General Procedure for Student’s t Distribution and Its Applications, Properties, Procedure Steps, Example Solved Problems

**STUDENT’S t DISTRIBUTION AND ITS APPLICATIONS**

If *X*~*N*(0,1) and *Y*~*χ*_{n}^{2}
are independent random variables, then

is said to have* t*-distribution with* n *degrees of
freedom. This can be denoted by* t _{n}*.

**Note 1: **The degrees of freedom of** ***t*** **is the same as the degrees of freedom of the
corresponding** **chi-square random variable.

**Note 2: **The** ***t*-distribution is used as the sampling distribution(s) of the
statistics(s) defined based on** **random sample(s) drawn from normal
population(s).

(i) If *X*_{1}, *X _{2}* , …,

(ii) If (*X _{1}*

(iii) If (*X _{1}*

Hence,

1*. t*–distribution is
symmetrical distribution with mean zero.

2. The graph of *t*-distribution is similar
to normal distribution except for the following two reasons:

(i) The normal distribution curve is higher in
the middle than *t*-distribution curve.

(ii) *t*–distribution has a
greater spread* *sideways than the normal distribution curve. It means
that there is more area in the tails of *t*-distribution.

3. The *t*-distribution curve is asymptotic
to *X*-axis, that is, it extends to infinity on either side.

4. The shape of *t*-distribution curve
varies with the degrees of freedom. The larger is the number of degrees of
freedom, closeness of its shape to standard normal distribution (fig. 2.1).

5. Sampling distribution of *t* does not depend on population
parameter. It depends on degrees of freedom (*n*–1).

The *t*-distribution has the following important applications
in testing the hypotheses for small samples.

1. To test significance of a single population mean, when
population variance is unknown, using *T*_{1}.

2. To test the equality of two population means when population
variances are equal and unknown, using *T*_{2}.

3. To test the equality of two means – paired *t*-test, based
on dependent samples, *T*_{3}.

**Step 1 : **Let** ***µ*** **and** ***σ*^{2}** **be
respectively the mean and variance of the population under study, where** ***σ*^{2}*
*is unknown. If* µ*_{0}* *is an admissible value of* µ*,
then frame the null hypothesis as

*H _{0}: µ = µ_{0} *and choose the suitable
alternative hypothesis from

(i) *H _{1}: µ ≠ µ_{0}* (ii)

**Step 2 : **Describe the sample/data and its descriptive measures. Let (** ***X*_{1},**
***X*_{2}, …,** ***X _{n}*) be a random

**Step 3 : **Specify the level of significance, α.

**Step 4 : **Consider the test statistic , under
*H*_{0}, where and
*S* are the sample mean and sample
standard deviation respectively. The approximate sampling distribution of the test
statistic under *H*_{0} is the *t*-distribution with (*n*–1) degrees of freedom.

**Step 5 : **Calculate the value of *t*
for the given sample ( x_{1} , x_{2} ,... x_{n} ) as .
here is the sample mean and s = is the sample standard
deviation.

**Step 6 : **Choose the critical value,** ***t _{e}*,
corresponding to

**Step 7 : **Decide on** ***H*_{0}** **choosing the suitable
rejection rule from the following table corresponding** **to *H*_{1}.

The average monthly sales, based on past experience of a
particular brand of tooth paste in departmental stores is ₹ 200. An
advertisement campaign was made by the company and then a sample of 26
departmental stores was taken at random and found that the average sales of the
particular brand of tooth paste is ₹ 216 with a standard deviation of ₹8. Does the campaign have helped in promoting
the sales of a particular brand of tooth paste?

**Step 1 : ****Hypotheses**

**Null Hypothesis ***H*_{0}:** ***µ*** **= 200

*i.e., *the average monthly sales of a particular brand of tooth paste is
not significantly* *different from ₹ 200.

**Alternative Hypothesis ***H*_{1}:** ***µ*** **> 200

*i.e., *the average monthly sales of a particular brand of tooth paste are
significantly* *different from ₹ 200. It is one-sided (right) alternative
hypothesis.

**Step 2 : ****Data**

The given sample information are:

Size of the sample (*n)* = 26. Hence, it is a small sample.

Sample mean ( ) = 216, Standard deviation of the
sample = 8.

**Step 3 : ****Level of significance**

*α *= 5%

**Step 4 : ****Test statistic**

The test statistic under *H*_{0}
is T =

Since *n* is small, the sampling distribution of *T* is
the *t*-distribution with (*n*–1) degrees of freedom.

**Step 5 : ****Calculation of test statistic**

The value of *T* for the given sample information is
calculated from

**Step 6 : ****Critical value**

Since *H*_{1} is one-sided (right) alternative
hypothesis, the critical value at *α* =0.05 is

t_{e} = t_{n-1, α} =t_{25,0.05} = 1.708

**Step 7 : ****Decision**

Since it is right-tailed test, elements of critical region are
defined by the rejection rule t_{0} >t_{e} = t_{n-1, α}
= t_{25,0.05} = 1.708. For the given sample information t_{0} =
10.20 > t_{e} =1.708. It* *indicates that given sample contains
sufficient evidence to reject *H*_{0}. Hence, the campaign has
helped in promoting the increase in sales of a particular brand of tooth paste.

