One Sample T Test Formula

 In statistics, hypothesis testing are utilized to get the probability for a particular hypothesis to be acceptable. Hypothesis is specified as statement which may or may not be acceptable. In statistics two hypothesis testing are applied. They are null and alternative hypotheses. The null and alternative hypotheses testing are conflicting to every other.  Various tests are used in hypothesis. T test is the important test using for hypothesis. Each test establish through a null hypothesis. Let us see about the one sample t test formula.

One Sample T Test Formula

Hypothesis testing has the following important steps

  • Null hypothesis
  • Alternative hypothesis
  • Test statistic
  • Level of significance
  • Inference

One sample t test formula

In null hypothesis we can consider that there is no significance distinction among the population mean with the sample mean.

In alternative hypothesis we can consider that there is a significant distinction among the population mean with the sample mean.

Compute the standard deviation using this method:

`S^2 = 1/(n-1)sum_(n=1)^(2n)(x_(i)-barx)^2 `

Compute the value of the one sample t-test, by applying this formula:

`t = (bar x -mu)/(s/sqrt(n)) ` ~ t(n-1)

Where, t represents the one sample t – test and μ represents the population mean.

Compare calculated value and table value

Calculated value > Table value

Null hypothesis is rejected. Otherwise alternative hypothesis is rejected.

Examples for One Sample T Test Formula

A random sample of 30 students scored an average of 27 marks and sum of the squares of the deviation taken from the mean is 90, can the sample be regarded as taken from the students taken 4.5 as average.

Solution

Null hypothesis:

H0: The known sample is taken from the population with mean 4.5

Alternative hypothesis:

H1: The known sample is not taken from the population with mean 4.5

Constructing the statistic:

Here ‘n’ is small

`t = (bar x -mu)/(s/sqrt(n)) ` ~ t(n-1)

Level of significance:

At 5% level for the 29 degrees of freedom ‘t’ table value of 2.045

Calculation:

μ =30,  `barx =27 `, n =30

`sum_(n=1)^(2n)(x_(i)-barx)^2 = 90`

`S^2 = 1/(n-1)sum_(n=1)^(2n)(x_(i)-barx)^2 `

=`1/29 xx90`

`S^2 =3.1`

s =1.7616 <br>

`t = (bar x -mu)/(s/sqrt(n))`~ t(n-1)

`t = (27- 30)/(1.7616/sqrt(30))`

`t = -3/0.3216`

t = 9.328

Calculated value = 9.328

Table value = 2.045

Calculated value > Table value

That is 9.328 > 2.045

Null hypothesis is rejected.

Thus the sample is not taken from the population with mean 4.5



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