EECM3724 STATISTICS FOR ECONOMICS
TEST: FIRST TEST MEMO
QUESTION 1 [14]
1.1) The table below contains data about the share of food spending in total spending (in percentage points,
𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒) and monthly per capita household income (𝑖𝑛𝑐𝑜𝑚𝑒) for a random sample of nine Free State
households.
Household A B C D E F G H I
𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒 49 49 35 26 42 21 47 48 53
𝑖𝑛𝑐𝑜𝑚𝑒 3361 3831 4145 4980 4162 6309 3216 4135 3223
1.1a) Estimate the sample means and sample standard deviations of 𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒 and 𝑖𝑛𝑐𝑜𝑚𝑒 for these 9
Free State households. [3]
Let 𝒙 = 𝒇𝒐𝒐𝒅𝒔𝒉𝒂𝒓𝒆 and 𝒚 = 𝒊𝒏𝒄𝒐𝒎𝒆
̅ = 𝟒𝟏. 𝟏𝟏𝟏𝟏
𝒙
̅ = 𝟒𝟏𝟒𝟏. 𝟑𝟑𝟑𝟑
𝒚
𝒔𝟐𝒙 = 𝟏𝟐𝟕. 𝟑𝟔𝟏𝟏 ∴ 𝒔𝒙 = 𝟏𝟏. 𝟐𝟖𝟓𝟒
𝒔𝒚 = 𝟗𝟕𝟓𝟖𝟏𝟓. 𝟖 ∴ 𝒔𝒚 = 𝟗𝟖𝟕. 𝟖𝟑𝟑𝟗
1.1b) Write down the interquartile range (IQR) for 𝑖𝑛𝑐𝑜𝑚𝑒. [3]
𝑰𝑸𝑹 = 𝑸𝟑 − 𝑸𝟏
→ 𝑸𝟏 = 𝟑𝟑𝟔𝟏; 𝑸𝟑 = 𝟒𝟏𝟔𝟐
∴ 𝑰𝑸𝑹 = 𝟒𝟏𝟔𝟐 = 𝟑𝟑𝟔𝟏 = 𝟖𝟎𝟏
1.1c) Report the largest z-score (in absolute terms) for 𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒. Are there any outliers present? Briefly
explain. [2]
Largest absolute z-value: either largest or smallest observation
𝒛𝒍𝒂𝒓𝒈𝒆𝒔𝒕 = 𝟏. 𝟎𝟓𝟑𝟓
𝒛𝒔𝒎𝒂𝒍𝒍𝒆𝒔𝒕 = −𝟏. 𝟕𝟖𝟐𝟎 ⇒ |−𝟏. 𝟕𝟖𝟐𝟎| = 𝟏. 𝟕𝟖𝟐𝟎
Largest z-value is 1.7820, which is less than 3, so no outliers.
1.1d) Estimate the sample correlation coefficient between 𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒 and 𝑖𝑛𝑐𝑜𝑚𝑒. Comment on the
direction and strength of the estimated relationship. [4]
𝒔𝒙𝒚 −𝟏𝟎𝟏𝟔𝟎. 𝟕𝟑𝟖𝟐
∴𝒓= = = −𝟎. 𝟗𝟏𝟏𝟒
𝒔𝒙 𝒔𝒚 𝟏𝟏. 𝟐𝟖𝟓𝟒 × 𝟗𝟖𝟕. 𝟖𝟑𝟑𝟗
𝟐 𝟐
= = 𝟎. 𝟔𝟔𝟔𝟕
√𝒏 √𝟗
This is therefore a strong (𝟎. 𝟗𝟏𝟒𝟒 > 𝟎. 𝟔𝟔𝟔𝟕), negative (−𝟎. 𝟗𝟏𝟒𝟒 < 𝟎) relationship
1
, 1.2) The mean monthly rental rate of two-bedroom flats in Bloemfontein is R3000, with a variance of 462400.
