Homework 3 Probability 1. As part of a Pick Your Prize promotion, a store invited customers to choose which of three prizes they’d like to win. They also kept track of respondents’ gender. The following contingency table shows the results: | MP3 Player| Camera| Bike| Total| Men| 62| 117| 60| 239|

Woman| 101| 130| 30| 261|
Total| 163| 247| 90| 500|
What is the probability that: a. a randomly selected customer would pick the camera? 247/500= 0.494= 49.4%b. A randomly selected woman would pick the camera? 130/261=0.498=49.8%c. A randomly selected customer would be both male and choose the bike? 60/500=0.12=12% d. A randomly selected man would choose either the camera or the bike? 177/239=0.74=74% 2. a. Which part(s) of question 1 above deal with joint probability? C,b. Which part(s) deal with conditional probability? B,D The Standard Normal Curve 3. In a recent year, about two-thirds of U.S. households purchased ground coffee. Consider the annual ground coffee expenditures for households who purchase coffee, assuming that these expenditures are approximately normally distributed with a mean of $45.16 and a standard deviation of $10.00. a. Find the probability that a household spent less than $25.00. Z=(25-45.16)/10= -2.016 =.4778=2.2% b. Find the probability that a household spent more than $50.00 (50-45.16)/10=.484=18.44% c. What proportion of households spent between $30.00 and $40.00? 20.09% d. 99% of households spent less than what amount?15.16 4. The breaking strength of plastic bags used for packaging produce is normally distributed, with a mean of 5 pounds per square inch and a standard deviation of 1.5 pounds per square inch. What proportion of bags have a breaking strength of: a. less than 3.17 pounds per square inch?11.12% b. at least 3.6 pounds per square inch? 67.67 c. between 5 and 5.5 pounds per square inch?12.93%...

...require that we know whether we have a sample or a population. 2. The following numbers represent the weights in pounds of six 7year old children in Mrs. Jones' 2nd grade class. {25, 60, 51, 47, 49, 45} Find the mean; median; mode; range; quartiles; variance; standard deviation. Solution: mean = 46.166.... median = 48 mode does not exist range = 35 Q1 = 45 Q2 = median = 48 Q3 = 51 variance = 112.1396 standard deviation =10.59 3. If the variance is 846, what is the standard deviation? Solution: standard deviation = square root of variance = sqrt(846) = 29.086 4. If we have the following data
34, 38, 22, 21, 29, 37, 40, 41, 22, 20, 49, 47, 20, 31, 34, 66 Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 3 4 5 6 | | | | | 219200 48714 0197 6
This distribution is right skewed (positively skewed) because the “tail” extends to the right. 5. What type of relationship is shown by this scatter plot?
45 40 35 30 25 20 15 10 5 0 0 5 10 15 20
Solution: Weak positive linear correlation 6. What values can r take in linear regression? Select 4 values in this interval and describe how they would be interpreted. Solution: the values are between –1 and +1 inclusive. -1 means strong negative correlation +1 means strong positive correlation 0 means no correlation .5 means moderate positive correlation etc. 7. Does correlation imply causation? Solution: No.
8. What do we call the r value. Solution: The correlation coefficient....

...Why We Don’t “Accept” the NullHypothesis
by Keith M. Bower, M.S. and James A. Colton, M.S.
Reprinted with permission from the American Society for Quality
When performing statistical hypothesis tests such as a one-sample t-test or the AndersonDarling test for normality, an investigator will either reject or fail to reject the nullhypothesis, based upon sampled data. Frequently, results in Six Sigma projects contain
the verbiage “accept the nullhypothesis,” which implies that the nullhypothesis has been
proven true. This article discusses why such a practice is incorrect, and why this issue is
more than a matter of semantics.
Overview of Hypothesis Testing
In a statistical hypothesis test, two hypotheses are evaluated: the null (H0) and the
alternative (H1). The nullhypothesis is assumed true until proven otherwise. If the
weight of evidence leads us to believe that the nullhypothesis is highly unlikely (based
upon probability theory), then we have a statistical basis upon which we may reject the
nullhypothesis.
A common misconception is that statistical hypothesis tests are designed to select the
more likely of two hypotheses. Rather, a test will stay with the...

...outcome xi according to its probability, pi. The mean also of a random variable provides the long-run average of the variable, or the expected average outcome over many observations.The common symbol for the mean (also known as the expected value of X) is , formally defined by
Variance - The variance of a discrete random variable X measures the spread, or variability, of the distribution, and is defined by
The standard deviation is the square root of the variance.
Expectation - The expected value (or mean) of X, where X is a discrete random variable, is a weighted average of the possible values that X can take, each value being weighted according to the probability of that event occurring. The expected value of X is usually written as E(X) or m.
E(X) = S x P(X = x)
So the expected value is the sum of: [(each of the possible outcomes) × (the probability of the outcome occurring)].In more concrete terms, the expectation is what you would expect the outcome of an experiment to be on average.
2. Define the following;
a) Binomial Distribution - is the discrete probabilitydistribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p. Therewith the probability of an event is defined by its binomial distribution. A success/failure experiment is also...

...Tutorial Note 5
• Chapter 17 Hypothesis Testing
• Chapter 18 Measures of Association
Objective: Computer laboratory session (II) – Please refer to the Computer Lab Notes
Discussion Question 1
H0 – NullHypothesis: a statement you want to reject (E.g Average = 50)
H1/HA – Alternative Hypothesis: a statement you want to prove (E.g Average is not 50)
What is related/dependent sample?
H0: Tutorial is NOT effective
H1: Tutorial is effective
|Respondent |Before tutorial |After tutorial | |After – Before |
|A |30 |35 | |+5 |
|B |70 |62 | |-8 |
|C |65 |69 | |+4 |
|D |80 |88 | |+8 |
What hypothesis testing procedure would you use in the following situations?
a) A test classifies applicants as accepted or rejected. On the basis of data on 200 applicants, we test the hypothesis that ad placement success is not related to gender.
H0: ad placement success is not related to gender. (Indep)
Ha: ad placement success is related to gender. (Dep)
2-independent...