A sample of 10 students from a school was selected. Their scores
in a particular subject are 72, 82, 96, 85, 84, 75, 76, 93, 94 and 93. Can we
support the claim that the class average scores is 90?

**Step 1 : ****Hypotheses**

**Null Hypothesis ***H*_{0}:** ***µ*** **= 90

*i.e., *the class average scores is not significantly different from 90.

**Alternative Hypothesis ***H*_{1}** **:** ***µ*** **≠ 90

*i.e., *the class means scores is significantly different from 90.

It is a two-sided alternative hypothesis.

**Step 2 : ****Data**

The given sample information are

Size of the sample (*n)* = 10. Hence, it is a small sample.

**Step 3 : ****Level of significance**

α= 5%

**Step 4 : ****Test statistic**

The test statistic under *H*_{0}
is T =

Since *n* is small, the sampling distribution of *T* is
the *t -* distribution with (*n*–1) degrees of freedom.

**Step 5 : ****Calculation of test statistic**

The value of *T* for the given sample information is
calculated from *t*_{0} = as under:

Sample mean

**Step 6 : ****Critical value**

Since *H*_{1}
is two-sided alternative hypothesis, the critical value at α = 0.05 is t_{e} = t_{n-1,
α/2} = t_{9,0.025} = 2.262

**Step 7 : ****Decision**

Since it is two-tailed test, elements of critical region are
defined by the rejection rule |t_{0}| > t_{e} = t_{ n-1, α/2} = t =2.262. For the given sample information |t_{0}| =
1.806 < t_{e} = 2.262.

It indicates that given sample does not provide sufficient
evidence to reject H_{0}. Hence, we conclude that the class average
scores is 90.

*Procedure:*

**Step 1 : **Let** ***μ _{X}*

Frame the null hypothesis as H_{0} : µ_{X} = µ_{Y}
and choose the suitable alternative hypothesis from (i) H_{1} : μ_{X}
≠ μ_{Y} (ii) *H*_{1} :
μ_{X} > μ_{Y} (iii) *H*_{1}
: μ_{X} < μ_{Y}

**Step 2 : **Describe the sample/data. Let (X_{1}, X_{2 }, …, X_{m})
be a random sample of m observations drawn from Population-1 and (Y_{1},
Y_{2} , …, Y_{n}) be a random sample of n observations drawn
from Population-2, where m and n are small (i.e., *m* < 30 and *n* < 30).
Here, these two samples are assumed to be independent.

**Step 3 : **Set up level of significance (*α*)

**Step 4 : **Consider the test statistic

where
*S*_{p} is the “pooled”
standard deviation (combined standard deviation) given by

The
approximate sampling distribution of the test statistic

is the t-distribution with *m+n*–2 degrees of freedom i.e., t ~ t_{m+n–2}.

**Step 5 : **Calculate the value of** ***T*** **for the given sample**
**(** ***x*_{1}** **,** ***x*_{2}** **,...**
***x _{m}*

**Step 6 : **Choose the critical value,** ***t _{e}*,
corresponding to

**Step 7 : **Decide on** ***H*_{0}** **choosing the suitable
rejection rule from the following table corresponding** **to *H*_{1}.

**Example 2.3**

The following table gives the scores (out of 15) of two batches of
students in an examination.

Test at 1% level of significance the average performance of the
students in Batch I and Batch II are equal.

*Solution:*

**Step 1 : ****Hypotheses:**** **Let** ***µ*_{X}** **and** ***µ*_{Y}** **denote respectively the average performance of
students in Batch I and Batch II. Then the null and alternative
hypotheses are :

**Null Hypothesis ***H*_{0}** **:** ***µ*** **_{X}** **=** ***µ*_{Y}

*i.e.*, the average performance of the students in Batch I and Batch II
are equal.

**Alternative Hypothesis ***H*_{1}** **:** ***µ*** **_{X}** **≠** ***µ*_{Y}

*i.e., *the average performance of the students in Batch I and Batch II
are not equal.

**Step 2 : ****Data**

The given sample information are:

Sample size for Batch I : *m* =10

Sample size for Batch II : *n* = 8

**Step 3 : ****Level of significance**

*α= *1%

**Step 4 : ****Test statistic**

The test statistic under *H*_{0} is

The sampling distribution of *T* under *H*_{0}
is the *t*-distribution with *m*+*n*–2 degrees of freedom *i.e.,
t ~ t _{m}*

**Step 5 : ****Calculation of test statistic**

To find sample mean and sample standard deviation:

**To find sample means:**

Let (*x*_{1} , *x*_{2} ,..., *x*_{10})
and (*y*_{1}, *y*_{2} ,..., *y*_{8})
denote the scores of students in Batch I and Batch II respectively.

