Question

A customer wants to estimate the average delivery time of a pizza from the local pizza parlor. Over the course of a few months, the customer orders 28 pizzas and records the delivery times. The average delivery time is 20.06 with a standard deviation of 5.271. If the customer estimates the time using a 95% confidence interval, what is the margin of error?

Question 1 options:

	1) 2.0439
	2) 1.6967
	3) 0.9961
	4) 0.6215
	5) 2.0405

Question 2 (1 point)

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 15 customers and find that the average dollar amount spent per transaction per customer is $86.485 with a standard deviation of $15.8647. Create a 99% confidence interval for the true average spent for all customers per transaction.

Question 2 options:

	1) ( 74.291 , 98.679 )
	2) ( 82.389 , 90.581 )
	3) ( 83.508 , 89.462 )
	4) ( -74.291 , 98.679 )
	5) ( 74.413 , 98.557 )

Question 3 (1 point)

The owner of a local phone store wanted to determine how much customers are willing to spend on the purchase of a new phone. In a random sample of 9 phones purchased that day, the sample mean was $338.73 and the standard deviation was $19.7969. Calculate a 95% confidence interval to estimate the average price customers are willing to pay per phone.

Question 3 options:

	1) ( -323.513 , 353.947 )
	2) ( 336.424 , 341.036 )
	3) ( 332.131 , 345.329 )
	4) ( 323.513 , 353.947 )
	5) ( 323.803 , 353.657 )

Question 4 (1 point)

Suppose you work for Fender Guitar Company and you are responsible for testing the integrity of a new formulation of guitar strings. To perform your analysis, you randomly select 40 'high E' strings and put them into a machine that simulates string plucking thousands of times per minute. You record the number of plucks each string takes before failure and compile a dataset. You find that the average number of plucks is 6,886.7 with a standard deviation of 117.85. A 99% confidence interval for the average number of plucks to failure is (6,836.2, 6,937.2). From the option listed below, what is the appropriate interpretation of this interval?

Question 4 options:

	1) We are 99% confident that the proportion of all 'high E' guitar strings fail with a rate between 6,836.2 and 6,937.2.
	2) We are 99% confident that the average number of plucks to failure for all 'high E' strings tested is between 6,836.2 and 6,937.2.
	3) We cannot determine the proper interpretation of this interval.
	4) We are 99% confident that the average number of plucks to failure for all 'high E' strings is between 6,836.2 and 6,937.2.
	5) We are certain that 99% of the average number of plucks to failure for all 'high E' strings will be between 6,836.2 and 6,937.2.

Question 5 (1 point)

The owner of a local golf course wants to determine the average age of the golfers that play on the course in relation to the average age in the area. According to the most recent census, the town has an average age of 23.44. In a random sample of 26 golfers that visited his course, the sample mean was 30.63 and the standard deviation was 8.771. Using this information, the owner calculated the confidence interval of (25.84, 35.42) with a confidence level of 99%. Which of the following statements is the best conclusion?

Question 5 options:

	1) We are 99% confident that the average age of all golfers that play on the golf course is greater than 23.44
	2) We are 99% confident that the average age of all golfers that play on the golf course is less than 23.44
	3) The average age of all golfers does not significantly differ from 23.44.
	4) The percentage of golfers with an age greater than 23.44 is 99%.
	5) We cannot determine the proper interpretation based on the information given.

Question 6 (1 point)

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 34.632 Newtons. In order to estimate the true mean within a margin of error of 9.703 Newtons with 99% confidence, how many samples would need to be tested?

Question 6 options:

	1) We do not have enough information to answer this question since we were not given the sample mean.
	2) 95
	3) 90
	4) 84
	5) 85

Question 7 (1 point)

A pharmaceutical company is testing a new drug to increase memorization ability. It takes a sample of individuals and splits them randomly into two groups. After the drug regimen is completed, all members of the study are given a test for memorization ability with higher scores representing a better ability to memorize. Those 28 participants on the drug had an average test score of 28.396 (SD = 4.142) while those 26 participants not on the drug had an average score of 40.736 (SD = 5.24). You use this information to create a 90% confidence interval for the difference in average test score. What is the margin of error? Assume the population standard deviations are equal.

