Question

A random sample of displaystyle{n}={400} students is selected from a population of displaystyle{N}={4000} students to estimate the average weight of t

Confidence intervals
ANSWERED
asked 2021-03-05
A random sample of \(\displaystyle{n}={400}\) students is selected from a population of \(\displaystyle{N}={4000}\) students to estimate the average weight of the students. The sample mean and sample variance are found to be \(\displaystyle{x}={140}{l}{b}{\quad\text{and}\quad}{s}^{2}={225}.\ \text{Find the}\ {95}\%{\left({z}={2}\right)}\) confident interval.

Answers (1)

2021-03-06
Step 1
The \(\displaystyle{\left({1}-\alpha\right)}{100}\%\) confidence interval formula for the population mean when population standard deviation is not known, is defined as follows:
\(\displaystyle{C}{I}=\overline{{x}}\pm{t}_{{\frac{\alpha}{{2}},{n}-{1}}}{\left(\frac{s}{\sqrt{{n}}}\right)}.\)
\(\displaystyle{t}_{{\frac{\alpha}{{2}},{n}-{1}}}\) is the critical value of the t-distribution with degrees of freedom of \(\displaystyle{n}-{1}\) above which, \(\displaystyle{100}{\left(\frac{\alpha}{{2}}\right)}\%{\quad\text{or}\quad}\frac{\alpha}{{2}}\) proportion of the observations lie, and below which, \(\displaystyle{100}{\left({1}-\alpha+\frac{\alpha}{{2}}={100}{\left({1}-\frac{\alpha}{{2}}\right)}\%{\quad\text{or}\quad}{\left({1}-\frac{\alpha}{{2}}\right)}\right.}\) proportion of the observation lie, \(\displaystyle\overline{{x}}\) is the sample mean, s is the sample standard deiation, and n is the sample size.
Step 2
The sample mean is \(\displaystyle\overline{{x}}={140}{l}{b}\), and the sample standard deviation is \(\displaystyle{s}={15}{l}{b}{\left(=√{225}\right)}\).
The sample size is \(\displaystyle{n}={400}\). The degrees of freedom is \(\displaystyle{399}{\left(={400}–{1}\right)}.\)
The confidence level is 0.95. Hence, the level of significance is \(\displaystyle{1}–{0.95}={0.05}.\)
Using Excel formula: \(\displaystyle={T}.{I}{N}{V}{.2}{T}{\left({0.05},{399}\right)}\), the critical value is, \(\displaystyle{t}_{{\frac{\alpha}{{2}}}}={t}_{{{0.025}}}\approx{1.96}.\)
Thus,
\(\displaystyle{C}{I}=\overline{{x}}\pm{t}_{{\frac{\alpha}{{2}}}},{n}-{1}{\left(\frac{s}{\sqrt{{n}}}\right)}\)
\(\displaystyle={140}\pm{\left({1.96}\right)}{\left(\frac{15}{\sqrt{{400}}}\right)}\)
\(\displaystyle={140}\pm{\left({1.96}\right)}{\left({0.75}\right)}\)
\(\displaystyle={140}\pm{1.47}\)
\(\displaystyle={\left({138.53},{141.47}\right)}\)
Thus, the required \(95\%\) confidence interval is (138.53 lb, 141.47 lb).
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