A survey of several 8 to 11-year-olds recorded the following amounts spent on a trip to the mall:$19.36, $29.95, $7.82, $8.36, $21.07, $14.06, $22.22Construct the 80% confidence interval for the average amount spent by 8 to 11-year-olds on a trip to the mall. Assume the population is approximately normal.Step 1 of 4: Calculate the sample mean for the given sample data. Round your answer to two decimal places.

The amounts spent on a trip to the mall is 19.36,29.95,7.82,8.36,21.07,14.06,22.22

Answered: A survey of several 8 to 11 year olds…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Given the following: Sample Mean = 84.3 Population Variance = 144 Sample size = 35 a. Calculate a 90% confidence interval for the mean. Interpret this interval. b. Calculate a 99% confidence interval for the mean. c. Explain the relationship between Confidence Levels and the resulting intervals.
The coach of a very popular men’s basketball team claims that the average distance the fans travel to the campus to watch a game is 35 miles. The team members feel otherwise. A sample of 16 fans who travel to games was randomly selected and yielded a mean of M= 36 miles and s= 5 miles. Test the coach’s claim at the 5% (.05) level of significance. one-tailed or two-tailed test: State the hypotheses: df= tα or t value for the critical region = sM = t (test statistic)= Decision:
A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $19.26, $21.87, $15.83, $20.57, $19.31, $24.74, $22.82 Construct the 95% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Copy Step 2 of 4: Calculate the sample standard deviation for the given sample data. Round your answer to two decimal places.
The monthly advertising expenditure of the company is normally distributed with astandard deviation of $200. If a sample of 36 randomly selected months yields a meanadvertising expenditure of $1360 monthly, what is a 94% confidence interval for the mean ofthe company’s monthly advertising expenditure?
A survey of 3737 randomly selected homeowners finds that they spend a mean of $73$73 per month on home maintenance with a standard deviation of $10$10 per month. Construct a 90%90% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. Round to the nearest cent.
A sample of 19 elements was used to construct a 99% confidence interval for the population variance, the chi-square value (s) used for this test are: Select one: a. 7.015 to 34.805 b. 7.633 to 36.191 O c. 6.844 to 38.582 d. 6.265 to 37.156 ge Next page ETIVITY NEXT ACTIVITY Jump to. Midterm 1
A survey of 73 randomly selected homeowners finds that they spend a mean of $74 per month on home maintenance. Construct a 95% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. Assume that the population standard deviation is $16 per month. Round to the nearest cent.
2) Construct a 95% confidence interval for the population mean, μ. Assume the population has a normal distribution. A sample of 20 part-time workers had mean annual earnings of $3120 with a standard deviation of $677. Round to the nearest dollar.
A survey of 4646 randomly selected homeowners finds that they spend a mean of $66$66 per month on home maintenance with a standard deviation of $12$12 per month. Construct a 95%95% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. Round to the nearest cent.

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