Question

suspects that when the temperature outside is higher more failures occur. data for 18 days and the temperatures in C

temp	failure
21.9	13
22.8	12
17.4	16
21.9	17
22.7	15
19.8	16
19.5	17
18.1	15
21.8	16
22.5	16
18.6	15
21.7	16
20.5	17
19.3	16
19.0	16
21.1	14
20.2	17
18.5	14

Is there evidence that more computer failures occur during hotter weather?

1. Null hypothesis ?

2. linear regression using R and interpret

Answer 1

1. Notice that here we want to test whether there is evidence that more computer failures occur during hotter weather. Note that the null hypothesis is always a hypothesis of no difference and what researcher wants to test or researcher's claim is alternative hypothesis. So

Null hypothesis Ho is

Ho: There is no relation between computer failures and temperature.

Note that the we want to test whether there is a significant correlation between temperature and computer failures. Hence the null hypothesis is as stated above.

2. Linear regression using R:

Observe that in this case it is believed that computer failures depends on temperature so

Independent variable(X) : Temperature recorded on a day.

Dependent Variable(Y) : Number of computer failures on a day.

We will use R software to fit the regression equation with above stated variables.

R software commands

>x=c(21.9,22.8,17.4,21.9,22.7,19.8,19.5,18.1,21.8,22.5,18.6,21.7,20.5,19.3,19,21.1,20.2,18.5)

>y=c(13,12,16,17,15,16,17,15,16,16,15,16,17,16,16,14,17,14)

>model=lm(y~x)

>summary(model)

We have created data vectors x and y. We used lm(y~x) function to fit the regression model. summary() function is used to get the summary statistics.

R software output

Observe the above output. Regression equation is

= 19.2865 - 0.1883 x

In words,

Computer failures= 19.2865-0.1883(temperature).

Observe above output. Multiple R-squared measures the percentage of variation in the dependent variable explained by independent variable and it is very helpful in determining whether the fitted regression model is good. Multiple R-squared is 0.05108 which is interpreted as 5.108% of variation in the dependent variable is explained by independent variable. Since only 5.108% of variation is explained which implies the fit is not good.

If you have any doubt, please do comment.

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