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 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. 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).