Question

A random sample of n₁ = 16 communities in western Kansas gave the following information for people under 25 years of age.

x₁: Rate of hay fever per 1000 population for people under 25

100	91	122	127	92	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

93	109	100	97	111	88	110
79	115	100	89	114	85	96

(i) Use a calculator to calculate x₁, s₁, x₂, and s₂. (Round your answers to two decimal places.)

x1	=
s1	=
x2	=
s2	=

(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 α = 0.05.
(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: μ₁ = μ₂; H₁: μ₁ < μ₂H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂    H₀: μ₁ = μ₂; H₁: μ₁ > μ₂H₀: μ₁ > μ₂; H₁: μ₁ = μ₂

(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 unknown standard deviations.The standard normal. We assume that both population distributions are approximately normal with known standard deviations.    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.

What is the value of the sample test statistic? (Test the difference μ₁ − μ₂. Round your answer to three decimal places.)


(c) Find (or estimate) the P-value.

P-value > 0.2500.125 < P-value < 0.250    0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005

Answer 1

a)

i)

x̅1=   109.750
s1 =    15.25
x̅2=   99.000
s2 =    11.48

ii)

Level of Significance ,    α =    0.05
H₀: μ₁ = μ₂; H₁: μ₁ > μ₂

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

difference in sample means =    x̅1-x̅2 =    109.7500   -   99.0   =   10.750  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    13.6316                  
std error , SE =    Sp*√(1/n1+1/n2) =    4.9887                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   10.7500   -   0   ) /    4.99   =   2.155

c)

Degree of freedom, DF=   n1+n2-2 =    28

0.005 < P-value < 0.025