Question

The measurements of a temperature sensor (in °F) are provided. Is the set of measurements well described by a normal distribution? Calculate the R^2 and R^2 adjusted values. Show all the calculations.

Order	Temperature (F)
1	118
2	119
3	120
4	121
5	121
6	122
7	123
8	123
9	124
10	124
11	124
12	124
13	125
14	125
15	125
16	125
17	125
18	126
19	127
20	128
21	128
22	129
23	129
24	129
25	130
26	131
27	131
28	131
29	133
30	133
31	136
32	137
33	137
34	140
35	140
36	141
37	142
38	142
39	151
40	160

Answer 1

I have used the R software to plot the values in order to check if its normal. I will attach the plot and the commands along with its output below.

>x=c(118,119,120,121,121,122,123,123,124,124,124,124,125,125,125,125,125,126,127,128,128,129,129,129,130,131,131,131,133,133,136,137,137,140,140,141,142,142,151,160)
> qqnorm(x)
> qqline(x)

If most of the values lie on the line, the set of values are said to follow a normal distribution. Since most of our values does not pass through our line, we can consider it not to be normal.

120 130 Sample Quantiles 140 150 160 Y T Theoretical Quantiles Normal Q-Q Plot

To find the R² and R² adjusted values,

Let us consider X as the order and Y as Fahrenheit and find the regression equation.

The regression equation is $\hat{Y}=\hat{b_{0}}+\hat{b_{1}}X$

Where, $\hat{b_{1}}=\frac{S_{xy}}{S_{xx}}$

$\hat{b_{0}}=\bar{y}-(\hat{b_{1}}*\bar{x})$

here, $S_{xx}=\sum_{i=1}^{40}X_{i}^{2}-\frac{(\sum_{i=1}^{40}X_{i})^{2}}{40}$

$S_{xy}=\sum_{i=1}^{40}X_{i}*Y_{i}-\frac{\sum_{i=1}^{40}X_{i}*\sum_{i=1}^{40}Y_{i}}{40}$

$\bar{x}=\frac{\sum_{i=1}^{40}X_{i}}{40} \: and \: \bar{y}=\frac{\sum_{i=1}^{40}y_{i}}{40}$

After substituting the values, we get S_xx=5330 and S_xy=3719.5

$\bar{x}=20.5\: and \: \bar{y}=129.975$

$\hat{b_{1}}=\frac{3719.5}{5330}=0.6978$

$\hat{b_{0}}=129.975-(0.6978*20.5)=115.6701$

The regression equation becomes $\hat{Y}=115.6701+0.6978X$

$R^{2}=1-\frac{SSE}{S_{yy}}$;

where SSE=sum of squares of errors = $\sum_{i=1}^{40}(y_{i}-\hat{y_{i}})^{2}$ =503.3502

$S_{yy}=\sum_{i=1}^{40}Y_{i}^{2}-\frac{(\sum_{i=1}^{40}Y_{i})^{2}}{40}=3098.975$

$R^{2}=1-\frac{503.3502}{3098.975}=0.8376$

$R^{2}_{adj}=1-\frac{n-1}{n-k-1}(1-R^{2})$

Here, k-number of independent variables = 1

$R^{2}_{adj}=1-\frac{39}{38}(1-0.8376)=0.8333$

Therefore, our model explains 83.33% of variability.