Question

PART A) You own a small storefront retail business and are interested in determining the average amount of money a typical customer spends per visit to your store. You take a random sample over the course of a month for 14 customers and find that the average dollar amount spent per transaction per customer is $109.477 with a standard deviation of $11.8571. Create a 90% confidence interval for the true average spent for all customers per transaction.

PART B) Researchers at a metals lab are testing a new alloy for use in high end electronics. The alloy is very expensive to make so their budget for testing is limited. The researchers need to estimate the average force required to bend a piece of the alloy to a 90 degree angle. From previous tests, the standard deviation is known to be 48.535 Newtons. In order to estimate the true mean within a margin of error of 12.979 Newtons with 95% confidence, how many samples would need to be tested?

PART C)

The owner of a local golf course wanted to determine the average age of the golfers that played on the course. In a random sample of 22 golfers that visited his course, the sample mean was 48.26 and the standard deviation was 5.207. Using this information, the owner calculated the confidence interval of (46.35, 50.17) with a confidence level of 90% for the average age. Which of the following is an appropriate interpretation of this confidence interval?

1)	We cannot determine the proper interpretation of this interval.

2)	We are 90% confident that the average age of all golfers that play on the golf course is between 46.35 and 50.17.

3)	We are 90% confident that the average age of the golfers surveyed is between 46.35 and 50.17.

4)	We are 90% confident that the proportion of the ages of all golfers is between 46.35 and 50.17.

5)

We are certain that 90% of the average ages of all golfers will be between 46.35 and 50.1

PART D) You are tracking Chipotle Mexican Grill's stock price, attempting to figure out a good price point for entry into the market. You look at 38 random days over the course of a few months and see that the closing price has an average of $292.47 with a standard deviation of $20.356. You construct a 99% confidence interval for the average stock price to be (283.503, 301.437). Of the following choices, what is the approporiate interpretation of this interval?

1)	We are certain that 99% of the daily closing prices will be between 283.503 and 301.437.

2)	We are 99% confident the average daily closing price for any day is between 283.503 and 301.437

3)	We are 99% confident that the proportion of all days that have a closing price between 283.503 and 301.437 is 99%

4)	We are 99% confident the average daily closing price for the days monitored is between 283.503 and 301.437.

5)

We cannot determine the proper interpretation of this interval. PART E) The owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on weekdays and weekends. The owner samples 26 weekdays and finds an average of 272.9 gallons of milk sold on those days with a standard deviation of 39.757. 26 Saturdays and Sundays are sampled and the average number of gallons sold is 393.424 with a standard deviation of 45.345. If a 95% confidence interval is calculated to estimate the difference between the average number of gallons sold on weekdays and weekends, what is the margin of error? Assume both population standard deviations are equal. PART F) It is believed that using a solid state drive (SSD) in a computer results in faster boot times when compared to a computer with a traditional hard disk (HDD). You sample 13 computers with an HDD and note a sample average of 20.81 seconds with a standard deviation of 6.395 seconds. A sample of 10 computers with an SSD show an average of 9.26 seconds with a standard deviation of 2.335 seconds. Construct a 90% confidence interval for the difference between the true average boot times of the two types of hard drives. Assume the difference will represent (HDD-SSD) and that the population standard deviations are statistically the same.

Answer 1

Solutionb:

From previous tests, the standard deviation is known to be 48.535 Newtons. In order to estimate the true mean within a margin of error of 12.979 Newtons with 95% confidence, how many samples would need to be tested?

MOE=E=12.979

z crit dor 95%=1.96

sd=48.535

n=z^2*sigma^2/E^2

=1.96^2*48.535^2/12.979^2

=53.72

n=54

required sample size=n=54

we need to sample 54

Solution2:

confidence interval allples for whole population-all

we are 90% confident that the true population mean of all lies in between 46.35 and 50.17

We are 90% confident that the average age of all golfers that play on the golf course is between 46.35 and 50.17.