Respuesta :
Answer:
The p-value of the test is 0.0734 < 0.1, which means that there is significant evidence, at the 0.1 level of significance, to conclude that the proportion of odd numbers is significantly different from 50%.
Step-by-step explanation:
Test if the proportion of odd numbers is significantly different from 50%.
At the null hypothesis, we test if the proportion is of 0.5, that is:
[tex]H_0: p = 0.5[/tex]
At the alternative hypothesis, we test if the proportion differs from 0.5, that is:
[tex]H_1: p \neq 0.5[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.5 is tested at the null hypothesis:
This means that [tex]\mu = 0.5, \sigma = \sqrt{0.5(1-0.5)} = 0.5[/tex]
Two thousand numbers are selected randomly; 960 were even numbers.
This means that [tex]n = 2000, X = \frac{960}{2000} = 0.48[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.48 - 0.5}{\frac{0.5}{\sqrt{2000}}}[/tex]
[tex]z = -1.79[/tex]
P-value of the test and decision:
The p-value of the test is the probability that the sample proportion differs from 0.5 by at least |0.48-0.5| = 0.02, which is P(|z|<1.79), which is 2 multiplied by the p-value of z = -1.79.
Looking at the z-table, z = -1.79 has a p-value of 0.0367
2*0.0367 = 0.0734
The p-value of the test is 0.0734 < 0.1, which means that there is significant evidence, at the 0.1 level of significance, to conclude that the proportion of odd numbers is significantly different from 50%.
