Question

The following table relates to the tourist arrivals during 1990 to 1996 in Botswana:

Years:                    1990   1991    1992   1993  1994  1995  1996

Tourist’s arrivals:    18       20        23       25      24      28       30

Fit a straight line trend by the method of least squares and estimates the number of tourists that would arrives in the year 2000.

Answer 1

Let $X=$ the number of years since 1990

$$\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & 18 & 20 & 23 & 25 & 24 & 28 & 30 \\ \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & x & y & xy & x^2 &y^2 \\ \hline & 0 & 18 & 0 & 0 & 324 \\ \hdashline & 1 & 20 & 20 & 1 & 400 \\ \hdashline & 2 & 23 & 46 & 4 & 529 \\ \hdashline & 3 & 25 & 75 & 9 & 625 \\ \hdashline & 4 & 24 & 96 & 16 & 576 \\ \hdashline & 5 & 28 & 140 & 25 & 784 \\ \hdashline & 6 & 30 & 180 & 36 & 900 \\ \hdashline Sum= & 21 & 168 & 557 & 91 & 4138 \\ \hdashline \end{array}$$

$$\bar{x}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nx_i=\dfrac{21}{7}=3$$

$$\bar{y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^ny_i=\dfrac{168}{7}=24$$

$$SS_{xx}=\displaystyle\sum_{i=1}^nx_i^2-\dfrac{1}{n}\bigg(\displaystyle\sum_{i=1}^nx_i\bigg)^2=91-\dfrac{21^2}{7}=28$$

$$SS_{yy}=\displaystyle\sum_{i=1}^ny_i^2-\dfrac{1}{n}\bigg(\displaystyle\sum_{i=1}^ny_i\bigg)^2=4138-\dfrac{168^2}{7}=106$$

$$SS_{xy}=\displaystyle\sum_{i=1}^nx_i\displaystyle\sum_{i=1}^ny_i-\dfrac{1}{n}\bigg(\displaystyle\sum_{i=1}^nx_i\bigg)\bigg(\displaystyle\sum_{i=1}^nx_i\bigg)$$

$$=557-\dfrac{21(168)}{7}=53$$

Based on the above calculations, the slope $m$ and the y-intercept $n$ are obtained as follows:

$$m=\dfrac{SS_{xy}}{SS_{xx}}=\dfrac{53}{28}=1.8929$$

$$y=\bar{y}-m\bar{x}=24-\dfrac{53}{28}(3)=\dfrac{513}{28}=18.3214$$

So the trend line is

$$y=18.3214+1.8929x$$

The number of tourists that would arrives in the year 2000 is

$$y=18.3214+1.8929(10)=37$$