Let two independent random samples, each of size 10, from two normal distributions N(mu_1, sigma_2) and N(mu_2, sigma_2) yield x = 4.8, s_1^2 = 8.64, y = 5.6, s_2^2 = 7.88. Find a 95% confidence interval for mu_1 − mu_2.

Question
Normal distributions
asked 2021-02-04
Let two independent random samples, each of size 10, from two normal distributions \(\displaystyle{N}{\left(\mu_{{1}},\sigma_{{2}}\right)}{\quad\text{and}\quad}{N}{\left(\mu_{{2}},\sigma_{{2}}\right)}\) yield \(\displaystyle{x}={4.8},{{s}_{{1}}^{{2}}}\)
\(\displaystyle={8.64},{y}={5.6},{{s}_{{2}}^{{2}}}\)
= 7.88.
Find a 95% confidence interval for \(\displaystyle\mu_{{1}}−\mu_{{2}}\).

Answers (1)

2021-02-05
Step 1
\(\displaystyle{A}{100}{\left({1}−\alpha\right)}\%\) confidence interval on the difference is
\(\displaystyle\overline{{x}}_{{1}}-\overline{{x}}_{{2}}-{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\le\mu_{{1}}-\mu_{{2}}\le\overline{{x}}_{{1}}-\overline{{x}}_{{2}},+{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\)
where \(\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{1}}}}}\)
Step 2
Considering 95% confidence interval,
\(\displaystyle\alpha={0.05}\)
\(\displaystyle\overline{{x}}_{{1}}={4.8}\)
\(\displaystyle\overline{{x}}_{{2}}={5.6}\)
\(\displaystyle{{s}_{{1}}^{{2}}}={8.64}\)
\(\displaystyle{{s}_{{2}}^{{2}}}={7.88}\)
\(\displaystyle{n}_{{1}}={10}\)
\(\displaystyle{n}_{{2}}={10}\)
Therefore
\(\displaystyle{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}={t}_{{\frac{{0.05}}{{2}},{10}+{10}-{1}}}\)
\(\displaystyle={t}_{{{0.025},{19}}}\)
=2.093
Step 3
To calculate \(\displaystyle{s}_{{p}}\)
\(\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{1}}}}}\)
\(\displaystyle{s}_{{p}}=\frac{\sqrt{{{\left({10}-{1}\right)}{8.64}+{\left({10}-{1}\right)}{7.88}}}}{{{19}}}{)}\)
\(\displaystyle{s}_{{p}}={2.797}\)
Step 4
Now, by substituting all the values in the equation, we get
\(\displaystyle\overline{{x}}_{{1}}-\overline{{x}}_{{2}}-{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\le\mu_{{1}}-\mu_{{2}}\le\overline{{x}}_{{1}}-\overline{{x}}_{{2}}\)
\(\displaystyle={t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\)
\(\displaystyle{4.8}-{5.6}-{2.093}\times{2.797}\times{0.447}\le\mu_{{1}}-\mu_{{2}}\le{4.8}-{5.6}+{2.093}\times{2.797}\times{0.447}-{3.424}\le\mu_{{1}}-\mu_{{2}}\le{1.82}\)
Hence 95% confidence interval in \(\displaystyle\mu_{{1}}-\mu_{{2}}{i}{s}-{3.424}\le\mu_{{1}}-\mu_{{2}}\le{1.82}\)
0

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