Question

Paired t-test Application (Part 1)An environmental engineer is tasked with determining whether a power plant cooling system is heating the water it uses more than allowed by environmental regulations. They measure water temperatures at the cooling system input and the cooling system output for several different days in several different seasons. A data file containing these measurements is Fin_PTA.csvPreview the document. Do a statistical analysis on this data to determine if the temperature change between the input and output of the cooling system is different than 6 degrees. What does your analysis indicate

Answer 1

Complete Question

The complete question is shown on the first and second  uploaded image

Answer:

Part 1

 The correct option is  B

Part 11

   The correct option is  H

Step-by-step explanation:

From the question we are told that

     The sample size is  n  =  15

Generally the sample mean for the input temperature is mathematically represented as

        $\= x _1 = \frac{\sum x_i}{n}$

        $\= x _1 = \frac{57.6 + 68.9 \cdots +60.4 }{15}$

      $\= x _1 = 62.57$

Generally the sample mean for the output temperature is mathematically represented as

        $\= x _2 = \frac{\sum x_i}{n}$

        $\= x _2 = \frac{65.1 + 74.4 \cdots +67.3 }{15}$

      $\= x _2 = 55.97$

Generally the difference between the mean of the input temperature and that of the output temperature is

      $d = \= x_1 - \= x_2$

=>    $d = 62.57 - 55.97$

=>    $d = 6.6$

Generally the standard deviation of the difference between the input temperature and the output temperature is mathematically represented as

     $s_d = \sqrt {\frac{1}{n-1 } \sum [d_i - d]^2}$

=>   $s_d = \sqrt{\frac{[(57.6 - 65.1) - 6.6]^2+[(68.9 - 74.4) - 6.6]^2+ \cdots +[(68.1 - 74.7) - 6.6]^2 }{15-1} }$

=>   $s_d = 1.732$

The null hypothesis is  $H_o : \mu_1 -\mu_2 = 6$

The alternative hypothesis is  $H_a : \mu _1 - \mu_2\ne 6$

Generally the test statistics is mathematically represented as

           $t = \frac{d - 6}{ \frac{s_d}{ \sqrt{n} } }$

          $t = \frac{6.6- 6}{ \frac{1.732}{ \sqrt{15} } }$

          $t = 1.342$

Generally the p-value is mathematically represented as

    $p-value = 2 * P(t > 1.342)$

From the student t distribution table( reference - danielsoper(dot)com(slash)statcalc(slash)calculator) at a degree of freedom of  df = n-1 =  15-1 = 14

       $P(t > 1.342) = t_{1.342 , 14} = 0.100478$

So    

     $p-value = 2 * 0.100478)$

     $p-value = 0.201$

From the values obtained we see that the $p-value > \alpha$ hence the decision rule is  fail to reject the null hypothesis

The  conclusion is  

The cooling system changes the temperature of the water by 6 degrees.