Additional Problems - Chapter 8 Sampling Techniques and Statistical Inference 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Problems
Choose the correct answer:
Question 1.
Non sampling error is reduced by
(a) Increasing sample size
(b) Decreasing sample size
(c) Reducing amount of data
(d) None of these
Answer:
(d) None of these
Question 2.
Any numerical value calculated from sample data is called
(a) Error
(b) Statistic
(c) Bias
(d) Mean
Answer:
(b) Statistic
Question 3.
In sampling with replacement a sampling unit can be selected
(a) Only once
(b) More than one time
(c) Less than one time
(d) None of above
Answer:
(b) More than one time
Question 4.
Standard deviation of sampling distribution of any statistic is called
(a) Sampling error
(b) Type-I error
(c) Standard error
(d) Non-sampling error
Answer:
(c) Standard error

Question 5.
The difference between statistic and parameter is called
(a) Random error
(b) Sampling error
(c) Standard error
(d) Bias
(e) Error
Answer:
(e) Error
Question 6.
In random sampling, the probability of selecting an item from the population is
(a) unknown
(b) known
(c) undecided
(d) zero
Answer:
(b) known
Question 7.
Match the following:

Answer:
(i) - (c)
(ii) - (d)
(iii) - (a)
(iv) - (e)
(v) -(b)
Question 8.
Fill in the blanks:
1. Any statement whose validity is tested on the basis of a sample is called _________
2. The alternative hypothesis is also called _______
3. The probability of rejecting the null hypothesis when it is true is called _______
4. The hypothesis $\mu \leq 10$ is a ____
5. If a hypothesis specifies the population distribution it is called _______
Answer:
1. Statistical hypothesis
2. Research hypothesis
3. Level of significance
4. Composite hypothesis
5. Simple hypothesis
Question 9.
Null and alternative hypothesis are statements about
(a) population parameters
(b) sample parameters
(c) sample statistics
(d) none of the above
Answer:
(a) population parameters

2 and 3 Mark Questions
Question 1.
A random sample of size 50 with mean $67.9$ is drawn from a normal population. If the S.E of the sample mean is $\sqrt{ } 0.7$, find a $95 \%$ confidence interval for the population mean. Solution:
$
\mathrm{n}=50, \bar{x}=67.9
$
$95 \%$ confidence limits for the population mean $\mu$ are
$
\bar{\alpha}-Z_{\alpha / 2}\left(\frac{\sigma}{\sqrt{n}}\right) \leq \mu \leq \bar{x}+Z_{\alpha / 2}\left(\frac{\sigma}{\sqrt{n}}\right)
$
$\frac{\sigma}{\sqrt{n}}$ is given as $\sqrt{0.7}$ and $Z_{\alpha / 2}=1.96$
$
\begin{aligned}
\Rightarrow 67.9-(1.96)(\sqrt{0.7}) & \leq \mu \leq 67.9+(1.96) \sqrt{0.7} \\
66.26 & \leq \mu \leq 69.54
\end{aligned}
$
Thus the $95 \%$ confidence intervals for estimating $\mu$ is given by $(66.26,69.54)$
Question 2.
A random sample of 500 apples was taken from large consignment and 45 of them were found to be bad. Find the limits at which the bad apples lie at $99 \%$ confidence level.
Solution:
Sample size $\mathrm{n}=500$
Proportion of bad apples $\mathrm{P}=\frac{45}{500}=0.09$
So proportion of good apples $\mathrm{Q}=1-0.09=0.91$
The confidence limits for population proportion are given by,
$
\begin{aligned}
& {\left[\mathrm{P}-\mathrm{Z}_{\alpha / 2} \sqrt{\mathrm{PQ} / n}, \mathrm{P}+\mathrm{Z}_{\alpha / 2} \sqrt{\mathrm{PQ} / n}\right]} \\
& =\left[0.09-2.58 \sqrt{\frac{(0.09)(0.91)}{500}}, 0.09+2.58 \sqrt{\frac{(0.09)(0.91)}{500}}\right] \\
& =(0.09-(2.58)(0.013), 0.09+(2.58)(0.013)) . \\
& =(0.09-0.034,0.09+0.034) \\
& =(0.056,0.124)
\end{aligned}
$
Thus the bad apples lie between $5.6 \%$ and $12.4 \%$

