A random sample of displaystyle{n}_{{1}}={16} communities in western Kansas gave the following information for

postillan4

postillan4

Answered question

2020-10-23

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
97911211299412311293125951251179712212788
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
9410999951138811079115100891148596
(i) Use a calculator to calculate x1,s1,x2,ands2. (Round your answers to two decimal places.)
(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<μ2
H0:μ1=μ2,H1:μ1>μ2
H0:μ1=μ2,H1:μ1μ2
H0:μ1>μ2,H1:μ1=μ12
(b) What sampling distribution will you use? What assumptions are you making?
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 unknown 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 known standard deviations,
What is the value of the sample test statistic? (Test the difference μ1μ2. Round your answer to three decimalplaces.)

What is the value of the sample test statistic? (Test the difference μ1μ2. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value >0.250
0.125<Pvalue<0,250
0,050<Pvalue<0,125
0,025<Pvalue<0,050
0,005<Pvalue<0,025
P-value <0.005
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value

Answer & Explanation

brawnyN

brawnyN

Skilled2020-10-24Added 91 answers

To calculate mean you just need to add all the data points in the sample and divide by sample size.
x1=97+91+121+..+127+8816=109.75
x2=94+109+99+..+85+9614=99
To calculate sample standard deviation, we use the following formula
Si=(xkxi)2ni1
S1=(xkx1)2n11
=(97109.75)2+(91109.75)2++(127109.75)2+(88109.75)2161
=15.36
S2=(xkx2)2n21
=(9499)2+(10999)2(9999)2++(8599)2+(9699)2141
=11.66
a) Level of significance =0.05
μ1 : mean of group under 25
μ2 : mean of group over 50
The claim is generally forms the Alternate hypotheses. As the claim is that the mean (rate of hay fever) of group over 50 is lower, the alternate hypotheses will reflect the claim and null hypotheses will be a null statement indicating no change or no difference.
H0:μ1=μ2
H0:μ1μ2
b)
With the assumption that the population for both groups are normally distributed and because the population standard deviations are unknown, we will use students' t test. Students t test. We assume that both population distributions are approximately normal with unknown standard deviations.
T=x1¯x2¯Standard error of diference=109.7599s12n1+s22n2=10.7514.74583+9.703297=2.174
C) p -value can be determined using t statistic and degrees of freedom
Formula for degrees of freedom is given by
df((S12/n1+S22/n2)2)(S12n1)2/n11+(S22n2)2/n21=27
Right tailed probability

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