Questions
Two pea plants are crossed. Both are heterozygous for purple blossom color, while one is homozygous...

Two pea plants are crossed. Both are heterozygous for purple blossom color, while one is homozygous for being short and the other is heterozygous for being tall. In pea plants, tall is dominant to short, and purple flowers are dominant to white.

Fill out the table below for the probability of each possible phenotype. Report probability as a decimal rounded to four places (e.g. 0.1250, not 1/8 or 12.5%).

Phenotype Probability
tall purple
short purple
tall white
short white

In a population of 150 pea plants, there are 53 tall-purple plants, 52 short-purple plants, 22 tall-white plants, and 23 short-white plants. In order to test if the two traits are experiencing independent assortment researchers would perform a chi squared test. The (null/alternative)  hypothesis states that the two genes are independently assorted while the (null/alternative)  hypothesis states the two genes are dependent.

What is your calculated Chi Squared statistic?

  • When performing a contingency table, do not round your expected values!!
  • Report your calculated X2 rounded to four decimal places

What is the corresponding P value?

  • Use the formula =1-(CHISQ.DIST(X2,df,TRUE)) to convert calculated X2into a P value
  • Report your answer rounded to 4 decimal places

Do you fail to reject or reject the null hypothesis?

As a result of this statistical analysis, it is possible to conclude that pea plant height and pea plant blossom color (are or are not)  linked traits.

In: Math

The following regression output was obtained from a study of architectural firms. The dependent variable is...

The following regression output was obtained from a study of architectural firms. The dependent variable is the total amount of fees in millions of dollars.

Predictor     

Coef

SE Coef      

T     

P

Constant

7.987

2.967

2.69

-

X1

0.12242

0.03121

3.92

0.0000

X2

-0.12166

0.05353

-2.27

0.028

X3

-0.06281

0.03901

-1.61

0.114

X4

0.5235

0.1420

3.69

0.001

X5

-0.06472

0.03999

-1.62

0.112

Analysis of Variance

Source         

DF

SS    

MS     

F    

P

Regression

5

3710.00

742.00

12.89

0.000

Residual Error

46

2647.38

57.55

Total

51

6357.38

X1 - # of architects employed by the company

X2 - # of engineers employed by the company

X3 - # of years involved with health care projects

X4 - # of states in which the firm operates

X5 - % of the firms work that is health care-related

  1. Write out the regression equation
  2. How large is the sample? How many independent variables are there?
  3. Conduct a global test of hypothesis to see if any of the set of regression coefficients could be different from 0. Use the .05 significance level. What is your conclusion?
  4. Conduct a test of hypothesis for each independent variable. Use the .05 significance level. Which variable would you consider eliminating first?
  5. Outline a strategy for deleting independent variables in this cas

In: Math

According to Harper's Index, 40% of all federal inmates are serving time for drug dealing. A...

According to Harper's Index, 40% of all federal inmates are serving time for drug dealing. A random sample of 16 federal inmates is selected.

(a) What is the probability that 11 or more are serving time for drug dealing? (Round your answer to three decimal places.)

(b) What is the probability that 5 or fewer are serving time for drug dealing? (Round your answer to three decimal places.)

(c) What is the expected number of inmates serving time for drug dealing? (Round your answer to one decimal place.)

In: Math

This case study concerns a bank's efforts to calculate credit risk scores (These are opposite of...

This case study concerns a bank's efforts to calculate credit risk scores (These are opposite of credit scores. Higher the value the riskier the customer) A loan officer at a bank needs to be able to identify characteristics that are indicative of people who are likely to default on loans and use those characteristics to identify good and bad credit risks. The loan officer also needs to be able to better quantify an individual’s credit risk level.

Information on 700 past customers is given in the file along with data for the following variables:

Age: customer age in years.

Employment: years that the customer has been with his/her current employer

Address: years that the customer has lived at his/her current address

Income: household annual income (in $1,000)

Debt _to _Income: debt to income ratio (x100)

Risk _Score: Credit risk score (the higher the score, the more risky)

Default _Indicator: an indicator of whether the customer had previously defaulted

Test the hypothesis that the average credit risk score of customers who have previously defaulted on their loans is higher than those who haven’t defaulted on their loans. (Divide the credit risk score into two groups those who previously defaulted and those who didn’t and then compare the two groups.)

You would like to build a regression model to predict the credit risk score based on all the other variables. What is the regression equation?

What is the predicted credit risk score of a customer aged 40, who has been with their employer for 10 years, has lived at the same address for 3 years, has an income of $200,000, debt to income ratio of 2.5 and has previously defaulted on a loan. Will the prediction be accurate? Explain

Compute the coefficient of determination (R2) and fully interpret its meaning.

Interpret the meaning of the coefficients of the independent variables income and address.

Which independent variables significant? Defend your answer.

