Price of eggs and milk: The following table presents the average price in dollars for a dozen eggs and a gallon of milk for each month from February through November

2001. Use a TI-84 calculator to answer the following.

If the price of eggs differs by $0.30 from one month to the next, by how much would you expect the price of milk to differ? Round your answer to at least two decimal places.

5. A hand of five cards is drawn without replacement from a standard deck.

(a) Compute the probability that the hand contains both the king of hearts and the king of spades.

(b) Let X = the number of kings in the hand. Compute the expected value E(X). Hint: consider certain random variables X1, . . . , X4.

(c) Let Y = the number of “face” cards in the hand. Given is that E(Y ) = 15/13. Find the variance V ar(Y ). Hint: consider certain random variables Y1, . . . , Y12 and use your result from part (a).

According to data from the city of Toronto, Ontario Canada there were more than 180,00 parking infractions in the city for December 2015, with fines totaling over 8,500,000 Canadian dollars, the fine (in Canadian dollars) for a random sample of 105 parking infractions in Toronto, Ontario Canada in December is listed below.

30, 30 30 30 40 60 40

15 50 150 40 30 30 30

40 30 40 30 30 30 40

40 40 30 60 60 30 150

40 30 250 40 30 30 30

30 30 30 40 30 40 30

50 15 40 40 30 40 30

40 30 30 40 30 30 30

100 30 40 30 30 30 40

30 30 30 40 100 30 40

30 40 30 40 40 40 40

30 30 30 60 30 40 40

30 40 15 60 30 15 150

150 40 40 30 30 160 60

30 40 60 30 40 40 30

Make frequency distribution for this data

An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "Intentional". Each subject was given a list of 50 words. Subjects in the "Sentences" group were told to form multiple sentences, each using at least two words from the list, and to keep forming sentences until all the words were used at least once. Subjects in the "Intentional" group were told to spend five minutes memorizing as many of the 50 words as possible. Subjects from both groups were then asked to write down as many words from their lists as they could recall. We are interested in drawing inference on the difference in the population average number of words recalled for subjects in the "sentences" group vs. subjects in the "intentional" group.

The data is in the table below.

Number of words recalled "Sentences" group 35 34 33 34 35 33 34 35

"Intentional" group 26 33 36 28 29 39 27 33

(For these questions, round all numeric answers to three decimal places)

a. Enter the values for the following statistics:

xsentences =

ssentences =

xintentional =

sintentional =

(xsentences - xintentional) =

standard error of (xsentences - xintentional) =

b. Construct an approximate 95% confidence interval for μsentences - μintentional

Lower bound =

Upper bound =

e. From these results, our statistical conclusion should be: (You have two attempts at this question.)

Fail to reject H0; we have good evidence that μsentences is greater than μintentional, because the 95% CI excludes zero.

Reject H0; have good evidence that μsentences is greater than μintentional, because the 95% CI excludes zero.

Fail to reject H0; we have don't have good evidence about whether μsentences is greater or smaller than μintentional, because the 95% CI excludes zero.

Reject H0; we have don't have good evidence about whether μsentences is greater or smaller than μintentional, because the 95% CI excludes zero.

Fail to reject H0; we have good evidence that μsentences is greater than μintentional, because the 95% CI contains zero.

Reject H0; we have good evidence that μsentences is greater than μintentional, because the 95% CI contains zero.

Fail to reject H0; we have don't have good evidence about whether μsentences is greater or smaller than μintentional, because the 95% CI contains zero.

Reject H0; we have don't have good evidence about whether μsentences is greater or smaller than μintentional, because the 95% CI contains zero.

f. What type of error *might* we have made? (You have two attempts at this question.)

We might have made a Type I error, because a Type I error is failing to reject a false H0.

We might have made a Type I error, because a Type I error is failing to reject a true H0.

We might have made a Type I error, because a Type I error is rejecting a true H0.

We might have made a Type I error, because a Type I error is rejecting a false H0.

We might have made a Type II error, because a Type II error is failing to reject a false H0.

We might have made a Type II error, because a Type II error is failing to reject a true H0.

We might have made a Type II error, because a Type II error is rejecting a true H0.

We might have made a Type II error, because a Type II error is rejecting a false H0.

We might have made either a Type I or a Type II error; both are possible.

We cannot possibly have made a Type I or a Type II error; neither are possible.

