ExamView+-+Psych+270+Exam+2+Fall+2018

.pdf

School

Institute of Health Profession Education & Research, KMU Peshawar *

*We aren’t endorsed by this school

Course

1F

Subject

Statistics

Date

Nov 24, 2024

Type

pdf

Pages

33

Uploaded by CountAntelope14965

Report
1 Name: ________________________ Class: ___________________ Date: __________ Exam 2 Please show all of your work!! Short Answer 1. For a population with = 40 and = 8, find the z-score corresponding to each of the following X values: (a) X=42, Z=__0.25__ (c) X=36, Z=__-0.5____ (b) X=52, Z=__1.5__ (d) X=32, Z=__-1____ Using ? = 𝑋−𝜇 𝜎 a. ? = 42−40 8 = 0.25 b. ? = 52−40 8 = 1.5 c. ? = 36−40 8 = -0.5 d. ? = 32−40 8 = -1 __________________________Show your Work___________________________
Name: ________________________ ID: A 2 2. For a sample with M = 38 and s = 12, find the X value that corresponds to each of the following z-scores: (a) Z= –0.25, X=__35___ (c) Z= 0.50, X=__44___ (b)Z= –1.50, X=__20___ (d) Z=2.00, X=__62___ Using ? = 𝑋−𝑀 𝑠 𝑋 = ?𝑠 + 𝑀 a. 𝑋 = (−0.25 ∗ 12) + 38 X= 35 b. 𝑋 = (−1.5 ∗ 12) + 38 X= 20 c. 𝑋 = (0.5 ∗ 12) + 38 X= 44 d. 𝑋 = (2 ∗ 12) + 38 X= 62
Name: ________________________ ID: A 3 __________________________Show your Work___________________________ 3. Use the following Population Scores to answer questions 1-5 below. Individuals A B C D E F G H I J Score 12 12 7 10 9 12 13 8 9 8 ____Calculate the z-score for the following X scores: (a) X=12, Z=__1___. (c) X=7, Z=_-1.5___. (b) X=10, Z=___0__. (d) X=8, Z=__-1___. 𝑀??𝑛 (𝜇) = 12+12+7+10+9+12+13+8+9+8 10 = 10
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 4 Variance ( 𝜎 2 ) = ∑(𝑋−𝜇) 2 𝑛 = (12−10) 2 +(12−10) 2 +(7−10) 2 +⋯……..+(8−10) 2 10 = 40 10 = 4 𝜎 = √4 = 2
Name: ________________________ ID: A 5 Using ? = 𝑋−𝜇 𝜎 a. ? = 12−10 2 = 1 b. ? = 10−10 2 = 0 c. ? = 7−10 2 = −1.5 d. ? = 8−10 2 = −1
Name: ________________________ ID: A 6
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 7 . a 4. For a normal distribution with a mean of µ = 100 and a standard deviation of = 10, find each of the
8 following probabilities: a. p(X > 102), p=____0.4207_____ ? = 102 − 100 10 = 0.2 P(X>102)= 0.4207 b. p(X < 85), p=___0.0668_______ ? = 85 − 100 10 = −1.5 P(X<85)= 0.0668 c. p(X < 130) p=____0.9987________ ? = 130 − 100 10 = 3 P(X<130)= 0.9987 d. p(95 < X < 105) p=_____0.3829_________ 95 − 100 10 < ? < 105 − 100 10 -0.5 < Z<0.5= 0.3829 5. A sample of n = 16 scores is selected from a normal population with = 60 and = 20. a. What is the probability that the sample mean will be greater than 50? Proportion=_________0.9773__________. b. What is the probability that the sample mean will be less than 56? Proportion=_________0.2119__________. c. What is the probability that the sample mean will be within 5 points of the
Name: ________________________ ID: A 9 population mean? That is, what is p(55< X < 65)? Proportion=________0.6827______________. σ/√n = 20/√16 = 20/4 = 5
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 10 a. Z = (X ̄ - μ) / (σ/√n) = (50 - 60) / (5) = -10 / 5 = -2 p(Z > -2)= 0.9773 b. Z = (X ̄ - μ) / (σ/√n) = (56 - 60) / (5) = -4 / 5 = -0.8 p(Z < -0.8)= 0.21186 c. Z1 = (55 - 60) / (5) = -1 Z2 = (65 - 60) / (5) = 1 p(-1 < Z < 1)= 0.6827
Name: ________________________ ID: A 11 ______________________Show your Work___________________________ Show all of your work 6. A researcher would like to determine whether an over-the-counter cold medication has an effect on mental alertness. A sample of n = 16 participants is obtained, and each person is given a standard dose of the medication one hour before being tested in a driving simulation task. For the general population, scores on the simulation task are normally distributed with = 60 and = 8. The individuals in the sample who took the drug had an average score of M = 56.5. a. Use a two-tailed test with = .05. Conduct the four steps for hypothesis testing and label each step: Step1, Step 2, Step 3, and Step 4. STEP 1: • Null Hypothesis (H0): The medication does not have an effect on mental alertness; μ = 60 (the population mean score remains the same). • Alternative Hypothesis (H1): The medication has an effect on mental alertness; μ ≠ 60 (the population mean score is different from 60). STEP 2: Significance level (α) = 0.05. This is a two-tailed test because we want to know if the sample mean is significantly different from the population mean in either direction. STEP 3: ? = 𝑀 − 𝜇 𝜎 √𝑛 = 56.5 − 60 8 √16 = −1.75 STEP 4: The critical z-values for a two-tailed test at a 0.025 level of significance (α/2) are approximately ±1.96. Since -1.75 falls between -1.96 and 1.96, we do not reject the null hypothesis. b. Calculate Cohen’s d. d= 56.5−60 8 = −0.4375 c. Are the data sufficient to conclude that their is a significant difference? Write your answer in the form of a sentence.
