Question

A random sample of n = 16 communities in western Kansas gave the following information for people under 25 years of age. X1Rate of hay fever per 1000 population for people under 25 97 91 122 127 94 123 112 93 125 95 125 117 97 122 127 88 A random sample of n2 = 14 regions in western Kansas gave the following information for people over 50 years old. X2: Rate of hay fever per 1000 population for people over 50 94 109 100 98 110 B8 110 115 100 89 114 8596 in USE SALT (i) Use a calculator to calculatex, S, , and S. (Round your answers to four decimal places.) Su X2 = 52 - (II) 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 -0.05. (a) What is the level of significance? State the null and alterate hypotheses. Hoi Hi-M2; : HIM CHOM - ; : * L2 c đt: I = 12, 1- 41 - 42 How > : N:41 - 12 (b) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known standard deviations. What is the value of the sample test statistic? (Test the difference 41 - H. Round your answer to three decimal places.)

Answer 1

i)

1755 ΣΧ, X1 109.69 n 16

51 Σ(X; - Χι)2 η –1 - 15.252

$\bg_white \overline{X_{2}} = \frac{\sum X_{i}}{n} = \frac{1387}{14} = 99.07$

$s_{2} = \sqrt{\frac{\sum (X_{i} - \overline{X_{2}})^2}{n-1}} = 11.358$

ii) a) The significance level is = 0.05

The null and the alternate hypothesis are

$H_{o} : \mu_{1} = \mu_{2}$

$H_{1} : \mu_{1} > \mu_{2}$

b) The student's t. We assume that both population distributions are approximately normal with unknown standard deviations.

The standard error of the sampling distribution is

$SE = \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}$

$SE = \sqrt{\frac{15.252^2}{16} + \frac{11.358^2}{14}} = 4.874$

The value of the sample test statistics is

$t = \frac{\overline{x_{1}} - \overline{x_{2}} - (\mu_{1} - \mu_{2})}{SE}$

$t = \frac{109.69 - 99.07 }{4.874} = 2.178$

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