Question

A well-known brokerage firm executive claimed that 50% 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 200 people, 44% of them said they are confident of meeting their goals. Test the claim that the proportion of people who are confident is smaller than 50% at the 0.005 significance level.

Answer 1

`p_v =P(Z<-1.697)=0.0448`

If we compare the p value obtained and using the significance level given
`\alpha=0.005` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults that said they are confident of meeting their goals is significantly lower than 0.5 or 50% .

Explanation:

1) Data given and notation

n=200 represent the random sample taken

X represent the people that said they are confident of meeting their goals

`\hat p=0.44` estimated proportion of adults that said they are confident of meeting their goals

`p_o=0.5` is the value that we want to test

`\alpha=0.005` represent the significance level

Confidence=99.5% or 0.995

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.5.:

Null hypothesis:
`p\geq 0.5`

Alternative hypothesis:
`p < 0.5`

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.44 -0.5}{\sqrt{(0.5(1-0.5))/(200)}}=-1.697`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.005`. The next step would be calculate the p value for this test.

Since is left tailed test the p value would be:

`p_v =P(Z<-1.697)=0.0448`

If we compare the p value obtained and using the significance level given
`\alpha=0.005` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults that said they are confident of meeting their goals is significantly lower than 0.5 or 50% .