Name: ________________________ ID: A 12 The data is not sufficient to conclude that there is a significant difference between the sample mean and the population mean. We fail to reject the null hypothesis at the 0.05 significance level. The sample mean of 56.5 is not significantly different from the population mean of 60. 7. Find the z-score for the sample mean for each of the following samples obtained from a population with = 50 and = 12. a. M = 56 for a sample of n = 4, Z=____1________. SE= 𝜎 √𝑛 = 12 √4 = 6 ? = 56 − 50 6 = 1 b. M = 42 for a sample of n = 9, Z=_____-2_______. SE= 𝜎 √𝑛 = 12 √9 = 4 ? = 42 − 50 4 = −2 c. M = 51 for a sample of n = 36, Z=____0.5_______.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 13 SE= 𝜎 √𝑛 = 12 √36 = 2 ? = 51 − 50 2 = 0.5 _______________________Show your Work___________________________ 8. For a normal distribution with µ = 500 and = 100, find the following values: a. What X value separates the highest 10% of the distribution from the rest of the scores? X=_______628___________.
Name: ________________________ ID: A 14 Using a standard normal distribution table or a calculator, you can find that the z-score corresponding to the 90th percentile is approximately 1.28. Now, use the z-score formula to find the X value: Z = (X - µ) / σ 1.28 = (X - 500) / 100 Now, solve for X: X - 500 = 1.28 * 100 X - 500 = 128 X = 500 + 128 X = 628 b. What X values form the boundaries for the middle 60% of the distribution? X=______(416, 584)___________. Using a standard normal distribution table or a calculator, you can find that the z-score corresponding to the 20th percentile is approximately -0.84, and the z-score corresponding to the 80th percentile is approximately 0.84. Now, use the z-score formula to find the X values:
Name: ________________________ ID: A 15 For the lower boundary: -0.84 = (X - 500) / 100 X - 500 = -0.84 * 100 X - 500 = -84 X = 500 - 84 X = 416 For the upper boundary: 0.84 = (X - 500) / 100 X - 500 = 0.84 * 100 X - 500 = 84 X = 500 + 84 X = 584
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 16 So, the X values that form the boundaries for the middle 60% of the distribution are 416 and 584. c. What is the probability of randomly selecting a score greater than X = 475? P=____0.5987_____________.
Name: ________________________ ID: A 17 Z = (475 - 500) / 100 Z = -25 / 100
Name: ________________________ ID: A 18 Z = -0.25 Now, you can find the probability by looking up the z-score of -0.25 in a standard normal distribution table or using a calculator. The probability can be written as p(Z > -0.25). From a standard normal distribution table, you can find that p(Z > -0.25) is approximately 0.5987.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 19 So, the probability of randomly selecting a score greater than X = 475 is approximately 0.5987.
Name: ________________________ ID: A 20
Name: ________________________ ID: A 21
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 22
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 23 _______________________Show your Work___________________________ 9. If a sample of n = 25 scores is selected from a normal population with = 100 and = 20, then the sample mean is expected to be around M = 100. What M values form the boundaries for the middle 80% of the distribution? __94.88___ to __105.12____. Using a standard normal distribution table or calculator, you can find the z-scores corresponding to the 10th and 90th percentiles. The z-score for the 10th percentile is approximately -1.28, and the z- score for the 90th percentile is approximately 1.28. Now, use these z-scores to find the M values: For the lower boundary: M = µ + (z * SE) = 100 + (-1.28 * 4) = 100 - 5.12 = 94.88 For the upper boundary: M = µ + (z * SE) = 100 + (1.28 * 4) = 100 + 5.12 = 105.12 What M values form the boundaries for the middle 99% of the distribution? __89.68___ to __110.32____. Using a standard normal distribution table or calculator, you can find the z-scores corresponding to the 0.5th and 99.5th percentiles. The z-score for the 0.5th percentile is approximately -2.58, and the z-score for the 99.5th percentile is approximately 2.58. Now, use these z-scores to find the M values: For the lower boundary: M = µ + (z * SE) = 100 + (-2.58 * 4) = 100 - 10.32 = 89.68 For the upper boundary: M = µ + (z * SE) = 100 + (2.58 * 4) = 100 + 10.32 = 110.32 What M value cuts off the top 10 %? __105.12_____.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 24 Using a standard normal distribution table or calculator, you can find the z-score corresponding to the 90th percentile, which is approximately 1.28. Now, use this z-score to find the M value: M = µ + (z * SE) = 100 + (1.28 * 4) = 100 + 5.12 = 105.12 What M value cuts off the top 2 %? ___108.2____. Using a standard normal distribution table or calculator, you can find the z-score corresponding to the 98th percentile, which is approximately 2.05. Now, use this z-score to find the M value: M = µ + (z * SE) = 100 + (2.05 * 4) = 100 + 8.2 = 108.2
