Answer to Question #265823 in Statistics and Probability for diwakar

Question

a sample of 60 units was inspected to find the number of defective,which was grouped in 0,1,2,3,4 or 5 defectives .if X is the random variable defining the number of defective units, then the frequency table of the X is as follows

X012345Frequency517121583

test at 1% level of significance whether this data can be considered to follow a binomial distribution.

Answer 1

$x=(0,1,2,3,4,5)$

observed frequencies:

$f=(5,17,12,15,8,3)$

relative frequencies are $r_x=f_x/60$

the sample mean is

$\mu=\sum xr_x=\frac{17+24+45+32+15}{60}=2.22$

binomial mean:

$\mu=np$ , where $n=6$ is the number of independent trials,

p is the success probability.

$p=2.22/6=0.37$

The chi-squared goodness-of-fit test:

the null hypothesis is that the distribution $Binom(6,0.37)$ is an appropriate model for the number of successes.

test statistic is:

$\chi^2=\sum\frac{(f_x-E_x)^2}{E_x}$

where expected frequencies are $E_x=60p_x$

$p_x=C^x_np^x(1-p)^{n-x}$

$p_0=0.0625,p_1=0.2203,p_2=0.3234,p_3=0.2533$

$p_4=0.1116,p_5=0.0262$

$E_0=3.75,E_1=13.22,E_2=19.40,E_3=15.20$

$E_4=6.70,E_5=1.57$

$\chi^2=\frac{(5-3.75)^2}{3.75}+\frac{(17-13.22)^2}{13.22}+\frac{(12-19.4)^2}{19.4}+\frac{(15-15.2)^2}{15.2}+\frac{(8-6.7)^2}{6.7}+\frac{(3-1.57)^2}{1.57}=5.8775$

$df=n-2=4$

critical value:

$\chi^2_{crit}=13.277$

Since $\chi^2<\chi^2_{crit}$ we accept the null hypothesis. Data can be considered to follow a binomial distribution.