Can you evaluate the expression and show the work ?
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
1) Data given and notation n
n=88 represent the random sample taken
X=35 represent the adults that said they are confident of meeting their goals
estimated proportion of adults that said they are confident of meeting their goals
is the value that we want to test
represent the significance level
z would represent the statistic (variable of interest)
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 that at least 41 % of investors are currently confident of meeting their investment goals.:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statisitc, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
4) Statistical decision
It's important to refresh the p value method or p value approach . "This methos 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 is not provided, but we can assume . The next step would be calculate the p value for this test.
Since is a unilateral test the p value would be:
Using the p value obtained and the significance level assumed we have
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of of investors that are currently confident of meeting their investment goals is not significantly lower than 0.41 or 41%.