In Exercises 10.25-10.30, hypothesis tests are proposed. For each

hypothesis test,

a. identify the variable.

b. identify the two populations,

c. determine the null and alternative hypotheses.

d. classify the hypothesis test as two tailed, left tailed, or right tailed.

Children of Diabetic Mothers. Samples of adolescent offspring of diabetic mothers (ODM) and nondiabetic mothers (ONM) were taken by N. Cho et al. and evaluated for potential differences in vital measurements, including blood pressure and glucose tolerance. The study was published in the paper "Correlations Between the Intrauterine Metabolic Environment and Blood Pressure in Adolescent Offspring of Diabetic Mothers" (Journal of Pediatrics, Vol. 136, Issue 5. pp. 587-592). A hypothesis test is to be performed to decide whether the mean systolic blood pressure of ODM adolescents exceeds that of ONM adolescents.

Short Answer

Expert verified

Part a. The variable in this study is systolic blood pressure.

Part b. The two populations in this study are adolescent offspring of diabetic mothers (ODM) and nondiabetic mothers (ONM).

Part c. Null Hypothesis H0:μ1=μ2

and alternative hypothesis Ha:μ1>μ2

Part d. The hypothesis test is right tailed test.

Step by step solution

01

Part (a) Step 1. Given Information

We have been proposed a hypothesis test.

A hypothesis test is to be performed to decide whether the mean systolic blood pressure of ODM adolescents exceeds that of ONM adolescents.

02

Part (a) Step 2. Identify the variable

A variable is an attribute or a characteristic that can be measured. The value of the variable may differ for each and every unit. That is, a variable is defined as the characteristic which is recorded for each case.

In the given study, systolic blood pressure was measured. It is clear that systolic blood pressure varies from person to person. Hence, the variable is systolic blood pressure

03

Part (b) Step 1. Identify the two populations

The population includes all the individuals of interest that are being examined. In other words, the collection of all people, items, or objects that are required for a specific study is defined as the population.

Here, the given study involves adolescent offspring of diabetic mothers (ODM) and nondiabetic mothers (ONM). Thus, the population is all adolescent offspring of diabetic mothers (ODM) and nondiabetic mothers (ONM).

04

Part (c) Step 1. Determine the null and alternative hypothesis

Let us assume that μ1denotes the mean systolic blood pressure of ODM adolescents and μ2denotes the mean systolic blood pressure of ONM adolescents.

The null hypothesis is defined as

H0: There is no significant difference between the mean systolic blood pressure of ODM and ONM adolescents.

role="math" localid="1652709845512" H0:μ1=μ2

The alternative hypothesis is defined as

Ha:The mean systolic blood pressure of ODM adolescents exceeds that of ONM adolescents.

Ha:μ1>μ2

05

Part (d) Step 1. Classification

As our alternative hypothesis suggest that the mean of the first population is greater than the mean of the second population.

So the hypothesis can be classified as right-tailed test.

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Most popular questions from this chapter

The primary concern is deciding whether the mean of Population 1 differs from the mean of Population 2 .

Identify the assumption for using the two means ztest and the two mean zinterval procedure that renders those procedures generally impractical.

V. Tangpricha et al. did a study to determine whether fortifying orange juice with Vitamin D would result in changes in the blood levels of five biochemical variables. One of those variables was the concentration of parathyroid hormone (PTH), measured in picograms/milliliter ( pg/ml ). The researchers published their results in the paper "Fortification of Orange Juice with Vitamin D: A Novel Approach for Enhancing Vitamin D Nutritional Health" (American Journal of Clinical Nutrition, Vol. 77, pp. 1478-1483). Concentration levels were recorded at the beginning of the experiment and again at the end of 12weeks. The following data, based on the results of the study, provide the decrease (negative values indicate an increase) in PTH levels, in pg/ml, for those drinking the fortified juice and for those drinking the unfortified juice.

At the 5% significance level, do the data provide sufficient evidence to conclude that drinking fortified orange juice reduces PTH level more than drinking unfortified orange juice? (Note: The mean and standard deviation for the data on fortified juice are 9.0pg/mL and 37.4pg/mL, respectively, and for the data on unfortified juice, they are 1.6pg/mLand34.6pg/mL, respectively.)

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