Miscellaneous Problems - Chapter 9 Applied Statistics 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On August 15, 2024

By SaraNextGen

Miscellaneous Problems
Question 1.
Using three yearly moving averages, Determine the trend values from the following data.

Solution:

Question 2.
From the following data, calculate the trend values using fourly moving averages.

Solution:

Question 3.
Fit a straight line trend by the method of least squares to the following data.

Solution:
Let $\mathrm{x}$ denote the years and $\mathrm{y}$ denote the sales. Since number of years is even, we take $\mathrm{X}=\frac{x-1983.5}{0.5}$

$
a=\frac{\sum Y}{n}=\frac{447.9}{8}=55.9875 \quad b=\frac{\sum X Y}{\sum X^2}=\frac{139.5}{168}=0.83
$
The straight line trend by method of least squares is given by
(i.e)
$
\begin{aligned}
& \mathrm{Y}=a+b \mathrm{X} \\
& \mathrm{Y}=55.9875+0.83 \mathrm{X} \\
& \mathrm{Y}=55.9875+0.83\left[\frac{x-1983.5}{0.5}\right]
\end{aligned}
$
The trend values are obtained by substituting the years for $\mathrm{x}$, and given in the table.
When $\mathrm{x}=1980, Y_{\mathrm{t}}=55.9875+0.83(-7)=50.1775$
When $\mathrm{x}=1981, \mathrm{Y}_{\mathrm{t}}=55.9875-5(0.83)=51.8375$
When $\mathrm{x}=1982, \mathrm{Y}_{\mathrm{t}}=55.9875-3(0.83)=53.4975$
When $\mathrm{x}=1983, \mathrm{Y}_{\mathrm{t}}=55.9875-(0.83)=55.1575$
When $\mathrm{x}=1984, \mathrm{Y}_{\mathrm{t}}=55.9875+(0.83)=56.8175$
When $\mathrm{x}=1985, \mathrm{Y}_{\mathrm{t}}=55.9875+3(0.83)=58.4775$
When $\mathrm{x}=1986, \mathrm{Y}_{\mathrm{t}}=55.9875+5(0.83)=60.1375$
When $\mathrm{x}=1987, \mathrm{Y}_{\mathrm{t}}=55.9875+7(0.83)=61.7975$
We find that $\Sigma Y \mathrm{Y}=\Sigma \mathrm{Y}=447.9$
Question 4.
Calculate the Laspeyre's, Paasche's and Fisher's price index number for the following data. Interpret the data.

Solution:

Laspeyre's price index number
$
\mathrm{P}_{01}^{\mathrm{L}} \frac{\sum p_1 q_0}{\sum p_0 q_0} \times 100 \Rightarrow \mathrm{P}_{01}^{\mathrm{L}}=\frac{239319}{479621} \times 100=49.9
$
Paasche's price index number
$
\mathrm{P}_{01}^{\mathrm{P}} \frac{\sum p_1 q_1}{\sum p_0 q_1} \times 100 \Rightarrow \mathrm{P}_{01}^{\mathrm{P}}=\frac{324518}{644867} \times 100=50.32
$
Fisher's price index number
$
\begin{aligned}
\mathrm{P}_{01}^{\mathrm{F}} & =\sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1}} \times 100 \\
\mathrm{P}_{01}^{\mathrm{F}} & =\sqrt{\frac{239319 \times 324518}{479621 \times 644867}} \times 100 \\
\mathrm{P}_{01}^{\mathrm{F}} & =\sqrt{0.2511} \times 100 \\
\mathrm{P}_{01}^{\mathrm{F}} & =50.1
\end{aligned}
$

From the index numbers, we conclude that for the same quantity, the price has reduced by $50.5 \%$ in the current year compared to the base year according to Laspeyre's index. By the Paasche's index number, we see that the price has reduced by $49.68 \%$ in the current year, and according to Fisher's index number it has reduced by $49.9 \%$ in the current year.
Question 5.
Using the following data, construct Fisher's Ideal Index Number and show that it satisfies the Factor Reversal Test and Time Reversal Test?

Solution:

$
\begin{aligned}
\text { Fisher's ideal index number } & =\sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1}} \times 100 \\
\mathrm{P}_{01}^{\mathrm{F}} & =\sqrt{\frac{1900 \times 1880}{1360 \times 1344} \times 100} \\
& =\sqrt{\frac{3,572,000}{1,827,840}} \times 100 \\
& =\sqrt{1.9542 \times 100}=139.8
\end{aligned}
$
Time reversal test: $P_{01} \times P_{10}=1$

This shows Fisher's ideal index number satisfies factor reversal test.
Question 6.
Compute the consumer price index for 2016 on the basis of 2015 from the following data.

Solution:
Consumer price index (CPI) is same as cost of living index. We use the aggregate expenditure method which gives $\mathrm{CPI}$ as $\mathrm{CPI}=\frac{\sum p_1 q_0}{\sum p_0 q_0} \times 100$

Consumer price index for the year 2016 is $=\frac{174}{146.50} \times 100=118.77$
On the basis of the year 2015 , The cost of living has increased by $18.77 \%$
Question 7.
An Enquiry was made into the budgets of the middle-class families in a city gave the following information.

