Questions
A random sample is drawn from a normally distributed population with mean μ = 18 and...

A random sample is drawn from a normally distributed population with mean μ = 18 and standard deviation σ = 2.3.

a. Are the sampling distributions of the sample mean with n = 26 and n = 52 normally distributed?

  • Yes, both the sample means will have a normal distribution.

  • No, both the sample means will not have a normal distribution.

  • No, only the sample mean with n = 26 will have a normal distribution.

  • No, only the sample mean with n = 52 will have a normal distribution.

b. Calculate the probabilities that the sample mean is less than 18.6 for both sample sizes. (Round answers to 4 decimal places.)

n Probability

26 ( _________)

52 ( _________)

In: Statistics and Probability

A laboratory procedure is used to measure the level of cadmium in soil. It is applied...

A laboratory procedure is used to measure the level of cadmium in soil. It is applied to a specimen that has a controlled level of 1.5 μg/g of cadmium. Three measurements are obtained by splitting the specimen into three parts of equal weight and applying the procedure to each part. Here are the results: 1.64, 1.85, and 1.94. (a) Does a 95% confidence interval give an indication that the laboratory procedure may be biased? b. What are your assumptions that you are making to produce the confidence interval? second part of question: For 95% confidence interval of (1.38, 1.92) is given for the mean of a population based on the normal distribution. Re-express this as a 90% confidence interva.

In: Statistics and Probability

The mean height of an adult giraffe is 17 feet. Suppose that the distribution is normally...

The mean height of an adult giraffe is 17 feet. Suppose that the distribution is normally distributed with standard deviation 0.9 feet. Let X be the height of a randomly selected adult giraffe. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(___,______)

b. What is the median giraffe height? _____ft.

c. What is the Z-score for a giraffe that is 19 foot tall?

d. What is the probability that a randomly selected giraffe will be shorter than 17.8 feet tall?

e. What is the probability that a randomly selected giraffe will be between 17.4 and 18.2 feet tall?

f. The 70th percentile for the height of giraffes is ________ft.

In: Statistics and Probability

According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23%...

According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? Why or why not? Test at the 5% level.

In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Nationally 1 in 88 children are diagnosed with ASD ("CDC features -," 2013).
Is there sufficient data to show that the incident of ASD is more in Arizona than nationally? Why or why not? Test at the 1% level.

In: Statistics and Probability

A laboratory procedure is used to measure the level of cadmium in soil. It is applied...

A laboratory procedure is used to measure the level of cadmium in soil. It is
applied to a specimen that has a controlled level of 1.5 μg/g of cadmium. Three measurements
are obtained by splitting the specimen into three parts of equal weight and applying the
procedure to each part. Here are the results: 1.64, 1.85, and 1.94.
(a) Does a 95% confidence interval give an indication that the laboratory procedure may be
biased?
b. What are your assumptions that you are making to produce the confidence interval?

second part of question: For 95% confidence interval of (1.38, 1.92) is given for the mean of a population
based on the normal distribution. Re-express this as a 90% confidence interval

In: Statistics and Probability

According to the Internal Revenue Service, income tax returns one year averaged $1,332 in refunds for...

According to the Internal Revenue Service, income tax returns one year averaged $1,332 in refunds for taxpayers. One explanation of this figure is that taxpayers would rather have the government keep back too much money during the year than to owe it money at the end of the year. Suppose the average amount of tax at the end of a year is a refund of $1,332, with a standard deviation of $725. Assume that amounts owed or due on tax returns are normally distributed.

(a) What proportion of tax returns show a refund greater than $2,200?
(b) What proportion of the tax returns show that the taxpayer owes money to the government?
(c) What proportion of the tax returns show a refund between $140 and $660?

(Round all the z values to 2 decimal places. Round your answers to 4 decimal places.)

(a) P(x > $2,200) =

(b) P(x < 0) =

(c) P($140 ≤ x ≤ $660) =

In: Statistics and Probability

Organization: Freestanding ER Services Title: ER Operations Manager Outline: Multiple Regression: Effect of Age and General...

Organization: Freestanding ER Services

Title: ER Operations Manager

Outline:

  1. Multiple Regression: Effect of Age and General Health Status Score on Total ER Cost.
    1. Purpose
    2. Importance
    3. Variables
    4. Sample Size
    5. Hypothesis
    6. Methodology
    7. Findings
    8. Interpretations/Implications

Description of each Section

Purpose: Identify the purpose of the analysis in this section. What questions do you hope to answer?

Importance: State and explain the importance of this analysis to the department/organization. How will the findings be used?