Michael measures the volume of several raindrops at Rainyville as 0.6 ml, 0.65 ml, 0.7 ml, 0.75 ml, 0.8 ml and 0.85 ml; and comes to believe that the raindrops at Rainyville are significantly larger than the national average of 0.65ml at the 5% level.

If Michael performs the appropriate statistical test, what will be his conclusion?

1. fail to reject the null hypothesis or 2. reject the null hypotheses

For this question assume that we have a random sample from a normal distribution with

unknown mean but known variance.

(a) Suppose that we have 36 observations, the sample mean is 5, and the population

variance is 9. Construct a 95% confidence interval for the population mean.

(b) Repeat the preceding with a population variance of 25 rather than 9.

(c) Repeat the preceding with a sample size of 25 rather than 36.

(d) Repeat the preceding but construct a 50% rather than 95% confidence interval.

(e) Repeat the preceding but construct a 99% rather than a 50% confidence interval.

An ordinary six-sided die is tossed once and one slip of paper is randomly drawn from a jar containing 5 slips, each of which is lettered U, W, X, Y, or Z. (a) Write the complete sample space that describes all possible outcomes. (b) Determine the probability that the die will show a 6 and that the drawn slip of paper is a U or W. (c) Let A represent the event that the die shows a number greater than 4. Let B represent the event that the drawn slip of paper is not Z. Determine the probability that at least one of these two events occur.

Is it possible to run Levene's test with multiple comparisons in minitab? I am trying to run equal variances, the levene's test, and minitab somehow doesn't like the data. I already made 3 columns in minitab sheet

prompt:

A Datamore Company designs applications software for a number of large firms. A study is conducted to assess the satisfaction of these firms with the supplied software. Datamore decides to determine if the type of industry [Human Resources, Information Technology, Financial, and Insurance] is relevant to satisfaction. Additionally, the company would like to assess the abilities of its sales force [Britknee Phanshear, Megan Hearts, & Jessika Wahlstrom]. They send questionnaires out a number of companies in each of the industries in they do business and ask them to evaluate the company and their contact. The satisfaction scores are found below

So I input the table above in minitab in 3 columns. but when I runt the equal variances test, it comes out an error. Could anyone tell me step by step on how to get the levene's test working?  

In what follows use any of the following tests/procedures: Regression, confidence intervals, one-sided t-test, or two-sided t-test. All the procedures should be done with 5% P-value or 95% confidence interval.

Open Brain data. SETUP: It is believed that the larger the head circumference the larger the brain volume should be. Given the data your job is to confirm or disprove this assertion.

DATA:

CCMIDSA: Corpus Collasum Surface Area (cm2)     FIQ: Full-Scale IQ      HC: Head Circumference (cm)     ORDER: Birth Order      PAIR: Pair ID (Genotype)        SEX: Sex (1=Male 2=Female)      TOTSA: Total Surface Area (cm2) TOTVOL: Total Brain Volume (cm3)        WEIGHT: Body Weight (kg)
8.42    96      57.2    1       6       1       1806.31 1079    61.236
7.44    88      57.2    1       7       1       2018.92 1104    79.38
6.84    85      57.2    1       8       1       2154.67 1439    99.792
6.48    97      57.2    1       9       1       1767.56 1029    81.648
6.43    124     58.5    1       10      1       1971.63 1160    72.576
7.62    101     57.2    2       6       1       1689.6  1173    61.236
6.03    93      57.2    2       7       1       2136.37 1067    83.916
6.59    94      55.8    2       8       1       1966.81 1347    97.524
7.52    114     56.5    2       9       1       1827.92 1100    88.452
7.67    113     59.2    2       10      1       1773.83 1204    79.38
6.08    96      54.7    1       1       2       1913.88 1005    57.607
5.73    87      53      1       2       2       1902.36 1035    64.184
6.22    101     57.8    1       3       2       2264.25 1281    63.958
5.8     103     56.6    1       4       2       1866.99 1051    133.358
7.99    127     53.1    1       5       2       1743.04 1034    62.143
7.99    89      54.2    2       1       2       1684.89 963     58.968
8.76    87      52.9    2       2       2       1860.24 1027    58.514
6.32    103     56.9    2       3       2       2216.4  1272    61.69
6.32    96      55.3    2       4       2       1850.64 1079    107.503
7.6     126     54.8    2       5       2       1709.3  1070    83.009

1. What test/procedure did you perform?

• a. One-sided t-test
• b. Two-sided t-test
• c. Regression
• d. Confidence interval

2. What is the P-value/margin of error?

• a. 9.30595E-20
• b. 0.304733709
• c. 0.022233887
• d. 1.86119E-19
• e. None of these

3. Statistical interpretation

• a. Since P-value is very small we are confident that the average of the first sample is smaller than the other
• b. Since P-value is very small we are very confident that the averages are different.
• c. Since P-value is very small we are confident that the slope of regression line is not zero.
• d. None of these.

