.14 Hepatitis C and bone density. Does chronic hepatitis C infection affect bone mineral density? In the general population of healthy young adults, bone mineral density is Normally distributed, and density scores are standardized to have a mean u of 0 and a standard deviation o of 1. A study examined the hone densities of a random sample of 28 adult men living with chronic hepati- r Their mean standardized bone density was -0.61. Does this tis C infection.? provide good evidence that the mean standardized bone density of adult men with chronic hepatitis C infection is less than 0? Follow the four-step process as illustrated in Example 14.9.

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14.14 

there evidence that the mean in this district differs from 100? Follow the four-step
ve mean u
00. Is
process, as illustrated in Example 14.9, in your answer.
bone mineral density? In the general population of healthy young adults, bone
mineral density is Normally distributed, and density scores are standardized to
bave a mean jµ of 0 and a standard deviation o of 1. A study examined the
bone densities of a random sample of 28 adult men living with chronic hepati-
tis C infection.' Their mean standardized bone density was -0.61. Does this
provide good evidence that the mean standardized bone density of adult men
with chronic hepatitis C infection is less than 0? Follow the four-step process as
illustrated in Example 14.9.
OL S
VISini sonsbiinoo ofc
21212
TESTS FROM CONFIDENCE INTERVALS
sampe mean x to estimate µ. Both rely on probabilities calculated from a Normal
sampiing distribution. In fact, a two-sided test at significance level a can be
carried out from a confidence interval with confidence level C =1 – a.
Both cests and confidence intervals for a population mean u start by using the
CHVBLEK
A level a two-sided hypothesis test rejects a hypothesis Ho: µ = µo
exactly when the value uo falls outside a level 1 – a confidence interval
for M.
SONFIDENCE INTERVALS AND TWO-SIDED TESTS
%3D
Transcribed Image Text:there evidence that the mean in this district differs from 100? Follow the four-step ve mean u 00. Is process, as illustrated in Example 14.9, in your answer. bone mineral density? In the general population of healthy young adults, bone mineral density is Normally distributed, and density scores are standardized to bave a mean jµ of 0 and a standard deviation o of 1. A study examined the bone densities of a random sample of 28 adult men living with chronic hepati- tis C infection.' Their mean standardized bone density was -0.61. Does this provide good evidence that the mean standardized bone density of adult men with chronic hepatitis C infection is less than 0? Follow the four-step process as illustrated in Example 14.9. OL S VISini sonsbiinoo ofc 21212 TESTS FROM CONFIDENCE INTERVALS sampe mean x to estimate µ. Both rely on probabilities calculated from a Normal sampiing distribution. In fact, a two-sided test at significance level a can be carried out from a confidence interval with confidence level C =1 – a. Both cests and confidence intervals for a population mean u start by using the CHVBLEK A level a two-sided hypothesis test rejects a hypothesis Ho: µ = µo exactly when the value uo falls outside a level 1 – a confidence interval for M. SONFIDENCE INTERVALS AND TWO-SIDED TESTS %3D
音号
Prae
have a µ of 0 and a o
bone In the of adults
14.14 C and Does C infection
expected a distribution.
he study findings are highly statistically significant (P~0). This doesn't
that the true mean body têmperature in the population of healthy adults is
fferent from 98.6 °F. Rather, it just says that we have very strong evidence
verject Ho. Because the sample size is rather large (n= 130), the sampling dis-
trion of x is quite narrow. The observed sample mean ī= 98.25 °F is therefore
TESTS FROM
uch further away a hypothetical population mean of 98.6 °F than would be
expected under such a narrow sampling distribution.
The computations performed by any statistical software in this example involve
the following steps:
from
1 Test statistic: z= (x – 0)(0//n) = (98.25 – 98.6)/(0.6//130) = -6.65
%3D
I P-value: P = 2P(Z < –6.65) ~ 0, computed using the standard Normal curve
This introductory chapter describes the computations involved in statistical infer-
ence when the “simple conditions" are met. As we have noted here, these "simple
conditions" are not easily met. In Chapter 15 we will discuss circumstances wher
some ci the conditions can be relaxed. In Chapter 17 we will describe a work
able alternative to the problem of not knowing the true value of the populatio
standard deviation ơ .
DOL
APPLY YOUR KNOWLEDGE
14.13 TQ test scores. Example 14.3 gives the IQ test scores of 31 seventh-grade g
in a Midwest school district. IQ scores follow a Normal distribution with stane
deviation o ==15. Treat these 31 girls as an SRS of all seventh-grade girls in
district. IQ scores in a broad population are supposed to have mean u = 10
there evidence that the mean in this district differs from 100? Follow the four
process, as illustrated in Example 14.9, in your answer.
%3D
%3D
of hot
bone mineral density? In the general population of healthy young adults
mineral density is Normally distributed, and density scores are standard
DOne densities of a random sample of 28 adult men living with chronic
beir mean standardized bone density was -0.61. D
mean standardized bone density of ad
O? Follow the four-step p
nave a mean u of 0 and a standard deviation o of 1. A study examin
Transcribed Image Text:音号 Prae have a µ of 0 and a o bone In the of adults 14.14 C and Does C infection expected a distribution. he study findings are highly statistically significant (P~0). This doesn't that the true mean body têmperature in the population of healthy adults is fferent from 98.6 °F. Rather, it just says that we have very strong evidence verject Ho. Because the sample size is rather large (n= 130), the sampling dis- trion of x is quite narrow. The observed sample mean ī= 98.25 °F is therefore TESTS FROM uch further away a hypothetical population mean of 98.6 °F than would be expected under such a narrow sampling distribution. The computations performed by any statistical software in this example involve the following steps: from 1 Test statistic: z= (x – 0)(0//n) = (98.25 – 98.6)/(0.6//130) = -6.65 %3D I P-value: P = 2P(Z < –6.65) ~ 0, computed using the standard Normal curve This introductory chapter describes the computations involved in statistical infer- ence when the “simple conditions" are met. As we have noted here, these "simple conditions" are not easily met. In Chapter 15 we will discuss circumstances wher some ci the conditions can be relaxed. In Chapter 17 we will describe a work able alternative to the problem of not knowing the true value of the populatio standard deviation ơ . DOL APPLY YOUR KNOWLEDGE 14.13 TQ test scores. Example 14.3 gives the IQ test scores of 31 seventh-grade g in a Midwest school district. IQ scores follow a Normal distribution with stane deviation o ==15. Treat these 31 girls as an SRS of all seventh-grade girls in district. IQ scores in a broad population are supposed to have mean u = 10 there evidence that the mean in this district differs from 100? Follow the four process, as illustrated in Example 14.9, in your answer. %3D %3D of hot bone mineral density? In the general population of healthy young adults mineral density is Normally distributed, and density scores are standard DOne densities of a random sample of 28 adult men living with chronic beir mean standardized bone density was -0.61. D mean standardized bone density of ad O? Follow the four-step p nave a mean u of 0 and a standard deviation o of 1. A study examin
Expert Solution
Step 1

Solution:

Given information:

n= 28 Sample size 

x = -0.61  Sample mean μ= 0  Population mean σ=1 Population standard deviation 

 

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