Variables: List all variables, explain how they will be measured, and how they will be collected.

Sample Size: State your sample size and when data is collected.

Hypothesis: What are your null and alternate hypothesis?

Methodology: State the statistical method used in this section.

Findings: Copy and Paste all relevant output from Excel into this section. State your major findings from the included tables/charts.

Interpretaions/Implications: Discuss what the department/organization needs to do based on the findings.

ID Insurance Location Wait Time Age GHSS Cost
126 Uninsured Moore 60 11 12 $12,000
107 Uninsured Moore 25 13 99 $7,800
110 Uninsured Moore 45 16 13 $478
141 Uninsured West 34 31 1 $135
160 Insurance Moore 2 47 14 $1,200
128 Uninsured Moore 22 47 16 $4,600
166 Insurance Moore 12 49 77 $4,400
121 Insurance Moore 25 52 15 $4,500
120 Insurance West 15 56 67 $4,450
124 Insurance Moore 22 57 61 $1,200
132 Insurance Moore 54 59 16 $1,200
130 Insurance Moore 55 60 25 $1,200
108 Uninsured Moore 10 60 26 $1,365
161 Insurance Moore 56 62 27 $1,200
115 Government Moore 15 63 66 $13,000
163 Insurance Moore 61 66 34 $1,400
158 Uninsured Moore 56 71 45 $1,300
138 Uninsured Moore 15 74 79 $4,900
113 Government Moore 25 78 56 $13,000
139 Uninsured West 13 78 77 $4,850
180 Government Moore 33 79 86 $12,000
182 Government Moore 34 80 57 $900
176 Government Moore 55 85 79 $1,245
177 Government Moore 60 87 49 $678
178 Government Moore 45 89 73 $450
133 Uninsured West 45 89 93 $9,850
149 Uninsured Moore 14 90 88 $4,500
193 Government Moore 34 90 99 $8,700
114 Government Moore 44 91 90 $5,000
102 Government Pelican 20 5 1 $680
165 Uninsured Pelican 11 5 2 $899
109 Uninsured Pelican 89 6 77 $12,000
152 Insurance Pelican 15 6 77 $14,000
140 Insurance Pelican 20 7 11 $9,000
192 Government West 20 7 13 $450
174 Government Pelican 20 7 15 $6,785
155 Insurance Pelican 20 7 24 $850
112 Government West 22 8 26 $450
169 Insurance Pelican 20 8 36 $960
137 Insurance Pelican 25 9 1 $6,000
194 Government West 22 9 11 $450
156 Insurance West 25 10 13 $11,000
197 Government Pelican 23 12 18 $195
143 Insurance Pelican 26 14 66 $650
164 Uninsured West 22 14 89 $4,500
170 Government West 24 15 99 $4,630
135 Insurance Pelican 31 16 14 $9,000
119 Insurance Pelican 34 17 16 $1,200
196 Government Pelican 24 18 19 $1,645
150 Uninsured Pelican 23 18 25 $879
134 Insurance Pelican 36 19 22 $950
185 Government Pelican 26 19 26 $1,200
145 Government West 28 19 88 $13,000
179 Government West 28 22 1 $456
157 Insurance West 36 22 44 $980
181 Government Pelican 29 24 13 $7,100
144 Insurance Pelican 44 24 36 $1,300
100 Uninsured Pelican 45 27 26 $1,500
159 Insurance Pelican 44 27 48 $15,000
131 Insurance Pelican 45 30 79 $1,500
125 Uninsured Pelican 56 30 99 $12,000
186 Government West 36 34 13 $156
136 Insurance Pelican 45 34 99 $4,500
116 Government West 36 36 16 $4,900
148 Insurance West 48 36 89 $14,800
106 Insurance West 55 38 36 $1,356
129 Insurance Pelican 55 43 46 $4,500
190 Government Pelican 36 44 16 $1,200
123 Insurance Pelican 56 44 36 $1,630
142 Uninsured Pelican 61 45 49 $4,680
117 Government Pelican 39 46 36 $4,950
104 Government Pelican 43 47 12 $4,977
154 Government Pelican 44 48 24 $1,200
103 Government West 46 50 56 $5,500
184 Government Pelican 24 51 57 $5,500
189 Government Pelican 46 55 23 $1,300
122 Government West 49 56 66 $1,230
153 Government West 15 56 99 $5,600
191 Government West 15 57 68 $1,340
183 Uninsured Pelican 9 58 63 $1,345
101 Government Pelican 18 59 89 $8,800
151 Insurance Pelican 14 64 89 $5,600
173 Government Pelican 13 66 23 $2,300
172 Government Pelican 55 67 69 $678
146 Government Pelican 14 67 88 $6,600
175 Government Pelican 24 74 37 $1,300
105 Government Pelican 15 74 88 $8,890
188 Government Pelican 4 76 36 $134
187 Government Pelican 3 78 69 $7,400
162 Insurance Pelican 14 88 90 $2,000
147 Government Pelican 13 88 99 $9,450
168 Government Pelican 13 91 73 $8,700
118 Insurance Pelican 13 91 94 $10,000
167 Insurance Pelican 13 93 93 $8,999
127 Insurance Pelican 14 94 74 $550
171 Government Pelican 14 98 74 $15,000
111 Insurance Pelican 12 99 73 $900
197 Uninsured West 55 80 59 $780
197 Government West 21 19 26 $1,450
197 Uninsured West 14 29 88 $8,900
197 Government West 19 36 44 $1,200
197 Uninsured West 15 36 55 $1,300
197 Government West 17 43 99 $900
197 Uninsured West 16 44 46 $4,400
197 Government West 35 45 78 $7,780
197 Uninsured Pelican 19 45 86 $4,465
197 Uninsured West 60 47 23 $1,200
197 Government Pelican 47 48 22 $1,430
197 Government Pelican 14 55 88 $12,800
197 Insured Pelican 10 65 10 $1,200
197 Government West 46 67 67 $650
197 Insured Pelican 21 69 79 $4,458
197 Insured West 22 70 15 $1,200
197 Insured Pelican 27 73 66 $4,600
197 Insured Pelican 28 74 78 $7,748
197 Insured Pelican 36 76 19 $1,200
197 Insured Pelican 25 77 48 $1,400
197 Government Pelican 13 77 79 $12,000
197 Insured Pelican 44 78 79 $9,900
197 Government Pelican 25 78 99 $1,800
197 Insured Pelican 76 81 89 $4,500
197 Insured Pelican 38 82 79 $5,000
197 Government Pelican 19 89 44 $3,000
197 Government Moore 55 89 96 $5,750
197 Insured Moore 37 89 99 $12,000
197 Government Moore 56 90 79 $10,000
197 Insured Moore 44 94 86 $55
51.18898 53.24409 4486.15

