 $800– $999 9 $1000– $1199 5 $1200– $1399 3 The sample mean amount of savings is $nothing. (Round to the nearest dollar as needed.) The sample standard deviation is $nothing. (Round to the nearest dollar as needed.)
A survey was conducted that asked 963 people how many books they had read in the past year. Results indicated that x=14.7 books and s=17.8 books. Construct a 90% confidence interval for the mean number of books people read. Interpret the interval.Construct a 90% confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice.(Use ascending order. Round to two decimal places as needed.)A. If repeated samples are taken, 90% of them will have a sample mean between ___ and ____ B. There is a 90% probability that the true mean number of books read is between ____ and ___ C. There is a 90% confidence that the population mean number of books read is between ___ and ___