Use Chebychev’s rule to find the interval that contains at least 60.94% of the monthly rental rates of two-
bedroom flats in Bloemfontein. [2]
𝒌 = 𝟏. 𝟔𝟎𝟎𝟏
𝝁 = 𝟑𝟎𝟎𝟎, 𝝈 = √𝟒𝟔𝟐𝟒𝟎𝟎 = 𝟔𝟖𝟎
𝝁 ± 𝒌𝝈 = 𝟑𝟎𝟎𝟎 ± 𝟏. 𝟔𝟎𝟎𝟏(𝟔𝟖𝟎) = (𝟏𝟗𝟏𝟏. 𝟗𝟑𝟐, 𝟒𝟎𝟖𝟖. 𝟎𝟔𝟖)
QUESTION 2 [13]
2.1) Sixty-two percent of South African mineworkers are permanent employees. Use the binomial
distribution function to find the probability that at least half of a random sample of six South African
mineworkers are permanent employees. [3]
𝒏 𝟔
𝒑(𝒙) = ( ) 𝝅𝒙 (𝟏 − 𝝅)𝒏−𝒙 = ( ) (𝟎. 𝟔𝟐)𝒙 (𝟎. 𝟑𝟖)𝟔−𝒙
𝒙 𝒙
At least half of 6 is ≥ 𝟑, so find 𝑷(𝑿 ≥ 𝟑) = 𝒑(𝟑) + 𝒑(𝟒) + 𝒑(𝟓) + 𝒑(𝟔)
While 𝑷(𝑿 < 𝟑) = 𝒑(𝟎) + 𝒑(𝟏) + 𝒑(𝟐) and 𝑷(𝑿 ≥ 𝟑) = 𝟏 − 𝑷(𝑿 < 𝟑)
𝒑(𝟎) = 𝟎. 𝟎𝟎𝟑
𝒑(𝟏) = 𝟎. 𝟎𝟐𝟗𝟓
𝒑(𝟐) = 𝟎. 𝟏𝟐𝟎𝟐
𝑷(𝑿 < 𝟑) = 𝟎. 𝟏𝟓𝟐𝟕 ∴ 𝑷(𝑿 ≥ 𝟑) = 𝟏 − 𝟎. 𝟏𝟓𝟐𝟕 = 𝟎. 𝟖𝟒𝟕𝟑
2.2) Nine equally qualified B.Com. (Economics) students apply for five summer internship positions at Big
Bank. Four of these students are women. Because the candidates are equally qualified, Big Bank decides
to use a random lottery to decide which applicants will fill these five internships. Use the hypergeometric
distribution function to find the probability that more than two of the summer interns appointed by Big
Bank will be women. [3]
𝟒 𝟓
(𝒙𝒓)(𝑵−𝒓 ) ( )( )
𝒑(𝒙) = 𝒏−𝒙
= 𝒙 𝟓 − 𝒙
(𝑵 ) 𝟗
𝒏 ( )
𝟓
So, find 𝑷(𝑿 > 𝟐) = 𝒑(𝟑) + 𝒑(𝟒) + 𝒑(𝟓)
𝒑(𝟑) = 𝟎. 𝟑𝟏𝟕𝟓
𝒑(𝟒) = 𝟎. 𝟎𝟑𝟗𝟕
𝒑(𝟓) = 𝟎
∴ 𝑷(𝑿 > 𝟐) = 𝟎. 𝟑𝟏𝟕𝟓 + 𝟎. 𝟎𝟑𝟗𝟕 + 𝟎 = 𝟎. 𝟑𝟓𝟕𝟐
2
TEST: FIRST TEST MEMO
QUESTION 1 [14]
1.1) The table below contains data about the share of food spending in total spending (in percentage points,
𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒) and monthly per capita household income (𝑖𝑛𝑐𝑜𝑚𝑒) for a random sample of nine Free State
households.
Household A B C D E F G H I
𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒 49 49 35 26 42 21 47 48 53
𝑖𝑛𝑐𝑜𝑚𝑒 3361 3831 4145 4980 4162 6309 3216 4135 3223
1.1a) Estimate the sample means and sample standard deviations of 𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒 and 𝑖𝑛𝑐𝑜𝑚𝑒 for these 9
Free State households. [3]
Let 𝒙 = 𝒇𝒐𝒐𝒅𝒔𝒉𝒂𝒓𝒆 and 𝒚 = 𝒊𝒏𝒄𝒐𝒎𝒆
̅ = 𝟒𝟏. 𝟏𝟏𝟏𝟏
𝒙
̅ = 𝟒𝟏𝟒𝟏. 𝟑𝟑𝟑𝟑
𝒚
𝒔𝟐𝒙 = 𝟏𝟐𝟕. 𝟑𝟔𝟏𝟏 ∴ 𝒔𝒙 = 𝟏𝟏. 𝟐𝟖𝟓𝟒
𝒔𝒚 = 𝟗𝟕𝟓𝟖𝟏𝟓. 𝟖 ∴ 𝒔𝒚 = 𝟗𝟖𝟕. 𝟖𝟑𝟑𝟗
1.1b) Write down the interquartile range (IQR) for 𝑖𝑛𝑐𝑜𝑚𝑒. [3]
𝑰𝑸𝑹 = 𝑸𝟑 − 𝑸𝟏
→ 𝑸𝟏 = 𝟑𝟑𝟔𝟏; 𝑸𝟑 = 𝟒𝟏𝟔𝟐
∴ 𝑰𝑸𝑹 = 𝟒𝟏𝟔𝟐 = 𝟑𝟑𝟔𝟏 = 𝟖𝟎𝟏
1.1c) Report the largest z-score (in absolute terms) for 𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒. Are there any outliers present? Briefly
explain. [2]
Largest absolute z-value: either largest or smallest observation
𝒛𝒍𝒂𝒓𝒈𝒆𝒔𝒕 = 𝟏. 𝟎𝟓𝟑𝟓
𝒛𝒔𝒎𝒂𝒍𝒍𝒆𝒔𝒕 = −𝟏. 𝟕𝟖𝟐𝟎 ⇒ |−𝟏. 𝟕𝟖𝟐𝟎| = 𝟏. 𝟕𝟖𝟐𝟎
Largest z-value is 1.7820, which is less than 3, so no outliers.