In: Statistics and Probability

Calculate the standard error. May normality be assumed? (Round your answers to 4 decimal places.) Standard...

Calculate the standard error. May normality be assumed? (Round your answers to 4 decimal places.)

Standard Error Normality
(a) n = 20, ππ = .21 (Click to select)  No  Yes
(b) n = 48, ππ = .27 (Click to select)  Yes  No
(c) n = 117, ππ = .53 (Click to select)  Yes  No
(d) n = 593, ππ = .002 (Click to select)  Yes  No

In: Statistics and Probability

Question: You have five dice, like in a game of Yahtzee! Suppose you roll the five...

Question:

You have five dice, like in a game of Yahtzee! Suppose you roll the five dice once and sum the numbers the five dice show.

(a) What is the mean and the standard deviation of the sum of five dice?

(b) Suppose you average 60 of such rolls with the five dice. What is the distribution of this average?

(c) What is the chance the average of 60 such rolls is larger than 18?

In: Statistics and Probability

The average American man consumes 9.6 grams of sodium each day. Suppose that the sodium consumption...

The average American man consumes 9.6 grams of sodium each day. Suppose that the sodium consumption of American men is normally distributed with a standard deviation of 0.9 grams. Suppose an American man is randomly chosen. Let X = the amount of sodium consumed. Round all numeric answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N( , )

b. Find the probability that this American man consumes between 9.8 and 10.4 grams of sodium per day.

c. The middle 10% of American men consume between what two weights of sodium? Low: High:

In: Statistics and Probability

Statistical Error: Central tendency Definition: The tendency of managers to give all employees average ratings. This...

Statistical Error: Central tendency Definition: The tendency of managers to give all employees average ratings. This reduces the variance in the data to the point where it becomes useless.

1. Provide a Human Resource decision that has to be made as an example to the central tendency effect.

Errors typically occur because the data used to make the decision is flawed in some way.

What data should be used for this decision?

How would this help avoid the error?

In: Statistics and Probability

Question 24 2 pts (CO7) A restaurant claims the customers receive their food in less than...

