Review the case study on page 359 of the textbook.

The data on the number of chocolate chips per bag for 42 bags of Chips Ahoy! cookies were obtained by the students in an introductory statistics class at the United States Air Force Academy in response to the Chips Ahoy! 1,000 Chips Challenge sponsored by Nabisco, the makers of Chips Ahoy! Use the data collected by the students to answer the following questions and to conduct the analyses required in each part.

A. Obtain and interpret a point estimate for the mean number of chocolate chips per bag for all bags of Chips Ahoy! Cookies. (Note: The sum of the data is 52,986.) B .Construct and interpret a normal probability plot, boxplot, and histogram of the data. C. Use the graphs in part (b) to identify outliers, if any. D.Is it reasonable to use the one-mean t-interval procedure to obtain a confidence interval for the mean number of chocolate chips per bag for all bags of Chips Ahoy! cookies? Explain your answer. E. Determine a 95% confidence interval for the mean number of chips per bag for all bags of Chips Ahoy! cookies, and interpret your result in words. (Note: = 1261.6; s = 117.6.)

A. ͞X =∑Xi/n = 52986/42 = 1261.57
BASED ON THE SAMPLE DATA ,WE ESTIMATE THAT MEAN CHIPS PER BAG TO APPROXIMATLY 1261.57 CHIPS PER BAG.

B. GRAPHS ARE ATTACHED
C. I NOTICED MAY BE 1545 OR 1546 ARE OUTLINERS, THAT WHAT IT LOOKED LIKE AT BOX PLOT.

D. I THINK YOU CAN USE ONE MEAN T-INTERVAL PROCEDURE EVEN IF THE STANDARDIZED DEVIATION NOT GIVEN BY USING THE FOLLOWING

STEP 1. FOR A CONFIDENCE LEVEL OF 1-α, USE TABLE IV TO FINDTα/2 WITH DF= N-1 WHERE N IS SAMPLE SIZE WE WANT 95 % CONFIDENCE INTERVAL SO α= 1-0.95 =0.05
FOR N = 42 DF = 42-1= 41
FROM TABLE IV Tα/2=T0.05/2= T0.025=2.020
STEP 2 THE CONFIDENCE INTERVAL FOR µ IS FROM
͞X –T(α/2) * S/√N TO ͞X + T(α/2)*S/√N
͞X = 1261.57
S= ͞X-µ/(N-1/√N) = 1261.57-1000...

...Pages
C H A P T E
R
6
The NormalDistribution
Objectives
Outline
After completing this chapter, you should be able to
1
2
3
Identify distributions as symmetric or skewed.
4
Find probabilities for a normally distributed
variable by transforming it into a standard
normal variable.
Introduction
6–1
NormalDistributions
Identify the properties of a...

...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...

...
NormalDistributionNormaldistribution is a statistics, which have been widely applied of all mathematical concepts, among large number of statisticians. Abraham de Moivre, an 18th century statistician and consultant to gamblers, noticed that as the number of events (N) increased, the distribution approached, forming a very smooth curve.
He insisted that a new discovery of a...

...
PGEG371: Data Analysis & Geostatistics
NormalDistributions
Laboratory Exercise # 3
1st and 5th February, 2015
Read through this instruction sheet then answer the ‘pre-Lab’ quiz BEFORE starting the exercises!
1. Aim
The purpose of this laboratory exercise is to use a NormalDistribution to find information about a data population.
On successful completion of this exercise, you should be able to
Describe what a...

...or variability of the data about the measurements of central tendency.
MEASUREMENTS OF CENTRAL TENDENCY The appropriateness of using the mean, median, or mode in data analysis is dependent upon the nature of the data set and its distribution (normal vs non-normal). The mean (denoted by x) is calculated by dividing the sum of the individual data points (where Σ equals “sum of”) by the number of observations (denoted by n). It is the arithmetic...

...decimal places)
2. Find the value of z if the area 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...

...NORMALDISTRIBUTION
1. Find the
distribution:
a.
b.
c.
d.
e.
f.
following probabilities, the random variable Z has standard normal
P (0< Z < 1.43)
P (0.11 < Z < 1.98)
P (-0.39 < Z < 1.22)
P (Z < 0.92)
P (Z > -1.78)
P (Z < -2.08)
2. Determine the areas under the standard normal curve between –z and +z:
♦ z = 0.5
♦ z = 2.0
Find the two values of z in standard normaldistribution so that:
P(-z < Z...

...completion of this course will provide students with a working knowledge of the principles of statistics, the ability to analyze and solve problems involving probability, and a working knowledge of averages and variations, normal probability distributions, sampling distributions, confidence intervals and testing statistical hypotheses. The emphasis of the course will be on the proper use of statistical techniques and their implementation rather than...

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