Two thousand numbers are selected randomly; 960 were even numbers. At the 0.10 level of significance, determine whether the proportion of odd numbers is significantly different from 50%.

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:

At the alternative hypothesis, we test if the proportion differs from 0.5, that is:

The test statistic is:

In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.

0.5 is tested at the null hypothesis:

This means that

Two thousand numbers are selected randomly; 960 were even numbers.

This means that

Value of the test statistic:

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%.

over the argument its not good to fight.

its c or d   take a lucky guess my dude