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 25 10. Some researchers claim that herbal supplements such as ginseng or ginkgo biloba enhance human memory. To test this claim, a researcher selects a sample of n = 25 college students. Each student is given a ginkgo biloba supplement daily for six weeks and then all the participants are given a standardized memory test. For the population, scores on the test are normally distributed with = 70 and = 15.. The sample of n = 25 students had a mean score of M = 75. a. Use a two-tailed test with = .05. Conduct the four steps for hypothesis testing and label each step: Step1, Step 2, Step 3, and Step 4. STEP 1: • Null Hypothesis (H0): The ginkgo biloba supplement does not enhance human memory; μ = 70 (the population mean memory score remains the same). • Alternative Hypothesis (H1): The ginkgo biloba supplement enhances human memory; μ ≠ 70 (the population mean memory score is different from 70). STEP 2: Significance level (α) = 0.05. This is a two-tailed test because we want to know if the sample mean is significantly different from the population mean in either direction. STEP 3: Z= (75−70) 15 √25 = 1.67 STEP 4: The critical z-values for this significance level are approximately ±1.96. Since 1.67 falls within the range of -1.96 to 1.96, you fail to reject the null hypothesis. b. Calculate Cohen’s d. d= 𝑀−𝜇 𝜎 = 75−70 15 = 0.333 c. Are the data sufficient to conclude that thier is a significant difference? Write your answer in the form of a sentence.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 26 The data is not sufficient to conclude that there is a significant difference in human memory scores with the use of ginkgo biloba. The sample mean of 75 is not significantly different from the population mean of 70 at the 0.05 significance level.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 27 Show your work 11. For a normal distribution with µ = 100 and = 15, find the following values: a. What is the probability of selecting a sample of n=100 scores with a mean greater than 102? P=_______0.0918__________. SE= 15 √100 = 1.5 Z= 102−100 1.5 = 1.33 P(Z >102)= 0.0918 b. What is the probability of selecting a sample of n=144 scores with a mean greater than 98? P=________0.9452_________. SE= 15 √144 = 1.25 Z= 98−100 1.25 = −1.6 P(Z >144)= 0.9452 c. For a sample of n=81, What proportion of the samples will have means between 97 and 103? P=________0.9254_________.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 28 For the lower boundary: Z_lower = (97 - 100) / (15 / √81) = -3 / 1.67 = -1.79
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 29 For the upper boundary: Z_upper = (103 - 100) / (15 / √81) = 3 / 1.67 = 1.79
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 30 Now, use a standard normal distribution table or calculator to find the probability that -1.79 < Z < 1.79. This probability is approximately 0.9254.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 31 . 12. In a test of physical fitness, a group of men ages 65 and older from a local retirement community were told to do as many sit-ups as they could. It is known that the population mean is 20, and = 8. The scores for the men from the retirement community are given below. 24, 25, 10, 23, 18, 29, 47, 32, 19, 24, 20, 28, 21, 20, 27, 25 a. Use a two-tailed test with = .05. Conduct the four steps for hypothesis testing and label each step: Step1, Step 2, Step 3, and Step 4. STEP 1: • Null Hypothesis (H0): The average number of sit-ups for men ages 65 and older in the retirement community is the same as the population mean, μ = 20. • Alternative Hypothesis (H1): The average number of sit-ups for men ages 65 and older in the retirement community is different from the population mean, μ ≠ 20. STEP 2: Significance level (α) = 0.05. This is a two-tailed test because we want to know if the sample mean is significantly different from the population mean in either direction. STEP 3: M = 392/16= 24.5 SE= 𝜎 √𝑛 = 8 √16 = 2 Z= 𝑀−𝜇 𝑆𝐸 = 24.5−20 2 = 2.25 STEP 4: The Critical Z value is ± 1.96. Since 2.25 falls outside the range of -1.96 to 1.96, you can reject the null hypothesis.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 32 b. Calculate Cohen’s d. ? = 𝑀 − 𝜇 𝜎 = 24.5 − 20 8 = 0.5625 c. Are the data sufficient to conclude that thier is a significant difference? Write your answer in the form of a sentence. The data is sufficient to conclude that there is a significant difference in the average number of sit- ups for men ages 65 and older in the retirement community compared to the population mean. The sample mean of 24.5 is significantly different from the population mean of 20 at the 0.05 significance level.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Name: ________________________ ID: A 33 .
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Recommended textbooks for you
Text book image
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage
SEE MORE TEXTBOOKS
Recommended textbooks for you
Text book image
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage
SEE MORE TEXTBOOKS