STAT 350 - Sections A, B, C, D, and E Business and Economic Statistics I Course Policies Instructor: Dr. Abigail Jager E-mail: jager@ksu.edu Phone: 532-3585 Oce: 108B Dickens Oce Hours: MW 1:00pm-2:00pm, F 10:00am-11:00am or by appointment GTAs: Nadeesha
Mini-project 2 The le Shadow.xls (found in the Data Sets folder on K-State online) contains data on 20 shadow stocks. The variables included in the data set are the exchange on which the stock is traded, earnings per share and the ratio of price to earnin
Mini-project 1 1. What scale of measurement is most appropriate for the following? (a) Attitude toward legalization of marijuana (favor, neutral, oppose) (b) Number of children in a family (c) Religious aliation (Catholic, Jewish, Protestant, Muslim, othe
Review and Examples
Counting Outcomes of Experiments Assigning Probabilities Events Basic Probability Relationships Random Variables Discrete Probability Distributions Expected Value (Mean) Variance and Standard Deviation
Suppose to win a lottery you mus
Hypothesis Test for a Population Proportion Our population proportion is denoted by p. The hypothesized value of the population proportion we will denote by p0 . We estimate p with the sample proportion p.
We have three dierent sets of hypotheses about p
Mini-project 6 1. Given that z is a standard normal random variable, compute the following: (a) P (z > 1.43) (b) P (z < 2.32) (c) P (0.46 < z < 1.80) (d) Find z such that the area to the right of z is 0.33. (e) Find z such that the area to the left of z i
Review and Examples
Range IQR Variance and Standard Deviation Shape of the Distribution z -scores Empirical Rule Covariance Correlation
According to the 2003 Annual Consumer Spending Survey, the average monthly Bank of America Visa credit card charge was
Review and Examples Between-Treatments Estimate of Variance
Within-Treatments Estimate of Variance
The F Test
The ANOVA Table
Fishers LSD Procedure
Example: Job satisfaction was measured for a sample of 10 individuals from each of the following profes
Small
90.5 78.2 92.3 95.7 94.1 100 91.8 95
Medium Large
91.1 98.9 94.2 84.3 84.8 89.2 86.4 88.3 89.2 90.2 85.9 84.2 90.2 80.6 75.8 82.3
This file cont ains passenger r at ings for t hr ee differ ent sizes of c medium, and lar ge (higher r at ings mean hig
Review and Examples Hypothesis Tests for 1 2 consist of four steps: 1. State the hypotheses. The hypotheses should take one of these forms: H0 : 1 2 D0 Ha : 1 2 < D0 H0 : 1 2 D0 Ha : 1 2 > D0 H0 : 1 2 = D0 Ha : 1 2 = D0
2. State the level of signicance. T
College High School 485 442 534 580 650 479 554 486 550 528 572 524 497 492 592 478 487 425 533 485 526 390 410 535 515 578 448 469
The College Board provided comparisons of SAT scores based on the highest level of education attained by the test taker's p
Mini-project 5 1. The 2002 New York City Housing and Vacancy Survey showed a total of 59,324 rent-controlled housing units and 236,263 rent-stabilized units built in 1947 or later. For these rental units, the probability distributions for the number of pe
Review and Examples The hypotheses for a hypothesis test take one of the following forms: H0 : 0 Ha : > 0 H0 : 0 Ha : < 0 H0 : = 0 Ha : = 0
The form chosen depends on the problem being asked. 0 is just a number. The hypotheses are statements about the pa
Review and Examples If you want to construct a condence interval for the population mean, you can determine the sample size you should use for your study before collecting your data. If you want your condence interval to have a margin of error of size E ,
Mini-project 9 1. A credit card company wants to determine the average amount their customers spend per month. They would like to have a margin of error of $200 for a 95% condence interval. Past data suggest that the population standard deviation is $1000