Question 3.
A sample of 400 students is found to have a mean height of $171.38 \mathrm{cms}$ can it be regarded as a sample from a large population with mean height $171.17 \mathrm{cms}$ and S.D $3.30 \mathrm{cms}$ ?
Solution:
Given
Sample size $\mathrm{n}=400$
Sample mean $\bar{x}=171.38 \mathrm{~cm}$
Population mean $\mu=171.17 \mathrm{~cm}$
Population SD $\sigma=3.30 \mathrm{~cm}$
$\mathrm{H}_0: \mu=171.17 \mathrm{~cm}$
$\mathrm{H}_1: \mu \neq 171.17 \mathrm{~cm}$
Test statistic:
$
\mathrm{Z}=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{171.38-171.17}{\frac{3.3}{\sqrt{400}}}=1.27
$
The table value of $z_{\alpha / 2}$ at $5 \%$ level is $1.96$. Since $\mathrm{Z}<z_{\alpha / 2}, \mathrm{H}_0$ is accepted. Therefore the given sample can be regarded as one from the population with mean $171.17 \mathrm{~cm}$.
Question 4.
An automatic machine fills tea in sealed tins with a mean weight of tea as $1 \mathrm{~kg}$ and S.D 1 gram. A random sample of 50 tins was examined and it was found that their mean weight was $999.5$ grams. Is the machine working properly?
Solution:
Given
Sample size $\mathrm{n}=50$
Sample mean $\bar{x}=999.5$ grams
Population mean $\mu=1000$ grams
Population SD $\sigma=1$ gram
$\mathrm{H}_0: \mu=1 \mathrm{~kg}$
$\mathrm{H}_1: \mu \neq 1 \mathrm{~kg}$
Test statistic

$
\mathrm{Z}=\frac{999.5-1000}{\frac{1}{\sqrt{50}}}=-3.54
$
The table value of $z_{\alpha / 2}$ at $1 \%$ level $=2.58$. Since $|\mathrm{Z}|>z_{\alpha / 2}, \mathrm{H}_0$ is rejected. Therefore the machine is not working properly.
5 Marks Questions
Question 1.
A simple random sample of size 100 has mean
(a) 15 , the population variance being 25 . Find an interval estimate of the population mean with a confidence level of $95 \%$ and $99 \%$
(b) If the population variance is not given, what should be done to find out the required estimates?
Solution:
(a) Given
Sample size $\mathrm{n}=100$
Sample mean $\bar{x}=15$
Population SD $\sigma=5$
The $95 \%$ confidence interval for the population mean is $\bar{x} \pm \mathrm{Z}_{\alpha / 2} \frac{\sigma}{\sqrt{n}}$
Here $z_{\alpha / 2}=1.96$. So we get
$
\begin{aligned}
& =15 \pm(1.96)\left(\frac{5}{\sqrt{100}}\right) \\
& =15 \pm(1.96)(0.5) \\
& =15 \pm 0.98 \\
& =14.02 \text { and } 15.98
\end{aligned}
$
Therefore $95 \%$ confidence interval for population mean $\mu$ is $(14.02,15.98)$
The $99 \%$ confidence interval is $\bar{x} \pm 2.58 \frac{\sigma}{\sqrt{n}}$
$
\begin{aligned}
& =15 \pm 2.58\left(\frac{5}{\sqrt{100}}\right) \\
& =15 \pm 2.58(0.5) \\
& =13.71 \text { and } 16.29
\end{aligned}
$
Therefore $99 \%$ confidence interval for the population mean $\mu$ is $(13.71,16.29)$
(b) If population S.D a is not known, then the sample S.D can be used in the place of o in estimating the confidence interval.

