The following table shows the length of stay distribution for guests staying at a beach resort, in days. The resort management makes a net profit of $250 per day per guest during the first 2 days of the stay, and $150 per day per guest after the first 2 days. How much profit will the resort owners make in a month (assuming 30 days in a month) if there are 100 guests arriving per day? [Hint: Note that a guest who only stays for two days is billed $500; find the average profit for one guest then work out for the entire month.]

Must be completed in Excel

Days	2	3	4	5	6	7	8	9	10
Prob (%)	5	10	12	12	11	15	14	12	9

Fill in the blank. In a drive thru performance study, the average service time for McDonald's is 203.21 seconds with a standard deviation of 5.67 seconds. A random sample of 90 times is taken. There is a 51% chance that the average drive-thru service time is less than ________ seconds.

1)

203.22

2)	There is not enough information to determine this.

3)

203.2

4)

203.07

5)

203.35

Men at a construction site were moving concrete blocks from a truck, a short distance to the base of a house. The first day the average man moved 62 blocks per hour.

The current day the numbers were : 70,63,76,86,86,62,97,70,77,81

1. Assume all assumptions are met. Provide descriptive statistics for the sample.

2. Conduct a statistical test to assess if the men were able to move more than 62 blocks per hour.

3. The men had more rest than they did the first day and were certain they could move atleast 5 more blocks per hour. Conduct a statistical test to see if they were able to, on average, move 5 more plants per hour than the previous mark of 62

A. Here is a bivariate data set.

x	y
25.7	52.5
29.4	64.7
21.8	54.9
35	63.4
31.4	74.7
21.6	46.5
40.8	69.7
37.9	77.7
16.7	41.1
29.6	78.1
13.1	45.5
36.1	78.4
32.1	68.2
45.2	76.8
36.1	57.9

Find the correlation coefficient and report it accurate to three decimal places.
r =
What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place.
r² = %

B. Based on the data shown below, calculate the correlation coefficient (to three decimal places)

x	y
1	4.97
2	4.04
3	3.51
4	3.78
5	5.15
6	7.22
7	6.69
8	5.76
9	6.73
10	6.6
11	9.77
12	8.24
13	6.91

A claim with an alpha =0.10 and a mu of 20. A sample size of 30 yields a sample mean of 17.5 and a sample standard deviation of 10. What is the upper confidence limit with 3 decimal places?

Distillation is a process for separating and collecting substances according to their reaction to heat. When heat is applied to a mixture, the substance that evaporates and is collected as it cools is the distillate. The unevaporated portion of the mixture is the residue. Oil obtained from orange blossoms through distillation is used in perfume. Suppose the oil yield is normally distributed. In a random sample of eleven distillations, the sample mean oil yield was 980.2 grams with sample standard deviation 27.6 grams. If possible, answer (a) to (d). If not possible, explain.

(a) Find the point estimate.

(b) Find the standard error.

(c) Find the margin of error at the 99% confidence level.

(d) Find and interpret the 99% confidence interval.

Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ² = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s² = 3.0. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence interval for the population variance. (a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 5.1; H₁: σ² ≠ 5.1

H_o: σ² = 5.1; H₁: σ² > 5.1    

H_o: σ² < 5.1; H₁: σ² = 5.1

H_o: σ² = 5.1; H₁: σ² < 5.1

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a normal population distribution.

We assume a uniform population distribution.    

We assume a binomial population distribution.

We assume a exponential population distribution.

(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.100

0.050 < P-value < 0.100    

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis. Since the P-value > α, we reject the null hypothesis.     Since the P-value ≤ α, we reject the null hypothesis. Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that the variance of age at first marriage is less than 5.1.

At the 5% level of significance, there is sufficient evidence to conclude that the that the variance of age at first marriage is less than 5.1.    

(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

lower limit
upper limit

Interpret the results in the context of the application.

We are 90% confident that σ² lies outside this interval.

We are 90% confident that σ² lies below this interval.    

We are 90% confident that σ² lies within this interval.

We are 90% confident that σ² lies above this interval.