Age in years Years with current employer Years at current address Household income in thousands Debt to income ratio (x100) Previously Defaulted Credit Risk Score
41 17 12 176 9.3 1 808.3943274
27 10 6 31 17.3 0 198.2974762
40 15 14 55 5.5 0 10.0361081
41 15 14 120 2.9 0 22.13828376
24 2 0 28 17.3 1 781.5883142
41 5 5 25 10.2 0 216.7089415
39 20 9 67 30.6 0 185.9601084
43 12 11 38 3.6 0 14.70865349
24 3 4 19 24.4 1 748.0412036
36 0 13 25 19.7 0 815.0570131
27 0 1 16 1.7 0 350.309226
25 4 0 23 5.2 0 239.0539023
52 24 14 64 10 0 9.790173473
37 6 9 29 16.3 0 364.4940475
48 22 15 100 9.1 0 11.87390385
36 9 6 49 8.6 1 96.70407786
36 13 6 41 16.4 1 212.0503906
43 23 19 72 7.6 0 1.404870603
39 6 9 61 5.7 0 104.1453903
41 0 21 26 1.7 0 91.9180135
39 22 3 52 3.2 0 4.373536462
47 17 21 43 5.6 0 3.047352362
28 3 6 26 10 0 293.9321797
29 8 6 27 9.8 0 106.7996198
21 1 2 16 18 1 629.7774553
25 0 2 32 17.6 0 861.3134014
45 9 26 69 6.7 0 16.46115799
43 25 21 64 16.7 0 1.437993467
33 12 8 58 18.4 0 276.7066755
26 2 1 37 14.2 0 503.3218674
45 3 15 20 2.1 0 76.41958523
30 1 10 22 10.5 0 433.6994251
27 2 7 26 6 0 288.7388759
25 8 4 27 14.4 0 231.1006843
25 8 1 35 2.9 0 74.95719559
26 6 7 45 26 0 950.0535168
30 10 4 22 16.1 0 211.9564036
32 12 1 54 14.4 0 335.9969153
28 1 8 24 17.1 1 643.9032953
45 23 5 50 4.2 0 2.268753579
23 7 2 31 6.6 0 132.8782071
34 17 3 59 8 0 31.76854323
42 7 23 41 4.6 0 31.90347933
39 19 5 48 13.1 0 28.07933138
26 0 0 14 7.5 1 511.04996
21 0 1 16 6.8 0 453.6168743
35 13 15 35 4.5 0 10.78188877
47 4 2 26 10.4 0 281.6573725
23 0 2 21 11.4 1 621.7847698

In: Math

Two plots at Rothamsted Experimental Station were studied for production of wheat straw. For a random...

Two plots at Rothamsted Experimental Station were studied for production of wheat straw. For a random sample of years, the annual wheat straw production (in pounds) from one plot was as follows.

5.70 7.03 5.84 7.03 7.31 7.18
7.06 5.79 6.24 5.91 6.14

Use a calculator to verify that, for this plot, the sample variance is s2 ≈ 0.411.

Another random sample of years for a second plot gave the following annual wheat production (in pounds).

6.19 7.59 7.66 8.15 7.22 5.58 5.47 5.86

Use a calculator to verify that the sample variance for this plot is s2 ≈ 1.117.

Test the claim that there is a difference (either way) in the population variance of wheat straw production for these two plots. Use a 5% level of signifcance.

(a) What is the level of significance?

State the null and alternate hypotheses.

Ho: σ12 = σ22; H1: σ12 > σ22

Ho: σ12 > σ22; H1: σ12 = σ22    

Ho: σ22 = σ12; H1: σ22 > σ12

Ho: σ12 = σ22; H1: σ12σ22



(b) Find the value of the sample F statistic. (Use 2 decimal places.)


What are the degrees of freedom?

dfN
dfD

What assumptions are you making about the original distribution?

The populations follow dependent normal distributions. We have random samples from each population.

The populations follow independent normal distributions.   

The populations follow independent chi-square distributions. We have random samples from each population.

The populations follow independent normal distributions. We have random samples from each population.


(c) Find or estimate the P-value of the sample test statistic. (Use 4 decimal places.)

p-value > 0.200

0.100 < p-value < 0.200    

0.050 < p-value < 0.100

0.020 < p-value < 0.050

0.002 < p-value < 0.020

p-value < 0.002


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

At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.    

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.


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

Fail to reject the null hypothesis, there is sufficient evidence that the variance in annual wheat production differs between the two plots.

Reject the null hypothesis, there is insufficient evidence that the variance in annual wheat production differs between the two plots.    

Reject the null hypothesis, there is sufficient evidence that the variance in annual wheat production differs between the two plots.

Fail to reject the null hypothesis, there is insufficient evidence that the variance in annual wheat production differs between the two plots.

