Question

A well-known brokerage firm executive claimed that 60% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 100 people, 69% of them said they are confident of meeting their goals. Test the claim that the proportion of people who are confident is larger than 60% at the 0.01 significance level.
The null and alternative hypothesis would be:________
a. H0:μ=0.6H0:μ=0.6
H1:μ≠0.6H1:μ≠0.6
b. H0:μ=0.6H0:μ=0.6
H1:μ<0.6H1:μ<0.6
c. H0:μ=0.6H0:μ=0.6
H1:μ>0.6H1:μ>0.6
d. H0:p=0.6H0:p=0.6
H1:p≠0.6H1:p≠0.6
e. H0:p=0.6H0:p=0.6
H1:p>0.6H1:p>0.6
f. H0:p=0.6H0:p=0.6
H1:p<0.6H1:p<0.6
The test is:________
a. two-tailed
b. left-tailed
c. right-tailed
The test statistic is:_______ (to 3 decimals)
The p-value is:_______ (to 4 decimals)
Based on this we:________
a. Fail to reject the null hypothesis
b. Reject the null hypothesis

Answer 1

We are testing a hypothesis. So, first we identify the null and the alternative hypothesis, then we find the test statistic, and with the test statistic, the p-value is found.

Null and alternative hypothesis:

Claim the the proportion is of 60%, thus, the null hypothesis is:

$H_0: p = 0.6$

Test if the proportion is greater than 60%, thus, the alternative hypothesis is:

$H_1: p > 0.6$

And the answer to the first question is given by option c.

Classification:

We are testing if the proportion is greater than a value, so it is a right-tailed test.

Test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

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

0.6 is tested at the null hypothesis:

This means that $\mu = 0.6, \sigma = \sqrt{0.4*0.6}$

Survey, conducted over a two week period, found that in a sample of 100 people, 69% of them said they are confident of meeting their goals.

This means that $n = 100, X = 0.69$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.69 - 0.6}{\frac{\sqrt{0.4*0.6}}{\sqrt{100}}}$

$z = 1.837$

The test statistic is z = 1.837.

p-value:

The p-value of the test is the probability of finding a sample proportion above 0.69, which is 1 subtracted by the p-value of z = 1.837.

Looking at the z-table, z = 1.837 has a p-value of 0.9669.

1 - 0.9669 = 0.0331

The p-value is 0.0331.

Decision:

The p-value of the test is 0.0331 > 0.01, and thus:

a. Fail to reject the null hypothesis

For another example of a problem of a test of hypothesis, you can take a look at:

brainly.com/question/24166849