Question

3. A random sample of 100 people was taken. There were sixty people who favored Brand A. We are interested in determining whether or not the proportion of the population in favor of Brand A is significantly more than 55%. At a 0.05 level of significance, it can be concluded that the portion of the population in favor of Brand A is ____. a. Significantly greater than 60% b. Not significantly greater than 55% c. Not significantly greater than 60% d. Significantly greater than 55% e. None of the above

Answer 1

Testing the hypothesis, it is found that the correct option is:

b. Not significantly greater than 55%.

At the null hypothesis, we test if the proportion is not significantly more than 55%, that is:

$H_0: p \leq 0.55$

At the alternative hypothesis, we test if the proportion is significantly more than 55%, that is:

$H_1: p < 0.55$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.
• p is the proportion tested at the null hypothesis.
• n is the sample size.

For this problem, the parameters are:

$p = 0.55, n = 100, \overline{p} = \frac{60}{100} = 0.6$

Hence, the value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.6 - 0.55}{\sqrt{\frac{0.55(0.45)}{100}}}$

$z = 1.01$

The p-value of the test is the probability of finding a sample proportion above 0.6, which is 1 subtracted by the p-value of z = 1.01.

Looking at the z-table, z = 1.01 has a p-value of 0.8438.

1 - 0.8438 = 0.1562.

The p-value of the test is 0.1562 > 0.05, which means that the proportion is not significantly greater than 55%, and option b is correct.

A similar problem is given at https://brainly.com/question/24330815