Answer to Question #326046 in Statistics and Probability for Hasmin

Question

The head of the Philippine University (PU) observes a decline on the alcoholic expenditures of learners from a monthly expenditure of Php 350 pesos in the previous year. To check on this, he randomly selected 10 PU learners who drink alcoholic beverages and asked the amount, in pesos, that they usually spend on alcoholic beverages in a month. It is known that the usual amount spent on alcoholic beverages by learners who drink alcoholic beverages follows the normal distribution with standard deviation of Php 10. The data collected are: 400, 235, 200, 250, 200, 300, 500, 430, 420, and 220.

Construct and interpret a 95% confidence interval for the true mean amount spent by the learners on alcoholic beverages.

Find a 90% confidence interval for the true mean amount spent by the learners on alcoholic beverages.

Answer 1

$n=10\\\bar{x}=315.5\\s^2=12446.94\\\gamma =0.95:\\\left( \bar{x}-\frac{s}{\sqrt{n}}t_{\frac{1+\gamma}{2},n-1},\bar{x}+\frac{s}{\sqrt{n}}t_{\frac{1+\gamma}{2},n-1} \right) =\\=\left( 315.5-\sqrt{\frac{12446.94}{10}}\cdot 2.2622,315.5+\sqrt{\frac{12446.94}{10}}\cdot 2.2622 \right) =\\=\left( 235.689,395.311 \right) \\The\,\,true\,\,mean\,\,is\,\,in\,\,the\,\,interval\,\,\left( 235.689,395.311 \right) \,\,with\,\,probability\,\,0.95\\\gamma =0.9:\\\left( \bar{x}-\frac{s}{\sqrt{n}}t_{\frac{1+\gamma}{2},n-1},\bar{x}+\frac{s}{\sqrt{n}}t_{\frac{1+\gamma}{2},n-1} \right) =\\=\left( 315.5-\sqrt{\frac{12446.94}{10}}\cdot 1.8331,315.5+\sqrt{\frac{12446.94}{10}}\cdot 1.8331 \right) =\\=\left( 250.828,380.172 \right)$