Question 7 options:

	1) 1.28060017
	2) 1.662
	3) 2.145
	4) 1.67468915
	5) 2.106

Question 8 (1 point)

In a consumer research study, several Meijer and Walmart stores were surveyed at random and the average basket price was recorded for each. It was found that the average basket price for 8 Meijer stores was $132.15 with a standard deviation of $24.701. 11 Walmart stores had an average basket price of $156.97 with a standard deviation of $19.049. Construct a 99% confidence interval for the difference between the true average basket prices (Meijer - Walmart). You can assume that the standard deviations of the two populations are statistically similar.

Question 8 options:

	1) (-50.62, 0.98)
	2) (-53.85, 4.21)
	3) (-34.84, -14.8)
	4) We only have the sample means, we need to know the population means in order to calculate a confidence interval.
	5) (-650.61, 600.97)

Question 9 (1 point)

Independent random samples are taken at a university to compare the average GPA of seniors to the average GPA of sophomores. Given a 95% confidence interval for the difference between the true average GPAs (seniors - sophomores) of (0, 1.13), what can you conclude?

Question 9 options:

	1) We are 95% confident that the difference between the two sample GPAs falls within the interval.
	2) We are 95% confident that the average GPA of seniors is greater than the average GPA of sophomores.
	3) There is no significant difference between the true average GPA for seniors and sophomores.
	4) We do not have enough information to make a conclusion.
	5) We are 95% confident that the average GPA of seniors is less than the average GPA of sophomores.

Question 10 (1 point)

The owner of a local golf course wants to estimate the difference between the average ages of males and females that play on the golf course. He randomly samples 24 men and 21 women that play on his course. He finds the average age of the men to be 37.722 with a standard deviation of 7.091. The average age of the women was 32.214 with a standard deviation of 5.243. He uses this information to calculate a 99% confidence interval for the difference in means, (0.436, 10.58). The best interpretation of this interval is which of the following statements?

Question 10 options:

	1) We are certain that the difference between the average age of all men and all women is between 0.436 and 10.58.
	2) We are 99% confident that the difference between the average age of the men and women surveyed is between 0.436 and 10.58
	3) We do not know the population means so we do not have enough information to make an interpretation.
	4) We are 99% sure that the average age difference between all males and females is between 0.436 and 10.58.
	5) We are 99% confident that the difference between the average age of all men and all women who play golf at the course is between 0.436 and 10.58

no need to show work

Answer 1

I am helping you with the first 4 questions ...

(1)

n = 28    

x-bar = 20.06    

s = 5.271    

% = 95    

Standard Error, SE = s/√n =    5.271/√28 = 0.996125369

Degrees of freedom = n - 1 =   28 -1 = 27

t- score = 2.051830493    

Margin of error = t * SE =     2.05183049297067 * 0.996125368615818 = 2.043880406

Option (a)

(2)

n = 15     

x-bar = 86.485     

s = 15.8647     

% = 99     

Standard Error, SE = s/√n =    15.8647/√15 = 4.096247926

Degrees of freedom = n - 1 =   15 -1 = 14   

t- score = 2.976842734     

Width of the confidence interval = t * SE =     2.97684273395329 * 4.09624792617179 = 12.19388588

Lower Limit of the confidence interval = x-bar - width =      86.485 - 12.1938858754957 = 74.29111412

Upper Limit of the confidence interval = x-bar + width =      86.485 + 12.1938858754957 = 98.67888588

The confidence interval is [74.2911, 98.6789]

Option (a)

(3)

n = 9     

x-bar = 338.73     

s = 19.7969     

% = 95     

Standard Error, SE = s/√n =    19.7969/√9 = 6.598966667

Degrees of freedom = n - 1 =   9 -1 = 8   

t- score = 2.306004133     

Width of the confidence interval = t * SE =     2.30600413329912 * 6.59896666666667 = 15.21724441

Lower Limit of the confidence interval = x-bar - width =      338.73 - 15.2172444088364 = 323.5127556

Upper Limit of the confidence interval = x-bar + width =      338.73 + 15.2172444088364 = 353.9472444

The confidence interval is [323.513, 353.947]

Option (d)

(4)

For this, the answer is Option (d).