Question

Question: The data below were obtained to predict the sound frequency to which a person’s ear will respond ...

The data below were obtained to predict the sound frequency to which a person’s ear will respond based upon the length of time a person has been exposed to a high level of noise. Here “length of exposure” is the amount of time in weeks that a person has been living close to a major airport and the “hearing range” is reported in thousand cycles per second.

Length of exposure In weeks	Hearing range In thousand cycles per second
47	15.1
56	14.1
116	13.2
178	12.7
19	14.6
75	13.8
160	11.9
31	14.8
12	15.3
164	12.6
43	14.7
74	14.0

1. Provide 95% confidence intervals for the means (e, ii).
2. Provide 95% prediction intervals for (e, ii).
3. Comment on your findings in (f) and (g).
4. What were the basic assumptions for the model in (c)?
5. Would you consider the model in (b) to be a good model? Why or why not?

Answer 1

a. Find the correlation coefficient r.

b. Identify our x, explanatory variable and y, response variable and construct a scatter plot of y vs x.

c. Determine least squares equation that can be used for predicting a value of y based on a value of x.

d. Does this slope (in c) differ significantly from 0. Use alpha=0.05.

e. Use your model to predict y when x is i)5, ii)100. f. Provide 95% confidence intervals for the means (e, ii). g. Provide 95% prediction intervals for (e, ii).

h. Comment on your findings in (f) and (g). i. What were the basic assumptions for the model in (c)?

j. Would you consider the model in (b) to be a good model? Why or why not?

# code starts here

x<- c(47,56,116,178,19,75,160,31,12,164,43,74)
y <- c(15.1,14.1,13.2,12.7,14.6,13.8,11.9,14.8,15.3,12.6,14.7,14)

model <- lm (y ~x)

summary(model)

predict(model,data.frame(x =100), interval= "confidence" )
predict(model,data.frame(x =100), interval= "prediction" )

#running above code

summary(model)

Call:
lm(formula = y ~ x)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.62191 -0.21749 -0.00311  0.15810  0.60064 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 15.321834   0.184512  83.040 1.57e-15 ***
x           -0.017499   0.001865  -9.381 2.85e-06 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.3645 on 10 degrees of freedom
Multiple R-squared:  0.898,     Adjusted R-squared:  0.8878 
F-statistic:    88 on 1 and 10 DF,  p-value: 2.847e-06

> predict(model,data.frame(x =100), interval= "confidence" )
       fit      lwr      upr
1 13.57188 13.32482 13.81895
> predict(model,data.frame(x =100), interval= "prediction" )
       fit      lwr      upr
1 13.57188 12.72298 14.42078

f)

95% confidence interval = (13.3248,13.8190)

g)

95 % prediction interval = (12.7230,14.4208)

h)

we see that prediction interval is wider than confidence interval

i)

• There must be a linear relationship between the outcome variable and the independent variables. Scatterplots can show whether there is a linear or curvilinear relationship.
• Multivariate Normality–Multiple regression assumes that the residuals are normally distributed.
• No Multicollinearity—Multiple regression assumes that the independent variables are not highly correlated with each other. This assumption is tested using Variance Inflation Factor (VIF) values.
• Homoscedasticity–This assumption states that the variance of error terms are similar across the values of the independent variables. A plot of standardized residuals versus predicted values can show whether points are equally distributed across all values of the independent variables.

j)

Yes, as R^2 = 0.898 which is a strong fit

p-value = 0.000

hence the model is significant