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 \(\displaystyle=={T}.{D}{I}{S}{T}.{R}{T}{\left({2.174},{27}\right)}={0.0193}\)