Question

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 HW_PTA 243.csv.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?
A) The cooling system changes the temperature of the water by an amount different than 6 degrees.
B) The cooling system changes the temperature of the water by 6 degrees.

Answer 1

Complete Question

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

Answer:

A

 The correct option is A

B

  The correct option is H

Step-by-step explanation:

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

The alternative  hypothesis is  $H_a  :  \mu_1 - \mu_2 \ne  6$      

The population mean difference is  $\mu_d  =  6$

Generally the sample size is  n =  15      

Generally the sample mean for 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 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 sample mean difference is mathematically represented as

   $\= d = 62.57 - 55.97$

=>$d = 6.6$

Generally the standard deviation is mathematically represented as

       $s_d  =  \sqrt{\frac{ \sum [d_1 - \= d]^2}{n-1} }$

=>    $s_d  =  \sqrt{\frac{ [ [57.6 -65.1]- 6.6]^2 +[68.9 -74.4]- 6.6]^2+\cdots +   [60.4 -67.3]- 6.6]^2}{15-1} }$

=>     $s_d  = 1.732$

Generally the test statistics is mathematically represented as

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

=>   $t = 1.342$

Generally the degree of freedom is mathematically represented as

    $df  = n -  1$

So

      $df  =  15  -  2$

=>    $df  =  14$

Generally the p-value  is mathematically represented as

     $p-value  = 2 *  P(t >  1.342 )$

Generally from the t -distribution table the probability of 1.342 at a degree of freedom of  $df  =  28$ is  

       $P(t >  1.342 ) = 0.10047$

So

     $p-value  = 2 * 0.10047$

=>   $p-value  = 0.201$

Generally from the value  obtained we see that $p-value  > \alpha$ 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