5
BIOL 375 – Exam 2 – FA2017 – Puedes Hacerlo:______
Instructions:
- You may use a hand-calculator for this test, but not a cell phone or laptop or PDA or ... .
- You may use a four-sided “cheat” sheet & distribution tables only.
- All steps must be clearly shown for computational problems.
1. (10 points)
The murder rates for U.S. cities has a mean of 20.8 and a standard deviation of 14.6 ...
(Statistical Abstract of the United States)
a)
Do you believe the distribution of murder rates could be well approximated by a normal distribution?
Why or Why Not
Draw Distribution – Normal Model Would Not Model The Rates Well, Notice Substantial Likelihood Below 0
b)
One of these cities, Washington, D.C. had murder rate of 75.2 ... Find its z-score.
If the distribution were roughly normally distributed, would this be an unusually high value? Why or Why Not
Z = (75.2 – 20.8) / 14.6 = 3.726 = 3.73 WoW! Very High! … nearly 4 stdev from mean! P(Z <= 3.73) approx. 1
c) What is the murder rate of a city that has a z-score of 0.0?
z = 0 … the rate is 0 stdev from the mean … so the murder rate is 20.8
2. (10 points)
Suppose that serum level of protein-bound iodine is normally distributed …
… with mean 50 mg/liter and standard deviation of 12 mg/liter.
a) What is the probability that the serum level is more than 35 mg/liter?
z = (35 – 50) / 12 = -1.25 ---
P(Z < -1.25) = 0.1056 giving probability more than 35 mg/liter = 1 – 0.1056 = .8944
b) What is the probability that the serum level is between 65 mg/liter and 83 mg/liter?
z = (83 – 50) / 12 = 2.75 --- P(Z < 2.75) = 0.9970
z = (65 – 50) / 12 = 1.25 --- P(Z < 1.25) = 0.8944
giving probability between 65 mg/liter and 83 mg/liter = 0.9970 - 0.8944 = 0.1026
3. (10 points)
It is reported that Americans consume an average of m = 115 pounds of fat per year with a standard deviation of s = 40 pounds and that a normal curve adequately describes how the amount of fat consumed varies.
a) What is the 96th percentile for this distribution?
P(Z < 1.75) = 0.9599 giving 115 + 1.75(40) = 185 pounds of fat as 96th percentile
b)
A random sample of n = 100 Americans is to be taken and the sample mean will be calculated.
What is the probability that the sample mean will be between 115 and 124 pounds?
By CLT Sample Mean Is Normal Model … mean = 115 and stdev = 40 / sqrt(100) = 4
z = (124 – 115) / 4 = 2.25 --- P(Z < 2.25) = 0.9878
z = (115 – 115) / 4 = 0.00 --- P(Z < 0.00) = 0.5000
giving probability sample mean will be between 115 and 124 pounds = 0.9878 – 0.5000 = 0.4878
4. (10 pts)
The developers of a training program designed to improve manual dexterity claim that people who complete the 6-week program will increase their manual dexterity. A random sample of 12 people enrolled in the training program was elected. A measure of each person’s dexterity on a scale from 1 (lowest) to 9 (highest) was recorded just before the start of and just after the completion of the 6-week program. The experiment resulted in the following data and analysis …
Person / A / B / C / D / E / F / G / H / I / J / K / LAfter
Program / 7.8 / 5.9 / 7.6 / 6.6 / 7.6 / 7.7 / 6.0 / 7.0 / 7.8 / 6.4 / 8.7 / 6.5
Before Program / 6.7 / 5.4 / 7.0 / 6.6 / 6.9 / 7.2 / 5.5 / 7.1 / 7.9 / 5.9 / 8.4 / 6.5
Paired T-Test for After - Before
N Mean StDev SE Mean
After 12 7.133 0.861 0.248
Before 12 6.758 0.887 0.256
Difference 12 0.375 0.367 0.106
T-Test of mean difference = 0 (vs > 0): T.S. T-Value = _____ P-Value = 0.002
Does the data provide evidence to indicate the mean manual dexterity got people who have completed the 6-week training program has significantly increased? Test using …
State the hypotheses, Ho & Ha
Ho – mean dexterity after = mean dexterity before
Ha – mean dexterity after > mean dexterity before
Calculate the value of the test statistic: ______Report the p-value: ______
TS t = (0.375 – 0) / 0.106 = 3.538
P-Value = 0.002
State a conclusion in the context of the problem – Justify Your Answer.
P-Value = 0.002 is small … below 0.1 … results unusual if Ho is true … Reject Ho
Sufficient evidence to indicate the manual dexterity is increased, on average, after the training program .
5. (10 pts)
A state highway patrol was interested in knowing whether frequent patrolling of highways substantially reduces the number of speeders. Two similar interstate highways were selected for the study – one heavily patrolled and the other only occasionally patrolled. After 1 month, random samples of 100 cars were chosen on each highway, and the number of cars exceeding the speed limit was recorded. This process was repeated on 5 randomly selected days. The data and a
signed-rank analysis are given below …
Day / Highway 1Heavily Patrolled / Highway 2
Occasionally Patrolled / Difference
Highway1 – Highway2
1 / 35 / 60 / -25 5(-)
2 / 36 / 40 / -4 2(-)
3 / 25 / 48 / -23 4(-)
4 / 54 / 38 / 16 3(+)
5 / 63 / 62 / 1 1(+)
Wilcoxon Signed Rank Test: diff
Test of median = 0.000000 versus median < 0.000000
N for Wilcoxon Estimated
N Test Statistic P Median
diff 5 5 + = 4 - = 11 0.209 -4.50
Fill in the blank in the above output and state a conclusion in context at the 10% significance level – Justify Your Answer.