Review and Examples
When you know the population standard deviation , you can use this condence interval for the population mean : x z/2 n When you do not know the population standard deviation , you rst estimate s (the sample standard deviation), and th
Review and Examples
Continuous Random Variables Continuous Probability Distributions The Normal Distribution The Standard Normal Distribution (z ) Computing Probabilities for Normal Distributions
1. Given that z is a standard normal random variable, comp
Review and Examples Hypothesis Tests for means (with unknown ) consist of four steps: 1. State the hypotheses. The hypotheses should take one of these forms: H0 : 0 Ha : < 0 H0 : 0 Ha : > 0 H0 : = 0 Ha : = 0
2. State the level of signicance. This is denot
Solutions to Homework 13 Problem 13.5 First, SSE = 1800 1200 = 600. The degrees of freedom for the Treatments is k 1 = 3 1 = 2. The degrees of freedom for the Error is nT k = 47 3 = 44. The degrees of freedom for the Total is nT 1 = 47 1 = 46. We then hav
Solutions to Homework 12 Problem 10.5 a. The point estimate is x1 x2 = 135.67 68.64 = 67.03. b. The margin of error is M OE = z/2 c. The 99% condence interval is (1 x2 ) M OE = 67.03 17.08 = (49.95, 84.11). x Problem 10.8 a. The test statistic is z= (1 x2
Solutions to Homework 11 Problem 9.25 For all parts of the problem 0 = 45 and n = 36. This means we have n 1 = 36 1 = 35 degrees of freedom for our t distribution. a. The test statistic is t= 44 45 x 0 = 1.15. = s/ n 5.2/ 36
The p-value is between 0.10 an
Solutions to Homework 10 Problem 9.1 a. The hypotheses are H0 : 600 Ha : > 600 The alternative hypothesis should correspond to the claim that the bills have been increasing ( > 600). b. If H0 cannot be rejected then we would conclude that it is possible t
Solutions to Homework 9 Problem 8.29 a. We can compute n:
2 z/2 2 1.962 6.252 = = 37.5. E2 22 So we should sample at least 38.
n=
b. We can compute n:
2 z/2 2 1.962 6.252 n= = = 150.06. E2 12 So we should sample at least 151.
Problem 8.35 a. The point est
Solutions to Homework 8 Problem 8.3 a. The 95% condence interval for the population mean is 15 x z/2 = 80 1.96 n 60 = 80 3.80 = (76.20, 83.80) b. If there were 120 items in the sample the 95% condence interval is 15 = 80 1.96 x z/2 n 120 = 80 2.68 = (77.3
Solutions to Homework 7 Problem 7.13 a. A point estimate of the population mean number of units sold per month is x= xi 94 + 100 + 85 + 94 + 92 = = 93. n 5
b. A point estimate of the population standard deviation is s= Problem 7.17 a. The total number of
Solutions to Homework 6 Problem 6.11 a. P (z 1.0) = 0.1587 b. P (z 1) = 1 0.1587 = 0.8413 c. P (z 1.5) = 1 0.0668 = 0.9332 d. P (2.5 z ) = 1 0.0062 = 0.9938 e. P (3 < z 0) = 0.5000 .0013 = 0.4987 Problem 6.13 a. P (1.98 z 0.49) = 0.6879 0.0239 = 0.6640 b.
Solutions to Homework 5 Problem 4.4 a. The tree diagram:
b. The possible experimental outcomes are HHH,HHT,HTH,HTT,THH,THT,TTH,TTT c. Each outcome is equally likely, so the probability for each outcome is Problem 4.9 Here we are counting the number of pos
Solutions to Homework 4 Problem 3.17 a. The mean price for models with a DVD player is 450 + 300 + 400 + 500 + 400 = 410. 5 The mean price for models without a DVD player is 300 + 300 + 360 + 290 + 300 = 310. 5 So it is about $100 additional to perchase a
Solutions to Homework 3 Problem 3.9 a. The mean number of minutes is xi 615 + 135 + + 105 6330 = = = 422 minutes. n 15 15 b. To compute the median we rst put the day in ascending order. 105 135 395 420 180 430 210 615 245 690 250 830 265 1180 380
We have