...If a six sided die is tossed two times and “3” shows up both times, the probability of “3” on the third trial is
a. much larger than any other outcome
b. much smaller than any other outcome
c. 1/6
d. 1/216
13. If P(A) = 0.4, P(B| A) = 0.35, P(A B) =0.69, then P(B) =
a. 0.14
b. 0.43
c. 0.75
d. 0.59
14. Two events with nonzero probabilities
a. can be both mutually exclusive and independent
b. can not be both mutually exclusive and independent
c. are always mutually exclusive
d. are always independent
15. If A and B are independent events with P(A) = 0.05 and P(B) = 0.65, then
P(AB) =
a. 0.05
b. 0.0325
c. 0.65
d. 0.8
16. A description of the distribution of the values of a random variable and their associated probabilities is called a
a. probabilitydistribution
b. random variance
c. random variable
d. expected value
The following represents the probabilitydistribution for the daily demand of microcomputers at a local store.
Demand Probability
0 0.1
1 0.2
2 0.3
3 0.2
4 0.2
17. Refer to above information, the expected daily demand is
a. 1.0
b. 2.2
c. 2, since it has the highest probability
d. of course 4, since it is the largest demand level
18. Refer to above information, the probability of having a demand for at least two...

...Take Home Test 2
1. A. NullHypothesis: There are no relations or associations among the groups’ mean scores.
Alternate Hypothesis: There is a relation or association among the student’s grade point averages and “if they rather prefer to stay at home than go out with friends”.
Correlations |
| Grade Point Average | I would rather stay at home and read than go out with my friends |
Grade Point Average | Pearson Correlation | 1 | .233 |
| Sig. (2-tailed) | | .120 |
| Sum of Squares and Cross-products | 12.667 | 5.002 |
| Covariance | .281 | .111 |
| N | 46 | 46 |
I would rather stay at home and read than go out with my friends | Pearson Correlation | .233 | 1 |
| Sig. (2-tailed) | .120 | |
| Sum of Squares and Cross-products | 5.002 | 36.457 |
| Covariance | .111 | .810 |
| N | 46 | 46 |
Based on the results of our Correlate Bivariate we see that the significance value is more than the p-value of .05 which means that the groups have no relationship between them. The significance value is .120. This means that we are going to accept the NullHypothesis and reject the Alternate Hypothesis. “I would rather stay at home and read than go out with my friends” has no relationship with the persons GPA.
B. NullHypothesis: There is no relation or association between people who rarely forget their appointment if they have...

...areas (N = 84) hired an average of 6.80 females (s = 4.16). Using an alpha of .05, test the nullhypothesis that there is no difference between the means.
a. What is the nullhypothesis? What is the research hypothesis?
i. The nullhypothesis is that there is no difference between urban and rural jurisdictions in the hiring of female police officers last year.
ii. The researchhypothesis is that there is a difference between urban and rural jurisdictions in the hiring of female police officers last year.
b. What are the degrees of freedom?
Formula: df=N1+N2-2 Given: N1=77, N2=84
df= 77 + 84 -2
df= 159
c. What is the critical value?
i. Due to the fact that the degrees of freedom is 159 and the table goes from 100 to infinity, infinity is the closest value that can give the closest critical value being 1.96 in the two-tailed test category. Since the critical value decreases as the obtained value increases, the critical value has to be less than 1.96 which is the critical value of 100. (Page 372)
d. Do you reject or fail to reject the nullhypothesis?
Given:
Urban (Group 1): Nonurban (Group 2):
N1=77 N2=84
Mean (): 9.65 Mean (): 6.80
Standard Deviation (s1): 4.17 Standard Deviation (s2): 4.16
Variance (s12): 17.39 Variance (s22): 17.31
Formula:
Due to the critical value being 1.96 and...

...under a Standard Normal curve
a) to the right of z is 0.3632;
b) to the left of z is 0.1131;
c) between 0 and z, with z > 0, is 0.4838;
d) between -z and z, with z > 0, is 0.9500.
Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table)
b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table)
c ) the area between 0 to z is 0.4838, z = 2.14
d) the area to the right of +z = ( 1-0.95)/2 = 0.025, therefore z = 1.96
3. Given the Normally distributed variable X with mean 18 and standard deviation 2.5, find
a) P(X < 15);
b) the value of k such that P(X < k) = 0.2236;
c) the value of k such that P(X > k) = 0.1814;
d) P( 17 < X < 21).
Ans : X ~ N ( 18, 2.52)
a) P ( X < 15)
P ( Z < (15-18)/2.5) = P ( Z < -1.2) = 0.1151 ( 4 decimal places)
b) P ( X < k) = 0.2236
P ( Z < ( k – 18) / 2.5 ) = 0.2236
From normal table, 0.2236 = -0.76
(k-18)/2.5 = - 0.76, solve k = 16.1
c) P (X > k) = 0.1814
P ( Z > (k-18)/2.5 ) = 0.1814
From normal table, 0.1814 = 0.91
(k-18)/ 2.5 = 0.91, solve k = 20.275
d) P ( 17 < X < 21)
P ( (17 -18)/2.5 < Z < ( 21-18)/2.5)
P ( -0.4 < Z < 1.2) = 0.8849 – 0.3446 = 0.5403 ( 4 decimal places)...

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