Answer to Question #318074 in Statistics and Probability for Owen

Question

1.Given a finite population 12, 14, 15, 16, and 20.

a Find the mean and the standard deviation to estimate the parameters.

b. Set up a sampling distribution of the mean using a sample size of 2 without replacement.

c. Show that the sampling distribution of the sample means is an unbiased estimator of the population mean.

2. Given the following data n = 75,= 50, and a = 10, the sampled population is normally distributed. a Compute the standard error of the mean.

b. At 95% confidence level, what is the margin of error?

c. Find 90% confidence interval for the population mean.

Answer 1

$a:\\\mu =\frac{12+14+15+16+20}{5}=15.4\\\sigma =\sqrt{\frac{\left( 12-15.4 \right) ^2+\left( 14-15.4 \right) ^2+\left( 15-15.4 \right) ^2+\left( 16-15.4 \right) ^2+\left( 20-15.4 \right) ^2}{5}}=2.6533\\b:\\Samples\,\,are\\\left( 12,14 \right) ,\bar{x}=13\\\left( 12,15 \right) ,\bar{x}=13.5\\\left( 12,16 \right) ,\bar{x}=14\\\left( 12,20 \right) ,\bar{x}=16\\\left( 14,15 \right) ,\bar{x}=14.5\\\left( 14,16 \right) ,\bar{x}=15\\\left( 14,20 \right) ,\bar{x}=17\\\left( 15,16 \right) ,\bar{x}=15.5\\\left( 15,20 \right) ,\bar{x}=17.5\\\left( 16,20 \right) ,\bar{x}=18\\P\left( \bar{x}=13 \right) =P\left( \bar{x}=13.5 \right) =P\left( \bar{x}=14 \right) =P\left( \bar{x}=14.5 \right) =P\left( \bar{x}=15 \right) =\\=P\left( \bar{x}=15.5 \right) =P\left( \bar{x}=16 \right) =P\left( \bar{x}=17 \right) =P\left( \bar{x}=17.5 \right) =P\left( \bar{x}=18 \right) =0.1\\c:\\E\bar{x}=0.1\left( 13+13.5+14+14.5+15+15.5+16+17+17.5+18 \right) =15.4\\2.a:\\s=\frac{\sigma}{\sqrt{n}}=\frac{10}{\sqrt{75}}=1.1547\\b:\\E=\frac{\sigma}{\sqrt{n}}t_{n-1,\frac{1+\gamma}{2}}=\frac{10}{\sqrt{75}}\cdot 1.9925=2.30074\\c:\\\gamma =0.9: \left( \bar{x}-\frac{s}{\sqrt{n}}t_{n-1,\frac{1+\gamma}{2}},\bar{x}+\frac{s}{\sqrt{n}}t_{n-1,\frac{1+\gamma}{2}} \right) =\\=\left( 50-\frac{10}{\sqrt{75}}\cdot 1.6657,50+\frac{10}{\sqrt{75}}\cdot 1.6657 \right) =\left( 48.0766,51.9234 \right) \\\gamma =0.95: \left( \bar{x}-E,\bar{x}+E \right) =\left( 50-2.3007,50+2.3007 \right) =\\=\left( 47.6993,52.3007 \right)$