Use the given data set to complete parts​ (a) through​ (c) below.​ (Use

α=​0.05.)

Using the linear correlation coefficient found in the previous​ step, determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables. Choose the correct answer below.

A.There is sufficient evidence to support the claim of a linear correlation between the two variables.

B.There is insufficient evidence to support the claim of a nonlinear correlation between the two variables.

C.There is sufficient evidence to support the claim of a nonlinear correlation between the two variables.

D.There is insufficient evidence to support the claim of a linear correlation between the two variables.

c. Identify the feature of the data that would be missed if part​ (b) was completed without constructing the scatterplot. Choose the correct answer below.

A.

The scatterplot reveals a perfect​ straight-line pattern, except for the presence of one outlier.

B.

The scatterplot does not reveal a perfect​ straight-line pattern, and contains one outlier.

C.

The scatterplot does not reveal a perfect​ straight-line pattern.

D.

The scatterplot reveals a perfect​ straight-line pattern and does not contain any outliers.

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 99 with 38 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

90% C.I. =

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Diameters of mature Jeffrey Pine Trees are normally distributed with a mean diameter of 22 inches and a standard deviation of 6 inches. (Round your answers to four decimal places)

a. Find the probability that a randomly selected Jeffrey Pine tree has a diameter greater than 30 inches.

b. Find the probability that a randomly selected Jeffrey Pine tree has a diameter no more than 15 inches.

c. Find the tree diameter that defines the 60th percentile.

d. If 20 Jeffrey Pine trees are randomly selected find the probability that their mean diameter is at least 24 inches

e. If 55 Jeffrey Pine trees are randomly selected find the probability that their mean diameter is at between 22 and 24 inches.

If Punxsutawney Phil sees his shadow on Feb. 2, six more weeks of winter weather lay ahead; no shadow indicates an early spring. Phil, a groundhog, has been forecasting the weather on Groundhog Day for more than 120 years, but just how good is he at his job?

Hypothesis tests: Does Punxsutawney Phil predict the weather accurately? I believe Punxsutawney Phil’s accurate predictions will be less than 50%. From a sample of 100 data values we find that Phil predicted the weather accurately 36 times. At the 0.05 level of significance, can we conclude that Phil will predict the weather less than 50% of the time?

H₀:

H_a:

α=

Decision Rule:

Critical Value = ___________

Standardized Test Statistic = p - p0 /p(1-p)/n =.

At the _________ level of significance there __________ reason to believe ___________________

__________________________________________________________________

Confidence Interval: Construct a 95% confidence interval about the true portion.

We are _____ confident that Punxsutawney Phil will predict the weather between

__________ and ___________ percent of the time.

There are many definitions of probability. The main one is that it is a measure of the likelihood of occurrence of events and has a value between 0 and 1. The closer the value is to 1, the more likely the event is....So, an event with probability 0 that mean it will never occur.

In a random experiment, the set of all possible outcomes are defined as the sample space and sometimes can be listed and sometimes can't. Sometimes our interest is limited to certain outcomes that make a certain event. For example, if you roll 2 dice, there are 36 possible outcomes in total. But, you may be interested in the event of getting a sum of 10, rather than all 36 possible outcomes that make up the sample space. In this case, if you denote the event by A, then A= the event of getting a sum of 10 consists of three outcome, in particular A = { (4,6), (5,5), (6,4)}. Assuming that the dice are fair (not biased), then P(A) = The probability of the event A =#(A)/#(S) = 3/ 36 and so on. If you let the event B be at least one of the dice is 4,in this case   B ={ (1,4), (2,4), (3,4),(4,4), (5,4) ,(6,4), (4,1),(4,2), (4, 3),(4,5), (4,6)},

P(B)= 11/36, P(A and B) = 2/36, and P(A union B) =12/36. So, Find P(A given B)= P(A|B) = P(A and B)/ P(B) =(2/36)/ (11/36) =2/11, which is more than twice P(A). Right?

Obviously, A and B are not mutually exclusive but Are they independent? (Hint: Just look at the example and the definition of independence)

In general, if you have 2 events A and B, what does it mean for these 2 events to be independent? How about mutually exclusive? Can they be both? Explain.

It sampled 40 customers in San Francisco and 50 customers in San Diego to assess potential demand.

On a scale of 1-7 (7 = very likely to buy), San Diego customers had a mean of 3.5 with a standard deviation of 1.1. SF customers had a mean of 4.1 with a standard deviation of 2.3.

Are these markets statistically different?

1.Compute standard error

2.Compute t-calc  

3.Compare |t-calc| to 1.96 (95% confidence in our results) and 2.58 (99% confidence)

4.If |t-calc| > 1.96, reject the null with 95% confidence

●

Standard error for 2 means (sxs_x ̅ ) = ?12?1+?22?2√((s_1^2)/n_1 +(s_2^2)/n_2 )

T-calc for 2 means = ?1−?2??(x ̅_1-x ̅_2)/s_x ̅

Complete the following ANOVA summary table for a two-factor fixed-effects ANOVA, where there are five levels of factor A (school) and four levels of factor B (curriculum design). Each cell includes 13 students.