In: Math

Why are there no negative values in the z table? If you were given an individual’s...

  1. Why are there no negative values in the z table?
  2. If you were given an individual’s raw score and it was above the mean, what steps would you take to determine that individual’s percentile? What if their raw score was below the mean?
  3. If you were given an individual’s z score and it was less than zero, what steps would you take to determine that individual’s raw score?
  4. How is the process of determining percentiles different if you were given a sample mean rather than an individual raw score?

In: Math

A mixture of pulverized fuel ash and Portland cement to be used for grouting should have...

A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compressive strength of more than 1300 KN/m2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with σ = 66. Let μ denote the true average compressive strength.

(a) What are the appropriate null and alternative hypotheses?

H0: μ = 1300
Ha: μ > 1300H0: μ > 1300
Ha: μ = 1300    H0: μ = 1300
Ha: μ ≠ 1300H0: μ < 1300
Ha: μ = 1300H0: μ = 1300
Ha: μ < 1300


(b) Let

X

denote the sample average compressive strength for n = 14 randomly selected specimens. Consider the test procedure with test statistic

X

itself (not standardized). What is the probability distribution of the test statistic when H0 is true?

The test statistic has a gamma distribution.The test statistic has a normal distribution.    The test statistic has a binomial distribution.The test statistic has an exponential distribution.


If

X = 1340,

find the P-value. (Round your answer to four decimal places.)
P-value =

Should H0 be rejected using a significance level of 0.01?

reject H0do not reject H0   


(c) What is the probability distribution of the test statistic when μ = 1350?

The test statistic has an exponential distribution.The test statistic has a normal distribution.    The test statistic has a gamma distribution.The test statistic has a binomial distribution.


State the mean and standard deviation of the test statistic. (Round your standard deviation to three decimal places.)

mean        KN/m2
standard deviation      KN/m2


For a test with α = 0.01, what is the probability that the mixture will be judged unsatisfactory when in fact μ = 1350 (a type II error)? (Round your answer to four decimal places.)

In: Math

Test the claim that the proportion of men who own cats is smaller than the proportion...

Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .10 significance level.

left tailed right tailed two tailed

test statistic

critical value

reject or accept the null

In: Math

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in...

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment,

20

subjects had a mean wake time of

101.0

min. After​ treatment, the

20

subjects had a mean wake time of

92.4

min and a standard deviation of

21.1

min. Assume that the

20

sample values appear to be from a normally distributed population and construct a

99%

confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of

101.0

min before the​ treatment? Does the drug appear to be​ effective?

Construct the

99%

confidence interval estimate of the mean wake time for a population with the treatment.

_ min < u < min

The confidence interval

the mean wake time of

101.0

min before the​ treatment, so the means before and after the treatment

This result suggests that the drug treatment

a significant effect.

First blank: DOES NOT INCLUDE or INCLUDE

Second blank: ARE DIFFERENT or COULD BE THE SAME

Third Blank: HAS or DOES NOT HAVE

In: Math

Why is sampling distribution a theoretical distribution?

Why is sampling distribution a theoretical distribution?

In: Math

A staff psychologist at a police precinct developed a week-long training course designed to improve on...

A staff psychologist at a police precinct developed a week-long training course designed to improve on the job sensitivity of police officers. The psychologist designs a study where some policemen randomly get and complete the course. A month later the psychologist records the number of domestic disputes the police officers successfully resolved from their police reports. What can the psychologist conclude with an α of 0.05. The success data are below.

no
course
course
11.2
12.5
10.6
12.7
8.3
15.6
12.1
14.8
16.3
14.3
17.4
11.2
16.5
15.4


a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- no course police precinct domestic disputes completing the course job sensitivity
Condition 2:
---Select--- no course police precinct domestic disputes completing the course job sensitivity

c) Compute the appropriate test statistic(s) to make a decision about H0.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
p-value =  ; Decision:  ---Select--- Reject H0 Fail to reject H0

d) , compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r2 =  ;   ---Select--- na trivial effect small effect medium effect large effect

e) Make an interpretation based on the results.

Participants that received training had significantly less resolved domestic disputes than those that did not receive training.

Participants that received training had significantly more resolved domestic disputes than those that did not receive training.  

  Participants that received training did not differ significantly on resolved domestic disputes than those that did not receive training.

In: Math

As part of a long-term study of individuals 65 yars of age or older, sociologists and...