Question 2.
A factory is producing 50,000 pairs of shoes daily. From a sample of 500 pairs, $2 \%$ were found to be of sub-standard quality. Estimate the number of pairs that can be reasonably expected to be spoiled at $95 \%$ level of confidence.
Solution:
$
\mathrm{N}=50,000, \mathrm{n}=500, \mathrm{P}=\frac{2}{100}, \mathrm{Q}=\frac{98}{100}
$
The estimated percentage of spoiled pairs in daily production $=\frac{2}{100} \times 50,000=1000$ The limits for the number of spoiled pans at $95 \%$ level of confidence
$
\begin{aligned}
& =N[P \pm 1.96 \sqrt{\mathrm{PQ} / n}] \\
& =50,000\left[\frac{2}{100} \pm 1.96 \sqrt{\frac{2}{100} \times \frac{98}{100} \times \frac{1}{500}}\right] \\
& =500\left[2 \pm 1.96 \sqrt{\frac{196}{500}}\right] \\
& =500[2 \pm 1.96 \sqrt{0.392}] \\
& =500[2 \pm 1.227] \\
& =(500 \times 0.773,500 \times 3.227)=(387,1614)
\end{aligned}
$
Question 3.
A company that packages peanuts states that at a maximum $6 \%$ of the peanut shells contain no nuts. At random, 300 peanuts were selected and 21 of them were empty. With a significance level of $1 \%$ can the statement made by the company be accepted?
Solution:
The population proportion $\mathrm{P}=6 \%=0.06$
The null hypothesis: $\mathrm{H}_0: \mathrm{P} \leq 0.06$
Alternative hypothesis: $\mathrm{H}_1: \mathrm{P}>0.06$
For $\alpha=1 \%=0.01 . Z_\alpha=2.33$
The test statistic is $\mathrm{P}+2.33\left(\sqrt{\frac{(0.06)(0.94)}{300}}\right)=0.092$
(where $\mathrm{n}=300, \mathrm{P}=0.06, \mathrm{Q}=0.94)$
Since the calculated value is less than the table value, $0.092<2.33$, we accept the null hypothesis $\mathrm{H}_0$.
Hence the statement of the company can be accepted.

Question 4.
A school principal claims that the students in his school are above average intelligence. A random sample of 30 students IQ scores has a mean score of 112.5. The mean population IQ is 100 with an SD of 15 . Is there sufficient evidence to support the principal's claim?
Solution:
Given
Population mean $\mu=100$
Population SD $\sigma=15$
Sample size $\mathrm{n}=30$
Sample mean $\bar{x}=112.5$
Null hypothesis $\mathrm{H}_0: \mu=100$
(the students have average I.Q)
Alternative hypothesis $\mathrm{H}_1: \mu>100$
(the students have above average I.Q scores)

Let the significance level $\alpha=0.05$. The table value $Z_\alpha=1.645$, since this is one-tailed test.
Test Statistic:
$
\begin{aligned}
\mathrm{Z} & =\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \\
& =\frac{112.5-100}{\frac{15}{\sqrt{30}}}=4.56
\end{aligned}
$
Since $4.56>1.645$ (ie) $\mathrm{Z}>\mathrm{Z}_\alpha$ at $5 \%$ level, we reject the null hypothesis. Hence we conclude that the students have above average IQ scores. So the principal's claim is right.

Question 5.
Boys of a certain age are known to have a mean weight of 85 pounds. A complaint is made that the boys living in hostels are underfed. So a sample of 25 boys are weighed and ' found to have a mean weight of $80.94$ pounds. The population S.D is 11.6. What should be concluded about the complaint?
Solution:
Given $\mathrm{n}=25, \mu=85, \bar{x}=80.94$, and $\sigma=11.6$
Null hypothesis $\mathrm{H}_0: \mu=85$ (the boys are not underfed)
Alternative hypothesis $\mathrm{H}_1: \mu<85$ (the boys are underfed)
Test statistic:
$
\begin{aligned}
Z & =\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \\
& =\frac{80.9485}{\frac{11.6}{\sqrt{25}}}=-1.75
\end{aligned}
$
Let us take the significance level $\alpha=0.05$. The table value is $Z_\alpha=-1.645$

Now $-1.75<-1.645$ (i.e) $\mathrm{Z}<\mathrm{Z}_\alpha$. Therefore we reject the null hypothesis. Since the calculated value falls in the rejection region. Hence we conclude that the boys are underfed. So the complaint should be addressed immediately.