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

Answer:

$p_v =P(Z$  

If we compare the p value obtained and using the significance level given $\alpha=0.005$ we have $p_v$ 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% .  

Step-by-step 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:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

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

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:  

$p_v =P(Z$  

If we compare the p value obtained and using the significance level given $\alpha=0.005$ we have $p_v$ 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% .