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:

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

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