The weight of people in a small town in missouri is known to be normally distributed with a mean of 188 pounds and a standard de
1 answer:
You might be interested in
Answer:
a) The 99% confidence interval is given by (0.198;0.242).
b) Based on the p value obtained and using the significance level assumed we have
so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.
c)
Step-by-step explanation:
<em>Data given and notation </em>
n=2362 represent the random sample taken
X represent the people who says that they would watch one of the television shows.
estimated proportion of people rated as Excellent/Good economic conditions.
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) <em>
</em>
<em>Concepts and formulas to use </em>
We need to conduct a hypothesis in order to test the claim that 24% of people are rated with good economic conditions:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statistic, 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
.
Part a: Test the hypothesis
<em>Check for the assumptions that he sample must satisfy in order to apply the test </em>
a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.
b) The sample needs to be large enough
np = 2362x0.22=519.64>10 and n(1-p)=2364*(1-0.22)=1843.92>10
Condition satisfied.
<em>Calculate the statistic</em>
Since we have all the info requires we can replace in formula (1) like this:
The confidence interval would be given by:
The critical value using and
would be
. Replacing the values given we have:
So the 99% confidence interval is given by (0.198;0.242).
Part b
<em>Statistical decision </em>
P value method or p value approach . "This method consists on 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 is . The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
So based on the p value obtained and using the significance level assumed we have
so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.
Part c
The confidence level assumed was 99%, so then the signficance is given by