# Thread: The following fig show the distribution of digits ....Test the hypothesis

1. ## The following fig show the distribution of digits ....Test the hypothesis

Question:
The following figures show the distribution of digits in numbers chosen at random from a telephone directory

$Digits : \qquad \ 0 \qquad 1 \qquad 2 \quad \ 3 \quad \ 4 \qquad 5 \quad \ 6 \quad \ 7 \quad \ 8 \quad \ 9$
$Frequency : \ 1026 \ 1107 \ 997 \ 966 \ 1075 \ 933 \ 1107 \ 972 \ 964 \ 853
$

Test the hypothesis that the digits are uniformly distributed at 5% level of significance.

2. Originally Posted by zorro
Question:
The following figures show the distribution of digits in numbers chosen at random from a telephone directory

$Digits : \qquad \ 0 \qquad 1 \qquad 2 \quad \ 3 \quad \ 4 \qquad 5 \quad \ 6 \quad \ 7 \quad \ 8 \quad \ 9$
$Frequency : \ 1026 \ 1107 \ 997 \ 966 \ 1075 \ 933 \ 1107 \ 972 \ 964 \ 853
$

Test the hypothesis that the digits are uniformly distributed at 5% level of significance.
$\chi^2$ or chi squared test, look it up try to do it then let us know what problems you are having

CB

3. ## what shouldbe the Null hypothesis

Originally Posted by CaptainBlack
$\chi^2$ or chi squared test, look it up try to do it then let us know what problems you are having

CB

What should be the null hypothesis for this ?

4. Originally Posted by zorro
What should be the null hypothesis for this ?
That each digit occurs with equal probability (so the expected frequencies are all equal).

CB

5. ## Am i doing it correctly

Originally Posted by CaptainBlack
That each digit occurs with equal probability (so the expected frequencies are all equal).

CB

From the last post what i have come to understand is that all the frequencies are equal

$H_0 : \sigma^2 = 0$
$H_a : \sigma^2 \neq 0$
$\alpha : 0.05$

Am i correct???

6. ## I am stuck

Originally Posted by CaptainBlack
That each digit occurs with equal probability (so the expected frequencies are all equal).

CB

I have googled and have lots of documentation but dont know how to solve this question . I am studying for my exam and need to find some document that is close to the above question so that i could understand the chi square test and the hypothesis a little better.....

7. ## Is this correct?

Originally Posted by CaptainBlack
$\chi^2$ or chi squared test, look it up try to do it then let us know what problems you are having

CB

Null Hypothesis : Digits are uniformly distributed
$
\begin{array}{c|c|c|c|c|c|c|c|c|c}
Digits & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
Freqs & 933 & 933 & 933 & 933 & 933 & 933 & 933 & 933 & 933
\end{array}$

$x^2$ = $\sum \frac{f_o - f_e}{f_e}$

$x^2$ = 108.1 ...........Is my answer right??

no of degree of freedom = 9 - 1 = 8

$x_{0.05} ^2$ for 8 d.f. = 2.73

Since the calculated > tabulated

we reject the null hypothesis

8. Originally Posted by zorro
Null Hypothesis : Digits are uniformly distributed
$
\begin{array}{c|c|c|c|c|c|c|c|c|c}
Digits & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
Freqs & 933 & 933 & 933 & 933 & 933 & 933 & 933 & 933 & 933
\end{array}$

$x^2$ = $\sum \frac{f_o - f_e}{f_e}$

$x^2$ = 108.1 ...........Is my answer right??

no of degree of freedom = 9 - 1 = 8

$x_{0.05} ^2$ for 8 d.f. = 2.73

Since the calculated > tabulated

we reject the null hypothesis
No.

First I make the expected frequencies $1000$ (this is the sum of the observed frequencies divided by the number of bins which here is $10$). There were $10$ bins corresponding to digits $0$ through $9$ not $9$)

Second the formula for calculating the value of $\chi^2$ is:

$\chi^2=\sum_{i=1}^N\frac{(O_i-E_i)^2}{E_i}$

The value I get for this is $\chi^2=58.5$ with $\nu=9$. This exceeds both the $5$ and $1 \%$ critical values for the $\chi$ squared distribution with this number of degrees of freedom and so the observed frequencies differ significantly (at which of these levels you choose to report the results)

CB

9. Originally Posted by CaptainBlack
No.

First I make the expected frequencies $1000$ (this is the sum of the observed frequencies divided by the number of bins which here is $10$). There were $10$ bins corresponding to digits $0$ through $9$ not $9$)

Second the formula for calculating the value of $\chi^2$ is:

$\chi^2=\sum_{i=1}^N\frac{(O_i-E_i)^2}{E_i}$

The value I get for this is $\chi^2=58.5$ with $\nu=9$. This exceeds both the $5$ and $1 \%$ critical values for the $\chi$ squared distribution with this number of degrees of freedom and so the observed frequencies differ significantly (at which of these levels you choose to report the results)

CB