Question

A well-known brokerage firm executive claimed that 20% 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 500 people, 12% of them said they are confident of meeting their goals.

Test the claim that the proportion of people who are confident is smaller than 20% at the 0.01 significance level.

The null and alternative hypothesis would be:

The test statistic is:                              (to 3 decimals)

The p-value is:    (to 4 decimals)

Based on this we:

• Fail to reject the null hypothesis
• Reject the null hypothesis

Answer 1

Solution:

Claim: the proportion of people who are confident is smaller than 20%

Level of significance =

Sample size = n = 500

Sample proportion of people who are confident of meeting their goals =

Part a) The null and alternative hypothesis would be:

H0: p = 0.20       Vs H1: p < 0.20

Part b) The test statistic is:      

Part c) The p-value is:

p-value = P( Z < z test statistic value)

p-value = P( Z < -4.472 )

Use excel command:

=NORM.S.DIST( z , cumulative)

=NORM.S.DIST( -4.472 , TRUE)

= 0.0000

Thus p-value = 0.0000

( Since z = -4.472 is too small , so area under z = -4.472 is approximately = 0, thus we can assume directly

P( Z < -4.472) = 0.0000 , thus p-value = 0.0000 )

Part d) Based on this we:

Since p-value = 0.0000 < 0.01 significance level, we reject null hypothesis H0.

Thus we conclude that there is sufficient evidence to support claim that the proportion of people who are confident is smaller than 20% at the 0.01 significance level.