1.1d) Estimate the sample correlation coefficient between 𝑓𝑜𝑜𝑑𝑠ℎ𝑎𝑟𝑒 and 𝑖𝑛𝑐𝑜𝑚𝑒. Comment on the
direction and strength of the estimated relationship. [4]
𝒔𝒙𝒚 −𝟏𝟎𝟏𝟔𝟎. 𝟕𝟑𝟖𝟐
∴𝒓= = = −𝟎. 𝟗𝟏𝟏𝟒
𝒔𝒙 𝒔𝒚 𝟏𝟏. 𝟐𝟖𝟓𝟒 × 𝟗𝟖𝟕. 𝟖𝟑𝟑𝟗
𝟐 𝟐
= = 𝟎. 𝟔𝟔𝟔𝟕
√𝒏 √𝟗
This is therefore a strong (𝟎. 𝟗𝟏𝟒𝟒 > 𝟎. 𝟔𝟔𝟔𝟕), negative (−𝟎. 𝟗𝟏𝟒𝟒 < 𝟎) relationship
1
, 1.2) The mean monthly rental rate of two-bedroom flats in Bloemfontein is R3000, with a variance of 462400.
Use Chebychev’s rule to find the interval that contains at least 60.94% of the monthly rental rates of two-
bedroom flats in Bloemfontein. [2]
𝒌 = 𝟏. 𝟔𝟎𝟎𝟏
𝝁 = 𝟑𝟎𝟎𝟎, 𝝈 = √𝟒𝟔𝟐𝟒𝟎𝟎 = 𝟔𝟖𝟎
𝝁 ± 𝒌𝝈 = 𝟑𝟎𝟎𝟎 ± 𝟏. 𝟔𝟎𝟎𝟏(𝟔𝟖𝟎) = (𝟏𝟗𝟏𝟏. 𝟗𝟑𝟐, 𝟒𝟎𝟖𝟖. 𝟎𝟔𝟖)
QUESTION 2 [13]
2.1) Sixty-two percent of South African mineworkers are permanent employees. Use the binomial
distribution function to find the probability that at least half of a random sample of six South African
mineworkers are permanent employees. [3]
𝒏 𝟔
𝒑(𝒙) = ( ) 𝝅𝒙 (𝟏 − 𝝅)𝒏−𝒙 = ( ) (𝟎. 𝟔𝟐)𝒙 (𝟎. 𝟑𝟖)𝟔−𝒙
𝒙 𝒙
At least half of 6 is ≥ 𝟑, so find 𝑷(𝑿 ≥ 𝟑) = 𝒑(𝟑) + 𝒑(𝟒) + 𝒑(𝟓) + 𝒑(𝟔)
While 𝑷(𝑿 < 𝟑) = 𝒑(𝟎) + 𝒑(𝟏) + 𝒑(𝟐) and 𝑷(𝑿 ≥ 𝟑) = 𝟏 − 𝑷(𝑿 < 𝟑)
𝒑(𝟎) = 𝟎. 𝟎𝟎𝟑
𝒑(𝟏) = 𝟎. 𝟎𝟐𝟗𝟓
𝒑(𝟐) = 𝟎. 𝟏𝟐𝟎𝟐
𝑷(𝑿 < 𝟑) = 𝟎. 𝟏𝟓𝟐𝟕 ∴ 𝑷(𝑿 ≥ 𝟑) = 𝟏 − 𝟎. 𝟏𝟓𝟐𝟕 = 𝟎. 𝟖𝟒𝟕𝟑
2.2) Nine equally qualified B.Com. (Economics) students apply for five summer internship positions at Big
Bank. Four of these students are women. Because the candidates are equally qualified, Big Bank decides
to use a random lottery to decide which applicants will fill these five internships. Use the hypergeometric
distribution function to find the probability that more than two of the summer interns appointed by Big
Bank will be women. [3]
𝟒 𝟓
(𝒙𝒓)(𝑵−𝒓 ) ( )( )
𝒑(𝒙) = 𝒏−𝒙
= 𝒙 𝟓 − 𝒙
(𝑵 ) 𝟗
𝒏 ( )
𝟓
So, find 𝑷(𝑿 > 𝟐) = 𝒑(𝟑) + 𝒑(𝟒) + 𝒑(𝟓)
𝒑(𝟑) = 𝟎. 𝟑𝟏𝟕𝟓
𝒑(𝟒) = 𝟎. 𝟎𝟑𝟗𝟕
𝒑(𝟓) = 𝟎
∴ 𝑷(𝑿 > 𝟐) = 𝟎. 𝟑𝟏𝟕𝟓 + 𝟎. 𝟎𝟑𝟗𝟕 + 𝟎 = 𝟎. 𝟑𝟓𝟕𝟐
2
