14.14 there evidence that the mean in this district differs from 100? Follow the four-stepve mean u00. Isprocess, as illustrated in Example 14.9, in your answer.bone mineral density? In the general population of healthy young adults, bonemineral density is Normally distributed, and density scores are standardized tobave a mean jµ of 0 and a standard deviation o of 1. A study examined thebone 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 thisprovide good evidence that the mean standardized bone density of adult menwith chronic hepatitis C infection is less than 0? Follow the four-step process asillustrated in Example 14.9.OL SVISini sonsbiinoo ofc21212TESTS FROM CONFIDENCE INTERVALSsampe mean x to estimate µ. Both rely on probabilities calculated from a Normalsampiing distribution. In fact, a two-sided test at significance level a can becarried out from a confidence interval with confidence level C =1 – a.Both cests and confidence intervals for a population mean u start by using theCHVBLEKA level a two-sided hypothesis test rejects a hypothesis Ho: µ = µoexactly when the value uo falls outside a level 1 – a confidence intervalfor M.SONFIDENCE INTERVALS AND TWO-SIDED TESTS%3D 音号Praehave a µ of 0 and a obone In the of adults14.14 C and Does C infectionexpected a distribution.he study findings are highly statistically significant (P~0). This doesn'tthat the true mean body têmperature in the population of healthy adults isfferent from 98.6 °F. Rather, it just says that we have very strong evidenceverject 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 thereforeTESTS FROMuch further away a hypothetical population mean of 98.6 °F than would beexpected under such a narrow sampling distribution.The computations performed by any statistical software in this example involvethe following steps:from1 Test statistic: z= (x – 0)(0//n) = (98.25 – 98.6)/(0.6//130) = -6.65%3DI P-value: P = 2P(Z < –6.65) ~ 0, computed using the standard Normal curveThis introductory chapter describes the computations involved in statistical infer-ence when the “simple conditions" are met. As we have noted here, these "simpleconditions" are not easily met. In Chapter 15 we will discuss circumstances whersome ci the conditions can be relaxed. In Chapter 17 we will describe a workable alternative to the problem of not knowing the true value of the populatiostandard deviation ơ .DOLAPPLY YOUR KNOWLEDGE14.13 TQ test scores. Example 14.3 gives the IQ test scores of 31 seventh-grade gin a Midwest school district. IQ scores follow a Normal distribution with stanedeviation o ==15. Treat these 31 girls as an SRS of all seventh-grade girls indistrict. IQ scores in a broad population are supposed to have mean u = 10there evidence that the mean in this district differs from 100? Follow the fourprocess, as illustrated in Example 14.9, in your answer.%3D%3Dof hotbone mineral density? In the general population of healthy young adultsmineral density is Normally distributed, and density scores are standardDOne densities of a random sample of 28 adult men living with chronicbeir mean standardized bone density was -0.61. Dmean standardized bone density of adO? Follow the four-step pnave a mean u of 0 and a standard deviation o of 1. A study examin

.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.