P-Value 0.209 is not small … not below 0.1 … NEETR Ho
Insufficient evidence to indicate patrolling of highways reduces the number of speeders …
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6. (10 pts)
A study of ammonia levels near the exit ramp of a highway tunnel resulted in daily ammonia concentrations in ppm on eight randomly selected days during afternoon drive-time. Use the following data and sign test analysis to determine if the median daily ammonia concentration exceeds 1.5 ppm …
1.37 1.41 1.42 1.51 1.51 1.52 1.53 1.55 1.58 1.60 1.61 1.63
Sign Test for Median: PPM
Sign test of median = 1.500 versus > 1.500
N Below Equal Above P Median
PPM 12 __3__ __0__ __9__ 0.0730 1.525
Fill in the blanks in the above output and state a conclusion in context at the 5% significance level – Justify Your Answer.
P-Value = 0.0730 is not small … not below 0.05 … NEETR Ho
Insufficient evidence to indicate the median daily ammonia concentration exceeds 1.5 ppm.
7. Short Answer ... (10 pts)
How Inferences Are Made ... Mr. A and Mr. B ...
Mr. A and Mr. B are performing an experiment to test Ho: m = 50 vs. Ha: m ¹ 50.
Mr. A states “we should reject Ho if the P-value is small (perhaps, less than 0.05).”
Mr. B counters “NO WAY!!, we should reject Ho if the P-value is large (perhaps, greater than 0.95).”
Who is correct Mr. A or Mr. B? Justify/Explain your answer.
Define any statistical terms you might use.
Mr. A … reject Ho when results are unusual … when results are inconsistent with Ho.
Results unusual if improbable … improbable when probability P-Value is small.
The experiment was conducted and the sample mean, , turned out to be 50.
Mr. A states “LOOK!!, the sample mean, , is 50; the P-value will be large near 1.”
Mr. B counters “NO WAY!!, the sample mean, , is 50; the P-value will be small near 0.”
Who is correct Mr. A or Mr. B? Justify/Explain your answer.
Mr. A … note the observed equals the expected … so results very consistent with Ho.
P-Value will be large … close to 1.
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
An alien has just arrived to planet Earth from a far away planet which LACKS VARIATION.
(No Dice, No Cards, No Color, The Only Coin Has A *Head* On Each Side, ... you get the idea).
The alien wants to understand the “big picture” of Statistics.
The alien has already encountered Mr. A and Mr. B, but they were of no help because they can never agree.
Can you describe in a few sentences the role Statistics plays on planet Earth?
Statistical Inference
The process of making guesses about the truth from a sample.
Statistics recognizes the presence of variation in data and variation in sample to sample results.
Resulting in the use of probabilistic tools to quantify associated risk and uncertainty.
8. (10 pts)
The media reported on research that claimed to show a link between morning sickness during early pregnancy and the sex of the baby. Suppose a follow-up study is done, in which 100 expectant mothers are classified according to whether they experienced severe, moderate, or mild morning sickness. The table below shows the results.
Baby Girl / Baby BoySevere / 14 (20)(45)/100 = 9 / 6 (55)(20)/100 = 11
Moderate / 27 (60)(45)/100 = 27 / 33 (60)(55)/100 = 33
Mild / 4 (20)(45)/100 = 9 / 16 (55)(20)/100 = 11
Perform The Appropriate Chi-Square Test.
a) State the null hypothesis in words.
Ho: No Association B/W between morning sickness during early pregnancy and the sex of the baby.
Ho: YES Association B/W between morning sickness during early pregnancy and the sex of the baby.
b) Incomplete Output Is Given Below --- Fill In The 2 Blanks.
DF / Test StatisticValue / P-Value
Chi-Square / (3-1)(2-1) = 2 / 10.1 / 0.006
[(14-9)^2]/9 + … [(16-11)^2]/11 = 10.1
d) Decide whether to reject the null hypothesis at the 5% significance level.
P-Value = 0.006 is small … below 0.05 … results unusual if no association … reject
e) State the conclusion in plain language (no statistics jargon).
Evidence to indicate an association between morning sickness during early pregnancy and the sex of the baby.
9. (10 pts)
The pulse rate (in bpm) of a random sample of 30 Peruvian Indians was collected.
The mean pulse rate of the sample was 70.2, with a sample standard deviation of 10.51.
Compute and Interpret an appropriate 95% confidence interval for the population mean pulse rate …
df = 30 – 1 = 29 --- 95% confidence gives critical value = 2.045
sample mean = 70.2 sample stdev = 10.51 … stderr = 10.51 / sqrt(30) = 1.92
70.2 +- 2.045(1.92)
70.2 +- 3.92
At 95% confidence, the reasonable values for the pulse rate (in bpm) of Peruvian Indians are 66.28 to 74.12.
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10. (10 pts)
A turkey geneticist wished to select for breeding purposes hens whose eggs have an average weight of 100 grams. Hens with eggs less than or greater than this weight, on average, are unacceptable. The mean weight of a random sample from one hen was 97 grams, with a standard deviation of 2.5 grams. Should this hen be kept in the breeding program? Fill in the 2 blanks in the output below and answer the question of interest …
One-Sample T
Test of μ = 100 vs ≠ 100
N Mean StDev SE Mean 95% CI T P
20 97.000 2.500 ______(95.830, 98.170) _____ 0.000
SE Mean = 2.5 / sqrt(20) = 0.559
T.S. t = (97 – 100) / 0.559 = -5.37
P-Value = 0.000 is very small … Reject Ho
Sufficient evidence to indicate the mean weight of eggs differs from 100 grams … do not keep this hen.