The research firm conducted a survey to determine the average mean amount of money smokers spend...

The research firm conducted a survey to determine the average mean amount of money smokers spend on cigarettes per week. A sample of 19 smokers reveals a mean and standard deviation of $20 and $5 respectively. find the intercept the 99% confidence interval for the mean.

Solutions

Expert Solution

Solution: n = 19 , x̄ = 20 , s = 5, df = n-1 = 18

C = 99% , using t distribution table , tc = 2.878

Margin of error = E = tc(s/√n) = 2.878(5/√19) = 3.3013

Lower Limit = x̄ - E = 20 - 3.3013 = 16.6987

Upper Limit = x̄ - E = 20 + 3.3013 = 23.3013

The 99% confidence interval for the mean

(16.6987 , 23.3013)

orchestra answered 1 year ago

Next > < Previous

Related Solutions

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes...

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. They found the distribution of amounts spent per week, followed the normal distribution with the standard deviation of $5. A sample of 49 steady smokers reveal that mean=20. a). What is the point estimate of the population mean? explain what it indicates b). using the 95% level of confidence. determine the confidence interval for sample mean. explain what it indicates

6. A research firm conducted a survey to determine the mean amount steady smokers spend on...

6. A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. They found that the distribution of amounts spent per week followed a normal distribution with a standard deviation of $5. A sample of 49 steady smokers revealed a mean of $20 a) What is the point estimate for the population mean? b) Using the 95% level of confidence, determine the confidence interval for the population mean. c) Determine the 85%...

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes...

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. The population mean is $25 and the population standard deviation is $15. What is the probability that a sample of $100 steady smokers spend between $24 and $26? a. 0.495 b. 0.252 c. 0.248 d. 0.748

A research team conducted a survey in which the subjects were adult smokers. Each subject in...

A research team conducted a survey in which the subjects were adult smokers. Each subject in a sample of 200 was asked to indicate the extent to which he or she agreed with the statement. “I would like to quit smoking”. The results were as follows. Strongly Agree: 102 Agree: 30 Disagree: 60 Strongly disagree: 8 Can one conclude on the basis of these data that, in the sampled population, opinions are not equally distributed over the four levels of...

1. The mean amount of money that U.S adults spend on food in a week is...

1. The mean amount of money that U.S adults spend on food in a week is $140 and standard deviation is $42. Random samples of size 40 are drawn from this population and the mean of each sample is determined. a. Find the mean and standard deviation of the sampling distribution of sample means. b. What is the probability that the mean amount spent on food in a week for a certain sample is more than $143? c. What is...

According to a recent survey conducted at a local college, we found students spend an average...

According to a recent survey conducted at a local college, we found students spend an average of 19.5 hours on their smartphone per week, with a standard deviation of 3.5 hours. Assuming the data follows the normal distribution. a) How many percent of students in this college spend more than 15 hours on their smartphone per week? b) If we randomly select 12 students, what is the probability that the average of these students spending on the internet is more...

5. Scenario: A survey is conducted to determine the average number of people living in a...

5. Scenario: A survey is conducted to determine the average number of people living in a household in Scranton. An interviewer asks a sample of 16 Scranton shoppers, selected randomly in the Viewmont Mall (no two living in the same household), how many people live in their household. The results are as follows: 1 1 1 2 2 2 2 2 2 3 3 3 4 4 5 6 Assume the distribution of the number in all households is...

A movie theater manager believes the average amount of money customers spend on snacks is over...

A movie theater manager believes the average amount of money customers spend on snacks is over $15. To test the claim, a theater employee tracks the spending of 25 randomly selected customers. The sample mean is $17.08 with a sample standard deviation of $4.81. a) What is the Null Hypothesis and the Alternative Hypothesis for the manager’s claim? b) Assuming the population is normal, should we use the Normal z-test or the Student’s t-test?

The average amount of money that people spend at Don Mcalds fast food place is $7.9500...

The average amount of money that people spend at Don Mcalds fast food place is $7.9500 with a standard deviation of $1.8200. 43 customers are randomly selected. Please answer the following questions, and round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of X? X ~ N(,) _____ What is the distribution of ¯x? ¯x ~ N(,) _____ What is the distribution of ∑x? ∑x ~ N(,) _____ What is the...

The average amount of money that people spend at Don Mcalds fast food place is $7.8000...

The average amount of money that people spend at Don Mcalds fast food place is $7.8000 with a standard deviation of $1.8800. 17 customers are randomly selected. Please answer the following questions, and round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of X ? X ~ N(,) What is the distribution of x ? x ~ N(,) What is the distribution of ∑ x? ∑ x ~ N(,) What is...

Subjects