Business statistics nmims latest solved assignments

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Business Stats
April 2021 Examination
Question 1) Nemi Mehta is owning 50 Hectares of land near Junagadh. On his Farm, he cultivates Giloy, which is a medicinal plant in Ayurveda. The
below-given table shows the sales of ‘Giloy Vati’ that his team is preparing as per one Ayurveda script (book), and the amount they spend on the
Advertisement of it.
Nemi’s problem is to analyze the effect of Advertisement on sales. Firstly, He wants to understand the presence of a linear relationship between the
sales and ‘amount spent in advertisement’. He also wants to run a correlation and regression to know whether he should keepspending money on
Advertisements or not. If sales figures are not affected by advertisement, he should not spend money on it. So, Calculate Karl Person’s correlation,
RegressionModel (with bo and b1), R-square, ANOVA Table. Also, provide EXCEL- Generated Output Along with your calculations.
Region code Sales (1NR 000s) Advertising (TV spots per month) (in INR, 000s)
1 260.3 5
2 286.1 7
3 279.4 6
4 410.8 9
5 438.2 12
6 315.3 8
7 565.1 11
8 570.0 16
9 426.1 13

10 315.0 7
11 403.6 10
12 220.5 4
13 343.6 9
14 644.6 17
15 520.4 19
16 329.5 9
17 426.0 11
18 343.2 8
19 450.4 13
20 421.8 14
Ans 1.
To find Karl Person’s correlation the following calculations are need to be done.
Now, for regression output Excel has been used. Since,
Question 2) The table given below is the ‘single year age population’ (taken from census 2011). This table shows the population of people (age-wise)
living in Leh at the time of the census 2011 survey. Transform this ungrouped data into Grouped data by forming age groups then find out Mean,
Median, Variance, Standard deviation, Ogive, and Histogram. Write the summary based on your calculations.
Age in Years Population Age in Years Population Age in Years Population Age in Years Population Age in Years Population
0 992 11 1998 22 2839 33 2696 44 1652
1 1958 12 1916 23 2935 34 2781 45 1806
2 1725 13 2138 24 3601 35 2799 46 1460

3 1814 14 2139 25 4110 36 2450 47 1226
4 1768 15 2096 26 4089 37 2142 48 1225
5 1871 16 2044 27 3716 38 2114 49 1006
6 1888 17 2027 28 3702 39 1725 50 1454
7 1768 18 2065 29 3084 40 2218
8 1712 19 2013 30 3475 41 1802
9 1780 20 2459 31 2844 42 1751
10 1862 21 2594 32 2684 43 1659
Ans 2.
To transform the given data from ungrouped to grouped data. Consider we want 6 age groups then we need to find class width. The class width is equal to range
divided by number of classes. Now, to find range the minimum value inage column is 0 and maximum value is 50.
Since, Range = Maximum – Minimum hence, it becomes Range = 50 – 0 = 50
Now, the class width = Range/6 = 50/6 = 8.33.Hence, by rounding to whole
a) Mean
b) Median

c) Variance
The standard deviation is nothing but square root of variance.
d) Standard deviation
e) Ogive (Cumulative frequency graph)
Based on the data provided, we can construct the cumulative frequencies associated to each class, by adding all the frequencies up to that class, or equivalently,
Question 3a) The given table shows the rainfall of Gujarat Region. Forecast the rainfall using Exponential Smoothing. Use Alpha =0.2, 0.5 and 0.8. Data
is available from 1997 to 2016, use this series for the calculation and forecast the rainfall for the year 2017. To know, the extent the prediction is correct
the actual rainfall for 2017 (1024.4 millimeters) is provided. Find out which alpha values among the three suggestions are near to accurate value?
SUBDIVISION YEAR ANNUAL (in MM)
Gujarat Region 1997 1068.9
Gujarat Region 1998 1070

Ans 3a.
For
The exponential smoothing (ES) forecast with smoothing constant α=0.2 for the nth period is computed using the following formula:
Fn=Fn−1+α(An−1−Fn−1)
Question 3b) A new gas-electric hybrid car has recently hit the market. The distance traveled on 1 gallon of fuel is normally distributed with a mean of
65 miles and a standard deviation of 4 miles. Find the probability of the following events. (show the concerned region by z curve)
1. The car travels more than 70 miles per gallon.
2. The car travels less than 60 miles per gallon.
3. The car travels between 55 and 70 miles per gallon.
Ans 3b.
It is given that the distance travelled on 1 gallon of fuel is normally distributed with a mean of 65 miles and a standard deviation of 4 miles. In symbolic form the
mean and standard deviation are

And
a) To find the probability that car travels more than 70 miles per
b) To find the probability that car travels less than 60 miles per gallon i.e. P(X <60)
c) To find probability that the car travels between 55 and 70 miles per gallon i.e. P (55 < X < 70)
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