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 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 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% 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 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 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:
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 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 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 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 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 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 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
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 |
(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. 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