As part of a long-term study of individuals 65 yars of age or older, sociologists and physicians at the Institute of Mental Health Research in Ottawa,

investigated the relationship between geographic location and depression. A random of sample individuals, all in reasonably good health, were selected from Victoria, BC; Edmonton, Alberta; Winnipeg, Manitoba; and Halifax, Nova Scotia. Each of the individuals sampled was given a standardized test to measure depressionThe data collected has been provided to you; a higher test score indicates a higher level of depression. A second part of the study considered the relationship between geographic lomtion and depression for individuals 65 years of age or older who had a chronic health condition such as arthritis, hypertension, and/or heart disease. A sample of individuals, all suffering 130111 a chronic illness, were selected fmm Victoria, BC, Edmonton, Albetta; Winnipeg, Manitoba; and Halifax Nova Scotia. Each of the individuals sampled was given a standardized test to measure depression. The data collected has been provided to you; a higher test score indicates a higher level of
depression.

1) Use the appropriate descriptive statistics to summarize the data. Provide your preliminary observations.

2) Develop a 95% confidence interval for the average depression score among healthy people and the chronically ill. Discuss your findings.

3) Perform an analysis of variance to test for any significant differences due to location for healthy people. Use a 0.05 level of significance. Comment on your fmdings.

4) Perform an analysis of variance to test for any significant differences due to location for chronically ill. Use a 0.05 level of significance. Comment on your findings.

Please provide statistics of patients regardless of their health and null and alternative hypothesis for each test.

Healthy Chronically Ill
Victoria Edmonton Winnipeg Halifax Victoria Edmonton Winnipeg Halifax
6.00 5.54 7.56 4.38 5.04 4.40 12.61 9.03
5.39 7.02 9.57 6.06 5.57 4.14 8.44 3.74
6.47 7.56 5.87 5.41 6.76 5.00 8.67 1.45
6.51 4.43 4.15 4.02 7.35 7.55 8.78 11.74
1.58 8.28 10.01 4.90 10.71 8.75 6.49 4.77
5.55 5.19 9.27 6.76 8.90 7.00 7.79 7.80
1.32 6.92 10.29 8.28 6.90 7.56 8.61 6.22
1.62 7.98 4.66 4.20 4.27 9.84 8.46 6.07
5.30 4.23 6.59 4.80 5.25 7.76 6.80 7.85
4.33 3.95 10.72 10.32 3.84 8.92 6.43 8.92
3.58 3.18 6.28 6.35 7.51 10.50 8.26 4.77
8.08 7.48 10.83 4.33 11.80 7.95 7.92 7.75
7.08 6.56 5.55 9.58 14.46 6.41 10.14 6.05
6.06 8.91 4.80 6.85 7.74 10.51 7.46 6.62
6.46 4.34 10.48 3.86 6.12 12.23 7.77 0.21
5.15 8.17 7.84 0.88 5.92 6.84 6.75 8.27
4.92 3.88 6.20 1.91 10.21 4.85 11.23 7.30
6.49 9.27 6.31 3.14 5.80 6.66 4.55 7.66
4.89 2.92 9.60 3.88 10.01 8.22 3.76 6.52
5.84 6.44 6.33 2.38 11.67 5.40 4.21 1.08
7.13 4.81 6.60 9.83 0.95 5.94 11.39 8.24
0.33 6.50 3.39 4.23 4.57 4.32 9.37 7.61
8.03 7.64 5.37 1.07 5.15 3.94 7.19 9.12
7.27 2.70 7.17 5.93 7.57 8.73 6.21 7.52
2.62 3.50 3.87 5.93 8.12 7.08 8.59 4.90
5.69 7.54 6.39 4.97 13.68 8.14 3.76 4.22
4.30 6.47 11.04 5.06 9.28 11.81 9.76 9.31
4.62 1.07 9.88 7.63 7.72 6.91 4.44 10.65

In: Math

Explain how ANOVA technique avoids the problem of the inflated probability of making Type 1 error...

Explain how ANOVA technique avoids the problem of the inflated probability of making Type 1 error that would arise using the alternative method of comparing groups of two at a time using the t-test for independent groups

In: Math

Perform a hypothesis test for population mean: Nearly 30 years ago the mean height for women...

Perform a hypothesis test for population mean: Nearly 30 years ago the mean height for women 20 years old and older was 63.7 inches. A recent random sample of 45 women who are 20 years old and older had a mean of 63..9 inches. Perform a hypothesis test on the following hypotheses: Null Hypothesis - the population mean is equal to 63.7 inches and the Alternate Hypothesis - the population mean is greater than 63.7 inches. Use a level of significance of .10 or 10%. The Standard Deviation for the recent random sample of 45 women was .5 inches.

Please show work

In: Math

The following table shows ceremonial ranking and type of pottery sherd for a random sample of...

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

Ceremonial Ranking Cooking Jar Sherds Decorated Jar Sherds (Noncooking) Row Total
A 81 54 135
B 94 51 145
C 77 77 154
Column Total 252 182 434

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 not 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 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?

Yes

No     


What sampling distribution will you use?

normal

uniform     

binomial

Student's t

chi-square


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 > 0.100

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 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: Math