2020 Data SAC B Revision Solutions PDF

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The European countries (Italy and France) tend to have a higher GDP compared to Australia. Higher GDP would attract more investors to purchase shares within the company like seen in figure 3 (Max Roser 2020). This would result in increased economic growth. As levels of demand for imports and oversea...

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Task B Is there a relationship between a student’s average daily sleep time (hours) and Body mass index? i)

Create a scatterplot of Body mass index against average daily sleep time (hours) for the given data.

The scatterplot indicates there is a strong, neagative non-linear associations between Body mass index against average daily sleep time (hours). There are no apparent outliers.

ii)

Calculate the least squares regression line, correlation coefficient and discuss the relationship in terms of strength, direction and form.

BMI =50.06−2.93 ×( Average Daily Sleep Time) r=−0.8941

The correlation coefficient indicates the relationship between Body mass index against average daily sleep time (hours) is strong, negative and of a linear form.

iii)

Draw the least squares regression line on the scatterplot Page 1 of 4

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Most show the claculation of two points and plot these accurately on the scatterplot. x=0

y=50.06-2.93x0=50.06

(0,50.06)

x=12

y=50.06-2.93x13=11.97

(13,11.97)

iv)

Interpret the slope and vertical intercept in terms of Body mass index and average daily sleep time (hours). Calculate the coefficient of determination and discuss what this tell us.

Slope is -2.93. On average the BMI decreases by 2.96 for each additional hour of sleeper day. The vertical intercept tis 50.06 this tells us that when the average daily sleep time is 0 hours the expected body mass index is 50.06.

v)

Calculate the coefficient of determination and discuss what this tell us.

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r =0.80

80% of variation in the body mass index can be explained by the variation in average daily sleep time.

i)

Looking at your class data select a student. Using the least squares regression line predict the Body mass index that the student will receive for their average daily sleep time (hours) Compare this answer with the actual and calculate the residual value. Student selected: C (3.5,46) Predicted value:

BMI =50.06−2.93 ×(3.5) ¿ 39.805 For student C with an average daily sleep time of 3.5 hours the predicted BMI value (39.805) is below the actual BMI value (46) Residual=actual BMI− predicted BMI

Residual=46 −39.805=6.195

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i)

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Perform a residual analysis to test the linearity of the model. If not linear, choose 2 suitable transformations and apply them to the data. Discuss why the two transformations were selected. Calculate the least squares regression line and correlation coefficient for the transformed data. Discuss which least squares regression line is best suited for making predictions.

The residual plot shows a curved pattern, therefore the residual plot indicates the linear model for Body mass index and average daily sleep time (hours) is not the most suitable model for this data. The assumption of linearity is not supported by the residual plot. Looking at the shape of the original scatterplot the two transformations chosen to improve linearity are log y, log x, reciprocal y and reciprocal x (state any 2) as the data had a decreasing trend.

log y 1 x

Transformed equations log ( BMI ) =1.73−0.04 × ( average daily sleep time ) 1 BMI =14.64+81.65 × average daily sleep time

The equation that should be used for making predictions is BMI =14.64+81.65 ×

1 average daily sleep time Page 3 of 4

r -0.9259 0.9340

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Because the correlation coefficient is closest to 1/-1 the reciprocal x transformation (r=0.9340) compared to r=-0.9259 log y transformation and r=−0.8941 for original least squares regression line. i)

Using the most suitable least squares regression line predict the average daily sleep time (hours) for a student with a Body mass index of 31. 1 31=14.64 + 81.65 × ii) averagedaily sleep time Average daily sleep time= 4.99 hours

vi)

Using the most suitable least squares regression line predict a student’s Body mass index if their average daily sleep time (hours) was 3 hours. 1 BMI =14.64+81.65 × 3 BMI=20.92

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