Consider two independent populations that are normally distributions. A simple random sample of n_1=41 from the first population showed bar x_11=33, and a simple random of size n_2=48 from the second population showed bar x_2=32 Suppose s_1=9s_1=9 and s_2=10s_2=10, find a 98% confidence interval for mu_1−mu_2mu_1-mu_2. (Round answers to two decimal places.) margin of error-? lower limit-? upper limit-?

Consider two independent populations that are normally distributions. A simple random sample of n_1=41 from the first population showed bar x_11=33, and a simple random of size n_2=48 from the second population showed bar x_2=32 Suppose s_1=9s_1=9 and s_2=10s_2=10, find a 98% confidence interval for mu_1−mu_2mu_1-mu_2. (Round answers to two decimal places.) margin of error-? lower limit-? upper limit-?

Question
Normal distributions
asked 2021-03-11
Consider two independent populations that are normally distributions. A simple random sample of \(\displaystyle{n}_{{1}}={41}\) from the first population showed \(\displaystyle\overline{{x}}_{{11}}={33}\), and a simple random of size \(\displaystyle{n}_{{2}}={48}\) from the second population showed
\(\displaystyle\overline{{x}}_{{2}}={32}\)
Suppose \(\displaystyle{s}_{{1}}={9}{s}_{{1}}={9}{\quad\text{and}\quad}{s}_{{2}}={10}{s}_{{2}}={10}\), find a 98% confidence interval for \(\displaystyle\mu_{{1}}−\mu_{{2}}\mu_{{1}}-\mu_{{2}}\). (Round answers to two decimal places.)
margin of error-?
lower limit-?
upper limit-?

Answers (1)

2021-03-12
Step 1
From the provided information,
\(\displaystyle{n}_{{1}}={41}\)
\(\displaystyle\overline{{x}}_{{1}}={33}\)
\(\displaystyle{n}_{{2}}={48}\)
\(\displaystyle\overline{{x}}_{{2}}={32}\)
\(\displaystyle{s}_{{1}}={9}\)
\(\displaystyle{s}_{{2}}={10}\)
The confidence level = 98%
Level of significance \(\displaystyle{\left(\alpha\right)}={1}-{0.98}={0.02}\)
The degree of freedom \(\displaystyle={n}_{{1}}+{n}_{{2}}-{2}={41}+{48}-{2}={87}\)
The critical value of t at 87 degree of freedom with 0.02 level of significance from the t value table is 2.37.
Step 2
The pooled standard deviation can be obtained as:
\(\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{2}}}}}\)
\(\displaystyle=\sqrt{{\frac{{{\left({41}-{1}\right)}{\left({9}\right)}^{{2}}+{\left({48}-{1}\right)}{\left({10}\right)}^{{2}}}}{{{41}+{48}-{2}}}}}\)
=9.553
The required margin of error can be obtained as:
\(\displaystyle{E}={t}{\left({s}_{{p}}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\right)}\)
\(\displaystyle={\left({2.37}\right)}{\left({9.553}\sqrt{{\frac{{1}}{{41}}+\frac{{1}}{{48}}}}\right)}\)
\(\displaystyle\approx{4.81}\)
Thus, the margin of error is 4.81.
Step 3
The required 98% confidence interval for \(\displaystyle\mu_{{1}}-\mu_{{2}}\) can be obtained as:
\(\displaystyle{C}{I}={\left(\overline{{x}}_{{1}}-\overline{{x}}_{{1}}\right)}\pm{E}\)
\(\displaystyle={\left({33}-{32}\right)}\pm{4.81}\)
=1+-4.81
=(-3.81,5.81)
Thus, the lower limit of confidence interval is -3.81 and the upper limit is 5.81.
0

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