Question 1:

As a store manger you would like to find out the average time it takes to unload the truck which delivers the merchandise for your store. For this purpose, you have taken a random sample of 35 days and found the average unload time to be 257 minutes for the sample. You also found that the standard deviation is 41 minutes for the sample. Find the upper boundary of the 95% confidence interval for the average unload time.

Round the answer to two decimal places.

Question 2:

What is the proportion of customers who buy service plans when they buy a computer? To find out, you surveyed 134 customers and found that 46 purchased the service plan. What is the upper boundary of 90% confidence interval for the population proportion? Use Excel and round your answer to two decimal places. Give the answer as a decimal percent (that is, for 10%, give 0.1)

Consider the following test performed with the level of significance 0.05:

H0: U >_100
HA: U<100

A random sample of size 33 is obtained from a normally distributed population. The population standard deviation is equal to 38.7. The sample mean happened to be 98. For this hypothesis test, what will be the critical value (the relevant z-alpha)? Hint: Watch the sign - the critical value can be negative or positive.

Round the answer to three decimal places.

Question 3:

It is known that the average return on a stock is 28.2% with the standard deviation equal to 21.6%.

A simple random sample of 66 returns is to be selected and the average return of the sample is to be found. You can say that this sample average is a random variable. What is the standard deviation of this random variable?

Round your answer to 2 decimal places.

Question 4:

A poll is to be conducted to find out how many books, on average, Canadians read. You are the person who is to select a random sample and to ask the people of the sample about the number of books they read in the previous year. How many people are needed for your sample to estimate the number of books within 2 books with 90% confidence? Assume that it is known that the standard deviation for number of books read by Canadians is 11.7 books.

Question 5:

For the Student's t-distribution with 15 degrees of freedom, compute t0.04. Use Excel and round the answer to two decimal places.

Question 6:

In 2015, 56% of employed adults reported that mathematical skills were very important to their job. The supervisor of the job placement office at a college thinks that this percentage has increased due to increased use of technology in the workplace. She takes a random sample of 480 employed adults and finds that 297 of them feel that mathematical skills are very important to their job. Is there sufficient evidence to conclude that the percentage of employed adults who feel mathematical skills are very important to their job has increased at 0.05 level of significance?

Please type the solution and the answer. Use the formula editor to enter numeric values and formulas. This question will be marked manually by your professor

Question 7:

A market research study in one Canadian city showed that the mean fare charged by taxi drivers is $21 and the standard deviation is $3.5. A random sample of 15 fares has been selected. What is the probability that the mean fare is between $20 and $23?
Please type the solution and the answer. Use the formula editor to enter numeric values and formulas. This question will be marked manually by your professor.

Question 8:

Based on the analysis of a random sample consisting of 45 months, 96% confidence interval for the average monthly profit is [$100,566; $111,457]. Explain in less than 100 words what this means.

Question 9:

z is drawn from the standard normal distribution and t is drawn from the Student's t-distribution with 7 degrees of freedom. Using Excel, match the pairs.

Reft-tailed probability is 0.089

Right-tailed probability is 0.6

Right-tailed probability is 0.04

Left-tailed probability is 0.3
A. t-number is 0.549
B. z-number is 0.253
C. z-number is -1.35
D. t-number is -2.046

Question 10:

A manager, who works for your company, said at a business meeting:

We have asked employees who would volunteer to participate in the survey. We paid $100 to each employee for the participation, to be fair and compensate for the time spent answering the survey questions. Because of this, the participation was high and we could achieve the sample size of 130. Based on this sample, we found that the confidence interval for the proportion of happy employees in our company is [97.6%, 99.2%], with the confidence level of 95%. Please note that we had to use the Student's t-distribution with 95 degrees of freedom to find this confidence interval.

Thus, our analysis has shown that at least 97.6% of our company's employees are very happy.

What is wrong with the manager's statement? Keep in mind that this question allows multiple correct answers and incorrect answers will result in the mark reduction.

Even if the manager had computed the confidence interval correctly, the manager failed to draw the correct conclusion. The mentioned confidence interval does not mean that at least 97.6% of employees are happy.

The manager must have used 99% confidence level.

The sample which the manager used was biased and therefore it could not be used for the statistical analysis.

The manager used an incorrect probability distribution to find the confidence interval.

The sample size must have been bigger, because one cannot compute the confidence intervals with sample sizes with fewer than 150 items.

Not mentioning other errors, the manager used the incorrect number of degrees of freedom.

Select the true statements, using Excel. There may be more than one true statement. If you select a wrong statement, the mark will be reduced for this question. So, please do not guess.

If standard deviation of the population is known to be 21 and the sample size is 68, the margin of error is 4.38, rounded to two decimal places. The level of confidence is 91%.

If standard deviation of the population is known to be 15 and the sample size is 57, the margin of error is 4.08, rounded to two decimal places. The level of confidence is 96%.

If standard deviation of the sample is known to be 29 and the sample size is 6, the margin of error is 39.84, rounded to two decimal places. The level of confidence is 98%.

If standard deviation of the sample is known to be 17 and the sample size is 15, the margin of error is 7.95, rounded to two decimal places. The level of confidence is 93%.