0-499	22
500-999	201
1000-1499	1,645
1500-1999	9,365
2000-2499	92,191
2500-2999	569,319
3000-3499	1,387,335
3500-3999	988,011
4000-4499	255,700
4500-4999	36,766
5000-5499	3,994

0-499	22
500-999	201
1000-1499	1,645
1500-1999	9,365
2000-2499	92,191
2500-2999	569,319
3000-3499	1,387,335
3500-3999	988,011
4000-4499	255,700
4500-4999	36,766
5000-5499	3,994

D) Use the normal model to determine the proportion of babies in each class

How do I manually determine the normal mode? Please provide step by step manually (excel is what I am using, however I need to show steps.

Thank you.

6.20. A convenience store owner wants to know how long his customers spend browsing the store before making a purchase. It is found that time spent is normally distributed with an average of m = 5 minutes and a standard deviation of s=2 : 2 minutes. Using a random sample of 14 customers, what is the probability that a customer, on average, will spend less than 4 minutes browsing the store?

1a) Explain why we reject the null hypothesis when the p-value is less than the level of significance?

b) Explain to someone unfamiliar with statistics how to tell whether a statistical test is left, right, or two tailed. Explain what to look for in the wording of a hypothesis test and with the alternate hypothesis.

c) Why can we never truly accept the null hypothesis?

Question 4:

The times that a cashier spends processing each person’s transaction are independent and identically distributed random variables with a mean of µ and a variance of σ² . Thus, if X_i is the processing time for each transaction, E(X _i) = µ and Var(X_i) = σ² .

Let Y be the total processing time for 100 orders: Y = X₁ + X₂ + · · · + X₁₀₀

(a) What is the approximate probability distribution of Y , the total processing time of 100 orders? Hint: Y = 100X, where X = 1 100 P100 i=1 Xi is the sample mean.

(b) Suppose for Z ∼ N(0, 1), a standard normal random variable:

P(a < Z < b) = 100(1 − α)%

Using your distribution from part (a), show that an approximate 100(1 − α)% confidence interval for the unknown population mean µ is:

(Y − 10bσ)/100 < µ < (Y − 10aσ)/100

(c) Now suppose that the population mean processing time is known to be µ = 1.5 minutes, and the population standard deviation processing time is known to be σ = 1 minute. What is the probability that it takes less than 120 minutes to process the 100 orders? If you use R, please provide the commands used to determine the probability. Could you show all steps in the hand written working for this question please.

6. Digital streaming has shifted some of the focus from traditional TV commercials to online advertisements.
The Harris poll reported in 2012 that 53% of 2,343 American adults surveyed said they watched digitally
streamed content. Calculate and interpret a 99% score CI for the proportion of all adult Americans
who watched digitally streamed content.

A husband and wife, Ed and Rina, share a digital music player that has a feature that randomly selects which song to play. A total of 3476 songs have been loaded into the player, some by Ed and the rest by Rina. They are interested in determining whether they have each loaded different proportions of songs into the player. Suppose that when the player was in the random-selection mode, 38 of the first 58 songs selected were songs loaded by Rina. Let p denote the proportion of songs that were loaded by Rina.

(a) State the null and alternative hypotheses to be tested. How strong is the evidence that Ed and Rina have each loaded a different proportion of songs into the player? Make sure to check the conditions for the use of this test. (Round your test statistic to two decimal places and your P-value to three decimal places. Assume a 95% confidence level.)

z = 2.36

P-value = 0.018

Conclusion: There is strong evidence that the proportion of songs downloaded by Ed and Rina differs from 0.5.

(b) Are the conditions for the use of the large sample confidence interval met? Yes, the conditions are met.

If so, estimate with 95% confidence the proportion of songs that were loaded by Rina. Round your answers to 3 decimal places. _____ to ________

A survey of 1935 people who took trips revealed that 181 of them included a visit to a theme park. Based on those survey results, a management consultant claims that less than 10 % of trips include a theme park visit. Test this claim using the ?=0.01 significance level.

(a) The test statistic is

(b) The P-value is

(c) The conclusion is  

A. There is not sufficient evidence to support the claim that less than 10 % of trips include a theme park visit.
B. There is sufficient evidence to support the claim that less than 10 % of trips include a theme park visit.