Question

A well-known brokerage firm executive claimed that 40% 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, 38% of them said they are confident of meeting their goals. Test the claim that the proportion of people who are confident is smaller than 40% at the 0.025 significance level. The null and alternative hypothesis would be: H 0 : μ ≤ 0.4 H 1 : μ > 0.4 H 0 : p ≤ 0.4 H 1 : p > 0.4 H 0 : p = 0.4 H 1 : p ≠ 0.4 H 0 : μ = 0.4 H 1 : μ ≠ 0.4 H 0 : μ ≥ 0.4 H 1 : μ < 0.4 H 0 : p ≥ 0.4 H 1 : p < 0.4 1. The test is:______ a. left-tailed b. two-tailed c. right-tailed 2. The test statistic is:______ (to 3 decimals) 3. The p-value is:______ (to 4 decimals) 4. Based on this we:_______ a. Fail to reject the null hypothesis b. Reject the null hypothesis

Answer 1

1) Null hypothesis:
`p \geq 0.4`

Alternative hypothesis:
`p < 0.4`

2)
`z=\frac{0.38 -0.4}{\sqrt{(0.4(1-0.4))/(500)}}=-0.913`

3)
`p_v =P(z<-0.913)=0.1806`

4) For this case since the p value is higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true proportion is not significantly less than 0.4

a. Fail to reject the null hypothesis

Explanation:

Information given

n=500 represent the random sample taken

`\hat p=0.38` estimated proportion of if the people they are confident of meeting their goals

`p_o=0.4` is the value to test

represent the significance level

z would represent the statistic

`p_v` represent the p value

Part 1

We want to test if the true proportion is less than 0.4, the system of hypothesis are.:

Null hypothesis:
`p \geq 0.4`

Alternative hypothesis:
`p < 0.4`

a. left-tailed

Part 2

The statistic is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

Replacing the info given we got:

`z=\frac{0.38 -0.4}{\sqrt{(0.4(1-0.4))/(500)}}=-0.913`

Part 3

The p value for this case can be calculated like this:

`p_v =P(z<-0.913)=0.1806`

Part 4

For this case since the p value is higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true proportion is not significantly less than 0.4

a. Fail to reject the null hypothesis