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

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:

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

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:

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.

0.6 is tested at the null hypothesis:

This means that

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

Value of the test statistic:

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:

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363