A random sample of 28 customers at a gas station shows an average gas
purchase of 8.9 gallons with a standard deviation of 3.2 gallons. Find the
98%
confidence interval estimating the population mean number of gallons
purchased at this station. Interpret this result.
A random sample of 28 customers at a gas station shows an average gas purchase of...
A sample of 64 customers at Ted’s service station shows a mean number of gallons of gasoline purchased per customer to be 13.6 gallons. If σ is 3 gallons, what is the 95% confidence interval of the mean number of gallons purchased per customer?A sample of 64 customers at Ted’s service station shows a mean number of gallons of gasoline purchased per customer to be 13.6 gallons. If σ is 3 gallons, what is the 95% confidence interval of the...
1) The owner of a local gas station would like to estimate the mean number of gallons sold to his customers. Assume that the number of gallons sold follow a normal distribution with a population standard deviation 0f 2.5 gallons. He selects a random sample of 100 sales and finds the mean number of gallons sold equals 9.2 What is the point estimate of the population mean? a. 8.2 b. 2.5 c. 100 d. 9 2) The owner of a...
A random sample of 13 customers in a Kentucky fast food
restaurant revealed an average bill of OMR 18.75 per person. The
population standard deviation is OMR5.4. Estimate 98%
confidence
interval for the mean bill of all customers.
A random sample of ' n' customers in a Kentucky fast food restaurant revealed an average bill of OMR & per person. The population standard deviation is OMR 6. Estimate 98% confidence interval for the mean bill of all customers. n X...
the average selling price of a smartphone purchased by a random sample of 46 customers was $317. Assume the population standard deviation was $32. a) Construct a 95% confidence interval to estimate the average selling price in the population with this sample. b) What is the margin of error for this interval?
1. A random sample of 82 customers, who visited a department store, spent an average of $71 at this store. Suppose the standard deviation of expenditures at this store is O = $19. What is the e 98% confidence interval for the population mean? 2. A sample of 25 elements produced a mean of 123.4 and a standard deviation of 18.32 Assuming that the population has a normal distribution, what is the 90% confidence interval for the population mean? 3....
2. In estimating the mean number of television viewing hours per family per week, a random sample of 400 families yields a mean of 32.6 hours and a standard deviation of 9.9 hours (a) Find a 95% confidence interval for the average number of viewing hours per family per week in the population. Interpret. (b) Find a 99% confidence interval for the average number of viewing hours per family per week in the population. Interpret. (c) Suppose instead only 25...
A random sample of 51 college students shows that they spend, on average, 4.5 hours on video games every night, with a standard deviation of 1.25 hours. Construct a 98% confidence interval to estimate the mean number of hours that college students spend on video games every nigh
A. A random sample of 32 different juice drinks has a mean of 98 calories per serving and a standard deviation of 31.5 calories. Construct a 99% confidence interval of the population mean number of calories per serving, and interpret the 99% confidence interval in 1 sentence: B. A random sample of 50 standard hotel rooms in Philadelphia, PA, has a mean nightly cost of $189.99 and a standard deviation of $35.25. Construct a 95% confidence interval of the mean...
The average selling price of a smartphone purchased by a random sample of 43 customers was $315. Assume the population standard deviation was $34. a. Construct a 95% confidence interval to estimate the average selling price in the population with this sample. b. What is the margin of error for this interval? a. The 95% confidence interval has a lower limit of and an upper limit of. (Round to the nearest cent as needed.) b. The margin of error is....
5. A random sample of 625 customers of the restaurant xyz revealed the average monthly expenses in the amount of € 43.50 with a standard deviation of 19.20. a) Assuming that the dining expenses are normally distributed, construct a 95% confidence interval for the population mean expenses among xyz consumers( keep 2 decimals in your answer) b) How confident are you that the estimated 43.50 will be within 1.50 of the population mean? c) What would your margin of error...