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 list containing these measurements is below:
Sample Input Temp (deg F) Output Temp (deg F)
1 57.6 65.1
2 68.9 74.4
3 68.1 74.7
4 47.7 55.4
5 48.1 51
6 67.2 71.9
7 69.9 79.2
8 48.1 57.2
9 35 41.4
10 56 61
11 44.3 50.8
12 65 70.5
13 48.3 56.9
14 54.9 61.7
15 60.4 67.3

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.

1. 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.

2. What is the P-Value associated with your decision above?

a. 0.001
b. 0.026
c. 0.049
d. 0.070
e. 0.110
f. 0.131
g. 0.148
h. 0.201
i. 0.278
j. 0.289

Answer 1

Answer:

1

The correct option is  b

2

The correct option is  h

Step-by-step explanation:

From the question we are told that

  The sample size is  n  = 15

   The difference in population proportion is $d = 6$

Generally the sample mean of the input temperature is  

     $\= x_1 = \frac{\sum x_i }{n}$

=>  $\= x_1 = \frac{57.6 + 68.9 + \cdots +60.4 }{15}$

=>  $\= x_1 = 62.57$

Generally the sample mean of the output temperature is  

     $\= x_2 = \frac{\sum x_i }{n}$

=>  $\= x_2 = \frac{65.1 + 74.4 + \cdots +67.3 }{15}$

=>  $\= x_2 = 55.97$

Generally the difference of the sample 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 is mathematically represented as

     $s_d = \sqrt{\frac{\sum(d_1 - \= d )^2}{n} }$

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

=>  $s_d = 1.732 [/texGenerally the test statistics is mathematically represented as [tex]t = \frac{ d - \= d }{ \frac{s_d}{\sqrt{n} } }$

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

=>   $t = 1.3417$

Generally the degree of freedom is mathematically represented as

    $df = n - 1$

=>  $df = 15 - 1$

=>  $df = 14$

Generally the probability of t obtained from the t - distribution table at a degree of freedom of  $df = 14$ is  

      $P(t >1.3417 ) = 0.10052569$

Generally the p-value is mathematically represented as

      $p-value = 2 * P(t > 1.3417)$

=>    $p-value = 2 * 0.10052569$

=>   $p-value = 0.201$

From the values 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.