Answer:
The p-value of the test is 0.1469 > 0.05, which means that there is no reason to believe that the proportion of adults favoring capital punishment today has increased, using a 0.05 level of significance.
Step-by-step explanation:
Suppose that, in the past, 40% of all adults favored capital punishment. Test if the proportion has increased:
At the null hypothesis, we test if the proportion is still of 40%, that is:
![H_0: p = 0.4](https://tex.z-dn.net/?f=H_0%3A%20p%20%3D%200.4)
At the alternative hypothesis, we test if the proportion has increased, that is, is greater than 40%, so:
![H_1: p > 0.4](https://tex.z-dn.net/?f=H_1%3A%20p%20%3E%200.4)
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.
0.4 is tested at the null hypothesis:
This means that ![\mu = 0.4, \sigma = \sqrt{0.4*0.6}](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.4%2C%20%5Csigma%20%3D%20%5Csqrt%7B0.4%2A0.6%7D)
Random sample of 15 adults, 8 favor capital punishment.
This means that ![n = 15, X = \frac{8}{15} = 0.5333](https://tex.z-dn.net/?f=n%20%3D%2015%2C%20X%20%3D%20%5Cfrac%7B8%7D%7B15%7D%20%3D%200.5333)
Value of the test statistic:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \frac{0.5333 - 0.4}{\frac{\sqrt{0.4*0.6}}{\sqrt{15}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.5333%20-%200.4%7D%7B%5Cfrac%7B%5Csqrt%7B0.4%2A0.6%7D%7D%7B%5Csqrt%7B15%7D%7D%7D)
![z = 1.05](https://tex.z-dn.net/?f=z%20%3D%201.05)
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion of 0.5333 or more, which is 1 subtracted by the p-value of z = 1.05.
Looking at the z-table, z = 1.05 has a p-value of 0.8531.
1 - 0.8531 = 0.1469.
The p-value of the test is 0.1469 > 0.05, which means that there is no reason to believe that the proportion of adults favoring capital punishment today has increased, using a 0.05 level of significance.