Question 24 2 pts (CO7) A restaurant claims the customers receive their food in less than 16 minutes. A random sample of 39 customers finds a mean wait time for food to be 15.8 minutes with a standard deviation of 4.9 minutes. At α = 0.04, what type of test is this and can you support the organizations’ claim using the test statistic? Claim is the alternative, reject the null so support the claim as test statistic (-0.25) is in the rejection region defined by the critical value (-2.05) Claim is the alternative, fail to reject the null so cannot support the claim as test statistic (-0.25) is not in the rejection region defined by the critical value (-1.75) Claim is the null, reject the null so cannot support the claim as test statistic (-0.25) is in the rejection region defined by the critical value (-2.05) Claim is the null, fail to reject the null so support the claim as test statistic (-0.25) is not in the rejection region defined by the critical value (-1.75) Flag this Question

Question 25 2 pts (CO7) A manufacturer claims that their calculators are 6.800 inches long. A random sample of 39 of their calculators finds they have a mean of 6.812 inches with a standard deviation of 0.05 inches. At α=0.08, can you support the manufacturer’s claim using the p value? Claim is the alternative, fail to reject the null and support claim as p-value (0.067) is less than alpha (0.08) Claim is the alternative, reject the null and cannot support claim as p-value (0.134) is greater than alpha (0.08) Claim is the null, reject the null and cannot support claim as p-value (0.067) is less than alpha (0.08) Claim is the null, fail to reject the null and support claim as p-value (0.134) is greater than alpha (0.08)

In: Statistics and Probability

An experimental surgical procedure is being studied as an alternative to the old method. Both methods...

An experimental surgical procedure is being studied as an alternative to the old method. Both methods are considered safe. Five surgeons perform the operation on two patients matched by age, sex, and other relevant factors, with the results shown. The time to complete the surgery (in minutes) is recorded.

Surgeon 1 Surgeon 2 Surgeon 3 Surgeon 4 Surgeon 5
  Old way 39 59 33 43 56
  New way 28 38 20 37 49

%media:2excel.png%Click here for the Excel Data File

(a-1) Calculate the difference between the new and the old ways for the data given below. Use α = 0.025. (Negative values should be indicated by a minus sign.)

X1 X2 X1 - X2
Surgeon Old Way New Way Difference
1 39 28      
2 59 38      
3 33 20      
4 43 37      
5 56 49      

(a-2) Calculate the mean and standard deviation for the difference. (Round your mean answer to 1 decimal place and standard deviation answer to 4 decimal places.)

  
  Mean   
  Standard Deviation   

(a-3) Choose the right option for H0:μd ≤ 0; H1:μd> 0.

  • Reject if tcalc > 2.776445105

  • Reject if tcalc < 2.776445105

(a-4) Calculate the value of tcalc. (Round your answer to 4 decimal places.)

tcalc            

(b-1) Is the decision close? (Round your answer to 4 decimal places.)

The decision is   (Click to select)   close   not close  .

The p-value is  .

(b-2) The new way is better than the old.

  • No

  • Yes

(b-3) The difference is significant.

  • Yes

  • No

In: Statistics and Probability

You are interested in comparing the average, systolic blood pressure of women athletes during intense exercise...

  1. You are interested in comparing the average, systolic blood pressure of women athletes during intense exercise to the healthy, systolic blood pressure of the general population when at rest (i.e. not during exercise) (µ = 120).

Because exercise increases systolic blood pressure, you predict that the average, systolic blood pressure for women athletes during exercise will be significantly greater than the systolic, resting blood pressure (µ = 120). Your alpha level is 0.05. (50 points)

Note: Your test value in SPSS is 120 – not zero.

Sample of women athletes:

Systolic Blood Pressure

125

133

120

113

124

131

123

128

124

120

  1. Are you running a one-tailed or two-tailed test?
  1. Write your alternative and null hypotheses.
  1. Which statistical analysis will you use to run your test (e.g. one-sampled t-test, an independent-samples t-test, a paired t-test, or chi-square test)?
  1. Run your statistical analysis using SPSS. Write your conclusions.

(Remember, if you are running a one-tailed test, your alpha value is located in one-tail, meaning your p-value needs to be less than 0.05 to reject the null hypothesis.

If you are running a two-tailed test, your alpha value is divided in half, meaning your p-value needs to be less than 0.025 to reject the null hypothesis)

In: Statistics and Probability

A simple random sample of size n is drawn from a population that is normally distributed....

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar​, is found to be 111​, and the sample standard​ deviation, s, is found to be 10. ​(a) Construct a 95​% confidence interval about mu if the sample​ size, n, is 26.

In: Statistics and Probability