Suppose babies born in a large hospital have a mean weight of 4088 grams, and a variance of 55,696

If 128 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would be greater than 4132 grams? Round your answer to four decimal places.

1. In a group of 15 ballpoint pens on a shelf int he stationery department of Office Max. 2 are known to be defective. If a costumer selects 3 of these pens at random, what is the probability that:

a.) At least 1 is defective?

b.) No more than 1 is defective?

2. In a manufacturing plant, three machines, A, B, and C, produce 40%, 35%, and 25%, respectively, of the total production. The company's quality-control department has determined that 1% if the items produced by Machine A, 1.5% of the items produced by Machine B, and 2% of the items produced by Machine C are defective. If an item is selected at random and found to be defective, what is the probability by Machine B? (Hint: draw a tree diagram)

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 11 ounces. Use Table 1 in Appendix B.

a. The process standard deviation is 0.1, and the process control is set at plus or minus 2 standard deviations. Units with weights less than 10.8 or greater than 11.2 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

In a production run of 1000 parts, how many defects would be found (to the nearest whole number)?

b. Through process design improvements, the process standard deviation can be reduced to 0.08. Assume the process control remains the same, with weights less than 10.8 or greater than 11.2 ounces being classified as defects. What is the probability of a defect (to 4 decimals)?

In a production run of 1000 parts, how many defects would be found (to the nearest whole number)?

c. What is the advantage of reducing process variation?

Suppose you are constructing a 95% confidence interval for the mean of a single sample, whose population standard deviation is known to be σ = 5. You calculate the sample size with some specified width (error) E.

(a) Reducing your confidence level to 80%, and reducing your original width (error) E by a third ( 1 3 ), how much bigger will the new sample size be compared to the first sample size above? (Hint: find the scaled size using algebra).

b) Suppose instead that your increase the sample size by a factor of 10 and you allow the confidence level to be 85%, how will the width (error) have scaled in size compared to the original width (error) E?

The breaking strength of yarn supplied by two manufacturers is being investigated. We know from experience with the manufacturers’ processes that σ1 = 5 psi and σ2 = 4 psi. A random sample of 20 test specimens from each manufacturer results in x¯1 = 88 psi and x¯2 = 91 psi.

(a) Show all steps (including how to find cutoff value) and construct a two-sided 90% Confidence Interval for the true difference in means. Draw a conclusion on whether yarn 1 and yarn 2 are significantly different using only the confidence interval constructed.

(b) Show all steps (including how to find cutoff value) and construct a two-sided 95% Confidence Interval for the true difference in means. Draw a conclusion on whether yarn 1 and yarn 2 are significantly different using only the confidence interval constructed.

4. A small airport has flights to only two cities, A and B. Suppose they get an average of 40 customers per
hour who want to fly to city A and 30 customers per hour who want to fly to city B. If these are independent
Poisson processes, then find the probability that
a) (3 pts) there are 7 or more customers who want to fly to city A in the next 6 minutes. Give your answer
to three decimal places.
b) (3 pts) 5 out of the next 8 customers want to fly to city A. Give your answer to three decimal places.
c) (3 pts) if 15 customers who want to fly to city B arrive in the next 30 minutes, then find the probability
that exactly four of them arrived in the first 5 minutes. Give your answer to three decimal places.

Suppose a rehabilitation psychologist has developed a new job skills training program for people who have not been able to hold a job. Of the 14 people who agree to be in the study, the researcher randomly picks seven of these volunteers to be in a experimental group who will go through the special training program. The other seven volunteers are put in a control group who will go through an ordinary job skills training program. After finishing the training program (of whichever type), all 14 are placed in similar jobs. A month later, each volunteer’s employer is asked to rate how well the new employee is doing using a 9- point scale where a score of 1 indicates “very poor” and 9 indicates “Excellent.” The following ratings for the 14 employees are given below.

New Training Program: 6 4 9 7 7 3 6

Ordinary Training Program: 6 1 5 3 1 1 4

A. Identify the independent variable and its levels; also identify the dependent variable and its measurement scale.

B. State the null hypothesis for the study described.

C. Use Excel to conduct the appropriate statistical test for the study described (p=.05). Be sure to properly state your statistical conclusion regarding the null hypothesis.

D. Provide an interpretation of your answer in part C.

E. What type of statistical error might you have made in part C?

A certain retail store bases its staffing on the number of customers that arrive during certain time slots. Based on prior experience this store expects 32% of its customers between 8:00 am and 12:00 pm; 21% of its customers between 12:00 pm and 4:00 pm; 35% of its customers between 4:00 pm and 8:00 pm; and 12% of its customers between 8:00 pm and midnight. On a certain day, the store had 214, 198, 276, and 134 customers in those time slots, respectively. Should the store change its staffing? (Consider an alpha of 0.05.)

Solution:

Ho: The expected values match the observed values
Ha: The expected values do not match the observed values assign(“exp”,c(32,21,35,12)) assign(“obs”,c(214,198,276,134))
sum((obs-exp)^2/exp) = 5426.773
1-pchisq(5426.773,3) = 0
p < alpha, therefore RHo: the store should change its staffing.

What was wrong with this solution?