A research company desires to know the mean consumption of meat per week among males over age 23. A sample of 246 males over age 23 was drawn and the mean meat consumption was 4.3 pounds. Assume that the population standard deviation is known to be 0.8 pounds. Construct the 95 % confidence interval for the mean consumption of meat among males over age 23. Round your answers to one decimal place.

Q: We notice a mean loss of 7 pounds in a sample of 11 dieters, with a standard deviation of three…

A: We first find the critical value for 95% confidence interval

Q: When the standard deviation used to construct a confidence interval for a population mean increases,…

A: Given data: When the standard deviation used to construct a confidence interval for a population…

Q: A health care prifessional wishes to estimate the birth weights of infants. How large a sample must…

A: Given Information :  A health care professional wishes to estimate the birth weights of infants. He…

Q: A survey of 100 retailers revealed that the mean after-tax profit was $80,000. Assuming that the…

A: Given data,n=100σ=15,000x=80,000z-value at 95% confidence is Zc=1.96

Q: How large a sample should be selected to provide a 95% confidence interval with a margin of error of…

A: Population standard deviation, sigma is 60.The z-critical value for 95% confidence interval is…

Q: The population standard deviation for the height of students from STAT class is three inches. If we…

A: The confidence level is 98%, population standard deviation is 3, and margin of error is 1.

Q: A random sample of 54 four-year-olds attending day care centers provided a yearly tuition average of…

A:

Q: A company that produces detergents wants to estimate the mean amount of detergent in 64-ounce jugs…

A: Standard Deviation = σ = 0.28 Margin of Error = E = 0.06 Confidence Level = 90 Significance…

Q: A boxer would like to know the mean diameter of a navel orange. If the standard deviation is 0.3…

A: Given Standard deviations=0.3 Margin of error=0.1 At 95% confidence z=1.96

Q: A manager is interested in finding the average number of sales per day at a sporting goods store. If…

A: Obtain the sample size. The sample size is obtained below as follows: Use EXCEL Procedure for…

Q: A market researcher, Codie, wants to find a 97% confidence interval for the mean number of dollars…

A:

Q: A market researcher wants to evaluate car insurance savings at a competing company. Based on past…

A: Given that, A market researcher wants to evaluate car insurance savings at a competing company.…

Q: A random sample of 51 four-year-olds attending day care centers provided a yearly tuition average of…

A:

Q: The 92 million Americans of age 50 and over control 50 percent of all discretionaryincome. AARP…

A: a. It is given that the sample mean is $1873 and the sample standard deviation is $550. The sample…

Q: A weekly tabloid reported that the average income of its subscribers is $39,100. Assume this…

A:

Q: A marketing specialist wants to estimate the average amount spent by visitors to an online…

A:

Q: Miss Romero noted that the mean scores of a random samples of 15 Grade- 8 students who had taken a…

A: Given,   Sample mean x¯ = 80.5standard deviation σ = 3Sample size n = 8

Q: An advertising executive wants to estimate the mean amount of time that consumers spend with digital…

A: a) The sample size is 117 and it is obtained below: From the given information, the population…

Q: . A biologist is interested in studying the effect of vitamin supplements on deer mouse body length.…

A:

Q: What sample size would be needed to estimate the average body temperature of humans if you wished to…

A: Given data,E=0.1z-value at 99% confidence is Zc=2.576σ=0.6n=?

Q: A survey of 31 accountants shows the average salary is $65,000 with a population std deviation of…

A: Given : Sample size, n= 31 Mean = $65000 Population standard deviation = 7000

Q: A medical statistician wants to estimate the average weight loss of people who are on a new diet…

A:

Q: For a population with a standard deviation of σ = 5, what is the z-score corresponding to a score…

A: We have to find z score..

Q: A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose…

A:

Q: A group of automotive engineers decided to conduct a study of its school buses and found that for 20…

A:

Q: A medical statistician wants to estimate the average weight loss of people who are on a new diet…

A: Given,standard deviation(σ)=8margin of error(E)=2α=1-0.94=0.06α2=0.03Z0.03=1.8808       (from…

Q: that the scores have a standard deviation of 9. A psych Construct a 95% confidence interval for the…

A: An Existing inventory for a test measuring self-esteemσ=9 indicates that the scores have a standard…

Q: The average age of a sample of 17 South African women who attend yoga classes was found to be 31…

A:

Q: In a survey, 12 people were asked how much they spent on their child's last birthday gift. The…

A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…

Q: A nurse at a local hospital is interested in estimating the birth weight of infants. How large a…

A:

Q: It is known that the population mean for the verbal section of the SAT is 500 with a standard…

A: From the provided information, Sample size (n) = 400 Sample mean (x̅) = 511 Population standard…

Q: In a random sample of 8 cell​ phones, the mean full retail price was $539.00 and the standard…

A: From the available information,   The confidence interval is, C=90%=90/100=0.90 The level of…

Q: The US department of Agriculture claims that the mean annual consumption of tea by a person in the…

A:

Q: A poll of 2500 people shows that 50% approve of a smoking ban in bars and restaurants in Kentucky.…

A: Given, n = 2500 p = 50% = 0.50 At 95% Confidence level, the value of Z is 1.96

Q: Suppose the standard deviation of the losses had been $3600 instead of $1800. What would the larger…

A:

Q: Miss Romero noted that the mean scores of a random sample of 15 Grade 8 students who had taken a…

A: Given: Sample size, n=15 Let PE denote point estimate Sample mean, x¯=80.5sample standard deviation,…

Q: In a survey, 23 people were asked how much they spent on their child's last birthday gift. The…

A:

Q: A recent study found that 61 children who watched a commercial for potato chips featuring a…

A: Denote μ1, μ2 as the population mean weights of potato chips eaten by children who watched the​…

Q: The average daily volume of a computer stock in 2011 was μ=35.1 million​ shares, according to a…

A:

Q: A random sample of 16 employees taken from the management level of the company gives a mean monthly…

A: Given that  Sample size = 16 , Mean = 17000 , Standard deviation = 5500

Q: A medical statistician wants to estimate the average weight loss of people who are on a new diet…

A: Given Data: ME=4/2=2 σ=8   z-value corresponding to 97% of CI=2.17

Q: A sample data set, n = 115, shows human body temperatures with a mean of 99.6ºF. If the standard…

A: Given Data: n=115x¯=99.6σ=0.75   z-value corresponding to 99% Confidence Interval=2.576

Q: A medical statistician wants to estimate the average weight loss of people who are on a new diet…

A:

Q: You are interested in estimating the mean weight of the local adult population of female…

A: It was stated that I am interested in estimating the mean weight of the local adult population of…

Q: The College Board National Office recently reported that in 2011-2012, the547,038 high school…

A: The given information is:

Q: What is the correct conclusion?   A recent study found that 61 children who watched a commercial for…

A:

Q: A recent study of 25 motorists showed that they spent an average of P750 per week. The standard…

A: Given : s

Q: Jessica, the owner of a medium-scale native handicrafts shop, knows from experience that the time…

A:

Q: The 92 million Americans of age 50 and over control 50 percent of all discretionaryincome. AARP…

A: Given: n=80σ=550x=1873 (a) At 95% C.I, the margin of error will be as follows: 95% C.I or 0.95Then,…

Q: A nutritionist claims that the mean tuna consumption by a person is 3.6 pounds per year. A sample…

A: Given that  Claim :  Mean tuna consumption by a person is 3.6 pounds per year Sample size (n) = 90…

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.