What changes in the cost of living have taken place in the middle-class families of a city?
Solution:

$
\text { Cost of living index number }=\frac{\sum \mathrm{PW}}{\sum \mathrm{W}}=\frac{12610}{100}=126.1
$
For the middle-class families of the city, the cost of living has increased up to $26.1 \%$ in 2011 as compared to 2010 .
Question 8.
From the following data, calculate the control limits for the mean and range chart.

Solution:
Since the sample size is 5 , we used $\mathrm{A}_2=0.577, \mathrm{D}_3=0, \mathrm{D}_4=2.114$, from the table given.

$
\begin{aligned}
& \overline{\bar{X}}=\frac{\sum \overline{\mathrm{X}}}{10}=\frac{510}{10}=51 \\
& \overline{\mathrm{R}}=\frac{\sum \mathrm{R}}{10}=6.5
\end{aligned}
$
The control chart for mean $\overline{\mathrm{X}}$ is
$
\begin{aligned}
\mathrm{UCL} & =\overline{\bar{X}}+\mathrm{A}_2 \overline{\mathrm{R}}=51+(0.577)(6.5)=54.75 \\
\mathrm{CL} & =\overline{\bar{X}}=51 \\
\mathrm{LCL} & =\overline{\bar{X}}-\mathrm{A}_2 \overline{\mathrm{R}}=51-(0.577)(6.5)=47.25
\end{aligned}
$
The control chart for range $\mathrm{R}$ is
$
\begin{aligned}
\mathrm{UCL} & =\mathrm{D}_4 \overline{\mathrm{R}}=(2.114)(6.5)=13.74 \\
\mathrm{CL} & =\overline{\mathrm{R}}=6.5 \\
\mathrm{LCL} & =\mathrm{D}_3 \overline{\mathrm{R}}=(0)(6.5)=0
\end{aligned}
$
We see that one sample $\bar{X}$ value 47 is below the LCL of $\bar{X}$. To infer that the process is not totally out of control since ${ }^{\wedge}$ the difference is less. Further investigation is recommended.
Question 9.
The following data gives the average life (in hours) and a range of 12 samples of 5 lamps each. The data are

Construct control charts for mean and range. Comment on the control limits.
Solution:
In this question the number of observations is 5 for each sample. So we use $\mathrm{A}_2=0.577, \mathrm{D}_3=0, \mathrm{D}_4=2.114$ from the table of control chart constants.

$
\begin{aligned}
& \overline{\bar{X}}=\frac{\sum \overline{\mathrm{X}}}{12}=\frac{16410}{12}=1367.5 \\
& \overline{\mathrm{R}}=\frac{\sum \mathrm{R}}{12}=\frac{5130}{12}=427.5
\end{aligned}
$
The control limits for $\overline{\mathrm{X}}$ chart is
$
\begin{aligned}
\mathrm{UCL} & =\overline{\overline{\mathrm{X}}}+\mathrm{A}_2 \overline{\mathrm{R}}=1367.5+(0.577)(427.5)=1614.17 \\
\mathrm{CL} & =\overline{\overline{\mathrm{X}}}=1367.5 \\
\mathrm{LCL} & =\overline{\overline{\mathrm{X}}}-\mathrm{A}_2 \overline{\mathrm{R}}=1367.5-(0.577)(427.5)=1120.83
\end{aligned}
$
The control limits for $\mathrm{R}$ chart is
$
\begin{aligned}
\mathrm{UCL} & =\mathrm{D}_4 \overline{\mathrm{R}}=(2.114)(427.5)=903.735 \\
\mathrm{CL} & =\overline{\mathrm{R}}=427.5 \\
\mathrm{LCL} & =\mathrm{D}_3 \overline{\mathrm{R}}=(0)(427.5)=0
\end{aligned}
$
We observe that the sample $\bar{X}$ value 1080 is below the LCL. All sample range values are within the control limits for $R$. We say that process is out of control.
Question 10.
The following are the sample means and ranges for 10 samples, each of size 5 . Calculate the control limits for the mean chart and range chart and state whether the process is in control or not.

Solution:

$
\overline{\bar{X}}=\frac{\sum \overline{\mathrm{X}}}{10}=\frac{49.82}{10}=4.982 \quad \overline{\mathrm{R}}=\frac{\sum \mathrm{R}}{10}=\frac{3.6}{10}=0.36
$
The control chart for mean is
$
\begin{aligned}
\mathrm{UCL} & =\overline{\bar{X}}+\mathrm{A}_2 \overline{\mathrm{R}}=4.982+(0.577)(0.36)=5.19 \\
\mathrm{CL} & =\overline{\bar{X}}=4.982 \\
\mathrm{LCL} & =\overline{\overline{\mathrm{X}}}-\mathrm{A}_2 \overline{\mathrm{R}}=4.982-(0.577)(0.36)=4.774
\end{aligned}
$
The control chart for range $R$ is
$
\begin{aligned}
\mathrm{UCL} & =\mathrm{D}_4 \overline{\mathrm{R}}=(2.114)(0.36)=0.761 \\
\mathrm{CL} & =\overline{\mathrm{R}}=0.36 \\
\mathrm{LCL} & =\mathrm{D}_3 \overline{\mathrm{R}}=(0)(0.36)=0
\end{aligned}
$
We observe that all sample $\bar{X}$ values are within the control limits value of $\bar{X}$ chart. But the sample range value $0.8$ is above the UCL of the R chart. So we conclude that the process is out of control.