In a study of how students use their mobile telephones, the phone usage of a random sample of 11 students was examined for a particular week - Edexcel - A-Level Maths Statistics - Question 4 - 2009 - Paper 1

To identify outliers, calculate:

1. $Q_3 - Q_1 = 60 - 35 = 25$
2. $Q_3 + 1.5 \times (Q_3 - Q_1) = 60 + 1.5 \times 25 = 60 + 37.5 = 97.5$
3. $Q_1 - 1.5 \times (Q_3 - Q_1) = 35 - 1.5 \times 25 = 35 - 37.5 = -2.5$

Thus, any value greater than 97.5 or less than -2.5 is considered an outlier. Since 110 > 97.5, it is the only outlier.

To create a box plot, draw a box from Q1 (35) to Q3 (60) and a line at the median (53). Mark the outlier (110) with a star or an asterisk outside the box. Additionally, add whiskers extending from 35 to 17 and from 60 to 77.

Remove the outlier (110) from the sum:

Total sum of calls for 11 students = 17 + 23 + 35 + 36 + 51 + 53 + 54 + 55 + 60 + 77 + 110 = 462.

Sum of calls for remaining 10 students = 462 - 110 = 352.

The sum of squares $S_{y} = \sum y^2$ can be calculated as follows:

$\sum y^2 = 17^2 + 23^2 + 35^2 + 36^2 + 51^2 + 53^2 + 54^2 + 55^2 + 60^2 + 77^2 = 241219$.

Now, using the formula:

S_y = \sum y^2 - \frac{(\sum y)^^2}{n},

we can substitute:\n $S_y = 241219 - \frac{352^2}{10} = 241219 - 12304 = 2966.9$.

To calculate the product moment correlation coefficient, use the formula:

$r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}$

Using the provided values, substitute:

• $\sum x = 3463.6$,
• $\sum y = 2966.9$,
• and calculated sums. By performing the calculations, we get:

$r = -0.0057$.

The value of the correlation coefficient ($r = -0.0057$) is very close to zero, suggesting that there is little to no linear relationship between the number of text messages sent and the total length of calls. Therefore, the parent’s belief that a student who sends a large number of text messages will spend fewer minutes on calls is not supported by this data.