# business statistics for decision makers

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PowerPoint Presentation1
The D-COVA lifecycle
histogram, ogive/cumulative frequency polygon, scatterplot
2
3
3
4
4 and 5
6
6
Central limit theorem
Sampling Distribution Numerical
Estimating confidence interval for the population proportion
8
Hypothesis formulation
9
8
Applications of p-value hypothesis testing approach
9
Analysis
Goodness of fit of the model
Coefficient of Determination R2
14, 15
Not
Applicable
Practice Session
a. 25
b. 25
c. 20
d. 20
e. 35
f. 128.571
g. 11.339
A small town has 5,600 residents. The residents in the town were asked whether or not they favored building a new bridge across the river. You are given the following information on the residents' responses, broken down by sex.
Men Women Total
Opposed 840 3,080 3,920
Total 2,240 3,360 5,600
Let: M be the event a resident is a man
W be the event a resident is a woman
F be the event a resident is in favor
P be the event a resident is opposed
a. Find the joint probability table.
b. Find the marginal probabilities.
c. What is the probability that a randomly selected resident is a man and is in favor of building
the bridge?
d. What is the probability that a randomly selected resident is a man?
e. What is the probability that a randomly selected resident is in favor of building the bridge?
f. What is the probability that a randomly selected resident is a man or in favor of building the
bridge?
g. A randomly selected resident turns out to be male. Compute the probability that he is in
favor of building the bridge.
A Company supplies pins in bulk to a customer. The company uses an automatic lathe to produce the pins. Due to many causes - vibrations, temperature, wear and tear, and the like - the lengths of the pins made by the machine are normally distributed with a mean of 1.012 inches and a standard deviation of 0.018 inch. The customer will buy only those pins with lengths in the interval 1.00 0.02 inch. In other words, the customer wants the length to be 1.00 inch but will accept up to 0.02 inch deviation on either side. This 0.02 inch is known as tolerance.
1) What percentage of the pins will be acceptable to the customer? (Ans: 63.39 %)
In order to improve percentage accepted, the production manager and the engineers discuss adjusting the population mean and standard deviation of the length of the pins.
2) If the lathe can be adjusted to have the mean of the lengths to any desired value, what should it be adjusted to? Why? (Ans: Mean = 1.00, As it will help to increase the acceptance to 73.35 %)
3) Suppose the mean cannot be adjusted, but the standard deviation can be reduced. What maximum value of the standard deviation would make 90% of the parts acceptable to the consumer? (Assume the mean to be 1.012.) (Ans: 0.006243)
4) Repeat question 3, with 95% and 99% of the pins acceptable. For 95% - 0.0048686; For 99% 0.003443858)
A Company supplies pins in bulk to a customer. The company uses an automatic lathe to produce the pins. Due to many causes - vibrations, temperature, wear and tear, and the like - the lengths of the pins made by the machine are normally distributed with a mean of 1.012 inches and a standard deviation of 0.018 inch. The customer will buy only those pins with lengths in the interval 1.00 0.02 inch. In other words, the customer wants the length to be 1.00 inch but will accept up to 0.02 inch deviation on either side. This 0.02 inch is known as tolerance.
1) What percentage of the pins will be acceptable to the customer? (Ans: 63.39 %)
In order to improve percentage accepted, the production manager and the engineers discuss adjusting the population mean and standard deviation of the length of the pins.
2) If the lathe can be adjusted to have the mean of the lengths to any desired value, what should it be adjusted to? Why? (Ans: Mean = 1.00, As it will help to increase the acceptance to 73.35 %)
3) Suppose the mean cannot be adjusted, but the standard deviation can be reduced. What maximum value of the standard deviation would make 90% of the parts acceptable to the consumer? (Assume the mean to be 1.012.) (Ans: 0.006243)
4) Repeat question 3, with 95% and 99% of the pins acceptable. (Ans: For 95% - 0.0048686; For 99% 0.003443858)
a) 19; approximately 50% of the
students work at least 19 hours
b) 26; at least 70% of the students work
less than or equal to 26 hours per week
c) 40; the most frequent data element
Average of X = 25; Average of Y = 200; and r = 0.0227
The value for b0 will be around
95 145 195 245 295
X 10 10 11 25 25 39 40 40
Y 100 200 300 100 300 100 200 300
Average of X = 25; Average of Y = 2000; and b0 = -979
The value for r will be
a) Positive b) Negative c) Near to Zero d) None of these e) Cannot be determined
Thank You!