**To find combined sample standard deviation:**

Pooled
standard deviation is:

The
value of T is calculated for the given information as

**Step 6 : ****Critical value**

Since *H*_{1} is two-sided alternative hypothesis,
the critical value at *α* = 0.01 is t_{e} = t_{m+n-2, }_{α}_{/2} = t_{16,0.005}
= 2.921

**Step 7 : ****Decision**

Since it is two-tailed test, elements of critical region are
defined by the rejection rule
|*t*_{0}| < *t _{e}*

It indicates that^{2} given sample contains insufficient
evidence to reject *H*_{0}. Hence, the mean performance of the
students in these batches are equal.

Two types of batteries are tested for their length of life (in
hours). The following data is the summary descriptive statistics.

Is there any significant difference between the average life of
the two batteries at 5% level of significance?

**Step 1 : ****Hypotheses**

**Null Hypothesis ***H*_{0}** **:** ***μ _{X}*

*i.e., *there is no significant difference in average life of two types of
batteries* A *and* B*.

**Alternative Hypothesis ***H*_{0}** **:** ***μ _{X}*

*i.e., *there is significant difference in average life of two types of
batteries* A *and* B*. It is a* *two-sided alternative hypothesis

**Step 2 : ****Data**

The given sample information are :

*m *= number of batteries under type* A *= 14

*n *= number of batteries under type* B *= 13

=
Average life (in hours) of type *A* battery = 94

* *= Average life (in hours) of type* B *battery = 86* *

*s _{X} *= standard deviation of type

*s _{Y} *= standard deviation of type

**Step 3 : ****Level of significance**

α= 5%

**Step 4 : ****Test statistic**

The test statistic under *H*_{0} is

The sampling distribution of *T* under *H*_{0}
is the *t*-distribution with *m*+*n*–2 degrees of freedom *i.e.,
t ~ t _{m}*

**Step 5 : ****Calculation of test statistic**

Under null hypotheses *H*_{0}:

where
s is the pooled standard deviation given by,

The
value of T is calculated for the given information as

**Step 6 : ****Critical value**

Since H_{1} is two-sided alternative hypothesis, the
critical value at α = 0.05 is t_{e} = t_{m+n-2, α/2} = t_{25,
0.025} = 2.060.

**Step 7 : ****Decision**

Since it is a two-tailed test, elements of critical region are
defined by the rejection rule |t_{0}| < t_{e} = t_{m+n-2,α/2}
= t_{25, 0.025} = 2.060. For the given sample information |t_{0}|
= 1.15 < t_{e} = 2.060. It indicates that2 given sample contains
insufficient evidence to reject H0. Hence, there is no significant difference
between the average life of the two types of batteries.

**Step 1 : **Let** ***X*** **and** ***Y*** **be two
correlated random variables having the distributions respectively** ***N*(*μ _{X}
*,

Frame null hypothesis as

*H*_{0}* *:* μ _{D}= *0

And choose alternative hypothesis from

(i) *H*_{1} : *μ _{D}≠*
0 (ii)

**Step 2 : **Describe the sample/data. Let (*X*_{1}*, X*_{2}*,
…, X _{m}*) be a random sample of

**Step 3 : **Set up level of significance (*α*)

**Step 4 : **Consider the test statistic

The approximate sampling distribution of the test statistic *T*
under *H*_{0} is *t -* distribution with (*n*–1) degrees
of freedom.

**Step 5 : **Calculate the value of** ***T*** **for the given data
as

**Step 6 : **Choose the critical value,** ***t _{e}*,
corresponding to

**Step 7 : **Decideon** ***H*_{0}** **choosing the suitable
rejection rule from the following table corresponding** **to *H*_{1}.

A company gave an intensive training to its salesmen to increase
the sales. A random sample of 10 salesmen was selected and the value (in lakhs
of Rupees) of their sales per month, made before and after the training is
recorded in the following table. Test whether there is any increase in mean
sales at 5% level of significance.

**Step 1 : ****Hypotheses**

**Null Hypothesis ***H*_{0}** **:** ***μ _{D}=*

*i.e.*, there is no significant increase in the mean sales after the
training.

**Alternative Hypothesis ***H*_{1}** **:** ***μ _{D}>*

*i.e.*, there is significant increase in the mean sales after the
training. It is a one-sided* *alternative hypothesis.

**Step 2 : ****Data**

Sample size *n* = 10

**Step 3 : ****Level of significance**

α= 5%

**Step 4 : ****Test statistic**

Test statistic under the null hypothesis is

The sampling distribution of *T* under *H _{0}*
is

**Step 5 : ****Calculation of test statistic**

To find and *s*:

Let *x* denote sales before training and *y* denote
sales after training

Here instead of *d _{i} = x_{i} – y_{i} *it is assumed

**Step 6 : ****Critical value**

Since *H*_{0} is a one-sided alternative hypothesis,
the critical value at 5% level of significance is *t _{e}*

**Step 7 : ****Decision**

It is a one-tailed test. Since |*t*_{0}| = 1.6052
< *t _{e}*

Hence, there is no evidence that the mean sales has increased
after the training.

Tags : Properties, Procedure Steps, Example Solved Problems | Statistics , 12th Statistics : Chapter 2 : Tests Based on Sampling Distributions I

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12th Statistics : Chapter 2 : Tests Based on Sampling Distributions I : Student’s t Distribution and Its Applications | Properties, Procedure Steps, Example Solved Problems | Statistics

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