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population. Chapter 8 Imagine you are sleeping at night and need to get
up early in the morning to go to work. Or, you are reading something
with intense concentration, or watching your most favorite
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.56 .44 p p p p Finding the mean, standard deviation, and shape of
the sampling distribution of p. JWCL216_ch07_300339.qxd 11/24/09
8:31 AM Page 325 The mean of the sampling distribution of is The
st
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then we use the t distribution (explained in Section 8.4.1) to make the
confidence interval for Case II. If the following two conditions are
fulfilled: 1. The population standard deviation is not know
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the amounts of detergent in all such jugs is .20 ounce. How large a
sample should the company select so that the estimate is within .04
ounce of the population mean? 8.32 A department store manager
wa
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to Look for in Table IV z Value 90% .0500 and .9500 1.64 or 1.65 95% .
0250 and .9750 1.96 96% .0200 and .9800 2.05 97% .0150 and .9850
2.17 98% .0100 and .9900 2.33 99% .0050 and .9950 2.57 or 2.58
E
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your 50 sample means. What shape does it have? c. What does the
central limit theorem say about the shape of the sampling distribution
of What mean and standard deviation does the sampling distributio
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each sample mean, we can expect that 90% of these intervals will
include and 10% will not. In Figure 8.4 we show means and of three
different samples of the same size drawn from the same population.
A
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number of degrees of freedom for a t distribution is equal to the sample
size minus one, that is, The number of degrees of freedom is the only
parameter of the t distribution. There is a different t d
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examinees. Calculate the mean and standard deviation of and describe
the shape of its sampling distribution when the sample size is (a) 16 (b)
50 (c) 1000 Solution Let and be the mean and standard dev
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medians, and then calculate the average of these sample medians. Does
this average of the medians equal the population mean? If yes, why
does this make sense? If no, how could you change exactly two o
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average breaking capacity of all boxes of this type. 8.25 KidPix
Entertainment is in the planning stages of producing a new computeranimated movie for national release, so they need to determine the
p
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higher confidence in our interval. We always attach a probabilistic
statement to the interval estimation. This probabilistic statement is
given by the confidence level. An interval constructed based o
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scores = 153 Sampling distribution of x for n =1000 x = 4.838 x = = 1020
Figure 7.5 Thus, whatever the sample size, the sampling
distribution of is normal when the population from which the samples
a
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for each of those samples. Note that we have rounded the values of to
two decimal places. p Total number of samples 5C3 5! 3!15 32! 5 4 3 2
1 3 2 1 2 1 10 p 35 .60 JWCL216_ch07_300339.qxd 11/24/09 8:
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were observed not washing their hands after going to the bathroom.
Assume that the percentage of all U.S. men who do not wash their
hands after going to the bathroom is 43%. Let be the proportion in a
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automatically determined. Consequently, the number of degrees of
freedom for this example is We subtract 1 from n because we lose 1
degree of freedom to calculate the mean. Table V of Appendix C lists
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a normal distribution with a mean of 16 ounces and a standard
deviation of .18 ounce. a. Find the probability that the mean amount of
ice cream in a random sample of 16 such cartons will be i. between
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mean, ; and the sample proportion, is an estimator of the population
proportion, p, p. x, x, p, p, x. x, x. x, JWCL216_ch08_340380.qxd
12/3/09 7:03 PM Page 341 342 Chapter 8 Estimation of the Mean an
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confidence interval for the population mean, . When the population
standard deviation is not known, then we replace it by the sample
standard deviation s, which is its estimator. Consequently, for the
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which is 1.645, which we will not do in this text. Finding the point
estimate and confidence interval for : known, n 30, and population
normal. JWCL216_ch08_340380.qxd 12/3/09 7:03 PM Page 346 8.3
Es
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always equal to a. p b. c. 11. The condition for the standard deviation of
the sampling distribution of the sample proportion to be is a. np 5 and
nq 5 b. n 30 c. nN .05 12. The sampling distribution
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obtained from a sample, n 81 and . It is known that 4.8. a. What is the
point estimate of ? b. Make a 95% confidence interval for . c. What is
the margin of error of estimate for part b? 8.13 The stan
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value for the given confidence level to use in the confidence interval in
this section. In such a situation when n is too large (for example, 500)
and is not included in the t distribution table, ther
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the value of to obtain a confidence interval for . Thus, E zsx x
Confidence Interval for The (1 )100% confidence interval for under
Cases I and II is where The value of z used here is obtained from th
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more is written as This probability is given by the area under the normal
curve for to the left of as shown in Figure 7.14. We find this area as
follows: For z x m sx 3123 3173 37.50 x $3123: 1.33 x $
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between 181.05 and 190.95 mg/dL. Note that is a point estimate of in
this example, and 4.95 is the margin of error. EXAMPLE 86 Sixtyfour
randomly selected adults who buy books for general reading wer
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of .1 inch. The quality control inspector takes a sample of 25 nails once
a week and calculates the mean length of these nails. If the mean of
this sample is either less than 2.95 inches or greater th
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proportion, of television sets that are good for each sample. Prepare
the sampling distribution of d. For each sample listed in part c, calculate
the sampling error. 7.78 Investigation of all five maj
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proportion of voters in the sample who voted for candidate A. 1.
Suppose that 440 of the 800 voters included in the exit poll voted for
candidate A, which gives Assuming that each candidate received 5
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deviation of these 150 sample means. TA7.3 Refer to SelfReview Test
Exercise 18. In this assignment, we will explore properties of the
sampling distribution of a sample proportion for different sampl