SECTION 10.2 Inferences about the Difference of Two Means µ,-,513X,: Rate of hay fever per 1000 population for people over 50951101019711288110791151008911485 96(Reference: National Center for Health Statistics.)Use a calculator to verify that I 109.50, s, -15.41, x, 99.36, and11.57.(a) Assume that the hay fever rate in each age group has an approximatelynormal distribution. Do the data indicate that the age group over 50 has aS2lower rate of hay fever? Use a = 0.05.(b) Find a 90% confidence interval for u, - Hy. Explain the meaning of theconfidence interval in the context of the problem.Education: Tutoring In the journal American Journal on Intellectual and .Explain the meaning of the(a) Do the data indiregion is higher than that in New England? Use a =(b) Find a 98% confidence interval for uconfidence interval in the context of the problem.ain0.01.H. Explain the meaning of the16. Medicine: Hay Fever A random sample of n.Kansas gave the following information for people under 25 years of ace16 communities in western%3D1: Rate of hay fever per 1000 population for people under 25x,:12012892123112 9398 90127 88125 95 125 117 97 12214 regions in western Kansas gave the followingA random sample of n,information for people over 50 years old:

Hay fever A random sample of n1=16 communities in westerns Kansas gave the following information for people under 25 years of age. x1: Rate of hay fever per 1000 population for people under 25 98 90 120 128 92 123 112 93 125 95 125 117 97 122 127 88 The sample size, mean, standard deviation for first sample is n1 =16 x1 =109.50s1 = 15.41 x2: Rate of hay fever per 1000 population for people over 50 95 110 101 97 112 88 110 79 115 100 89 114 85 94 n2 =14x2 =99.36s2 =11.57 Here variance is unknown so used t tests for hypothesis testing The hypothesis tests is H0 :μ1 =μ2 Versus H1 : μ1 > μ2 The tests statistics is t = x1 -x2 -μ1 - μ2s21n1 +s22n2 Degree of freedom (d.f.) = n1 +n2 -2d. f. = 16+14-2d. f. = 28 The data indicates that the age group over 50 has a lower rate for a hay fever? Use alpha =0.05 Assume that the hay fever rate of each age group has an approximately normal distribution . t =109.50-99.36 -015.41216 + 11.57214t = 10.1414.8418 + 9.5618t = 10.144.940t = 2.0526 The critical value for d.f. = 28 with 0.05 level of significance is tα2,28 = 1.701 The critical value is less than tests statistics value. So reject the null hypothesis at 0.05 level of significance We have sufficient evidence to claim the age group over 50 has a over rate of hay fever. The 90% confidence interval for μ1 -μ2 t0.1/2, 28 =1.313 C. I. =x1 -x2 -tα2,df s21n1 +s22n2 ≤ μ1 -μ2 ≤ x1 -x2 +tα2,df s21n1 +s22n2C. I = (109.50-99.36) - 1.313×4.940 ≤ μ1 -μ2 ≤ (109.50-99.36) + 1.313×4.940 C. I. = 10.14-6.4862 ≤ μ1 -μ2 ≤ 10.14+6.4862 C. I. =3.6538 ≤ μ1 -μ2 ≤ 16.6262 We can see that the sample mean is not belong to the 90% confidence interval. So the hypothesis test is rejected.

Math

Statistics

(a) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use a b) Find a 90% confidence interval for u, - ,. Explain the meaning of the confidence interval in the context of the problem. 0.05.

(a) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use a b) Find a 90% confidence interval for u, - ,. Explain the meaning of the confidence interval in the context of the problem. 0.05.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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