4. Conclusion

• a. Yes, I am confident that the above assertion is correct.
• b. No, we cannot claim that the above assertion is correct.

A randomized trial was performed to evaluate the effectiveness of a new drug on controlling Type I diabetes in teenagers. A random sample of 100 patients were obtained from the pediatric diabetes clinic at Sick Kids in Toronto, Ontario; 50 were randomly assigned to the treatment group (new drug) and 50 were randomly assigned to the control group (existing drug). You may assume that basic factors such as validity of the inclusion criteria, blinding, etc. were performed appropriately. Baseline information such as age and gender were collected and key outcomes of A1C level and number of hypoglycemic events were measured after four weeks. A1C levels indicate what percentage of your hemoglobin is coated with sugar (glycated). Higher A1C levels indicate poorer blood sugar control and a higher risk for diabetes complications. A hypoglycemic event occurs when the plasma glucose levels become too low; this is a common and adverse effect of diabetes therapy which has been shown to negatively impact on quality of life.

Put the data into SPSS and find out what if there is a statistical significant difference in A1C levels between the treatment and control groups? Run the appropriate test at the 5% level of significance and decide on a 1-tail or 2-tail test.

I do not know what type of test to run with this data

Instructions: Read each scenario and use the checkbox list to identify the most likely measurement

issue. Briefly explain your choice and propose a solution.2. TheTupelo Housing Program assists economically disadvantaged and homeless individuals to move into safe, healthy, and affordable housing. National service volunteers counsel individuals on their housing needs, help them apply for housing assistance and follow up to provide continued assistance and to verify an individual’s housing status up to 9 months after initial service. National service volunteers find that it is more difficult to track homeless men than women over the 9-month period to verify their housing status, so outcome data are missing for many homeless male clients. What is the most likely measurement problem? ☐Reliability ☐Validity ☐Bias Briefly explain your answer. Briefly, how might the program address this measurement problem?

The following table shows ceremonial ranking and type of pottery sherd for a random sample of 434 sherds at an archaeological location.

Use a chi-square test to determine if ceremonial ranking and pottery type are independent at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Ceremonial ranking and pottery type are independent.
H₁: Ceremonial ranking and pottery type are not independent.H₀: Ceremonial ranking and pottery type are not independent.
H₁: Ceremonial ranking and pottery type are independent.    H₀: Ceremonial ranking and pottery type are not independent.
H₁: Ceremonial ranking and pottery type are not independent.H₀: Ceremonial ranking and pottery type are independent.
H₁: Ceremonial ranking and pottery type are independent.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

binomialchi-square    Student's tnormaluniform

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005

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

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 sufficient evidence to conclude that ceremonial ranking and pottery type are not independent.At the 5% level of significance, there is insufficient evidence to conclude that ceremonial ranking and pottery type are not independent.    

In a recent study, the Centers for Disease Control reported that diastolic blood pressures of adult women in the U.S. are approximately normally distributed with mean 80.8 and standard deviation 9.9.

What proportion of women have blood pressures lower than 70? The proportion is _________

What is the 80th percentile of blood pressures? The 80th percentile is_________

A woman has a blood pressure of 84 mm. What percentile is her blood pressure on? (Round up the final answer to the nearest whole number.) A score of 84 mm is on the____rd percentile, approximately.

A diastolic blood pressure greater than 90 is classified as hypertension (high blood pressure). What proportion of women have hypertension? The proportion is _________

Find the correlation coefficient r for the given table below and use it to determine if there is a Strong Negative or Strong Positive or Weak Negative or Weak Positive correlation. (Check your spelling!) Redshift | 0.0233 |0.0539 | 0.0718 | 0.0395 | 0.0438 | 0.0103
                                   =Distance| 0.32 | 0.75    | 1.00 | 0.55    | 0.61   | 0.14