Question

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

Modeling data distributions
ANSWERED
asked 2020-10-23

A random sample of \(\displaystyle{n}_{{1}}={16}\) communities in western Kansas gave the following information for people under 25 years of age.
\(\displaystyle{X}_{{1}}:\) Rate of hay fever per 1000 population for people under 25
\(\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}\)
A random sample of \(\displaystyle{n}_{{2}}={14}\) regions in western Kansas gave the following information for people over 50 years old.
\(\displaystyle{X}_{{2}}:\) Rate of hay fever per 1000 population for people over 50
\(\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}\)
(i) Use a calculator to calculate \(\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.\) (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 \(\displaystyle\alpha={0.05}.\)
(a) What is the level of significance?
State the null and alternate hypotheses.
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}\)
(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 \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimalplaces.)

What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value \(\displaystyle>{0.250}\)
\(\displaystyle{0.125}<{P}-\text{value}<{0},{250}\)
\(\displaystyle{0},{050}<{P}-\text{value}<{0},{125}\)
\(\displaystyle{0},{025}<{P}-\text{value}<{0},{050}\)
\(\displaystyle{0},{005}<{P}-\text{value}<{0},{025}\)
P-value \(\displaystyle<{0.005}\)
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value

Expert Answers (1)

2020-10-24

To calculate mean you just need to add all the data points in the sample and divide by sample size.
\(\displaystyle\overline{{x}}_{{1}}=\frac{{{97}+{91}+{121}+\ldots\ldots..+{127}+{88}}}{{16}}={109.75}\)
\(\displaystyle\overline{{x}}_{{2}}=\frac{{{94}+{109}+{99}+\ldots\ldots..+{85}+{96}}}{{14}}={99}\)
To calculate sample standard deviation, we use the following formula
\(\displaystyle{S}_{{i}}=\sqrt{{\frac{{\sum{\left({x}_{{k}}-\overline{{x}}_{{i}}\right)}^{2}}}{{{n}_{{i}}-{1}}}}}\)
\(\displaystyle{S}_{{1}}=\sqrt{{\frac{{\sum{\left({x}_{{k}}-\overline{{x}}_{{1}}\right)}^{2}}}{{{n}_{{1}}-{1}}}}}\)
\(\displaystyle=\frac{\sqrt{{{\left({97}-{109.75}\right)}^{2}+{\left({91}-{109.75}\right)}^{2}+\ldots\ldots+{\left({127}-{109.75}\right)}^{2}+{\left({88}-{109.75}\right)}^{2}}}}{{{16}-{1}}}\)
\(\displaystyle={15.36}\)
\(\displaystyle{S}_{{2}}=\sqrt{{\frac{{\sum{\left({x}_{{k}}-\overline{{x}}_{{2}}\right)}^{2}}}{{{n}_{{2}}-{1}}}}}\)
\(\displaystyle=\frac{\sqrt{{{\left({94}-{99}\right)}^{2}+{\left({109}-{99}\right)}^{2}{\left({99}-{99}\right)}^{2}+\ldots\ldots+{\left({85}-{99}\right)}^{2}+{\left({96}-{99}\right)}^{2}}}}{{{14}-{1}}}\)
\(\displaystyle={11.66}\)
a) Level of significance \(\displaystyle={0.05}\)
\(\displaystyle\mu_{{1}}\) : mean of group under 25
\(\displaystyle\mu_{{2}}\) : 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.
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}\succ\mu_{{2}}\)
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.
\(T=\frac{\bar{x_{1}}-\bar{x_{2}}}{\text{Standard error of diference}}=\frac{109.75-99}{\frac{\sqrt{s_{1}^{2}}}{n_{1}}+\frac{s_{2}^{2}}{n-2}}=\frac{10.75}{\sqrt{14.74583+9.703297}}=2.174\)
C) p -value can be determined using t statistic and degrees of freedom
Formula for degrees of freedom is given by
\(\displaystyle{d}\frac{ f{{\left({\left({{S}_{{1}}^{{2}}}\text{/}{n}_{{1}}+{{S}_{{2}}^{{2}}}\text{/}{n}_{{2}}\right)}^{2}\right)}}}{{{\left(\frac{{{S}_{{1}}^{{2}}}}{{n}_{{1}}}\right)}^{2}\text{/}{n}_{{1}}-{1}+{\left(\frac{{{S}_{{2}}^{{2}}}}{{n}_{{2}}}\right)}^{2}\text{/}{n}_{{2}}-{1}}}={27}\)
Right tailed probability \(\displaystyle=={T}.{D}{I}{S}{T}.{R}{T}{\left({2.174},{27}\right)}={0.0193}\)

27
 
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