Factorise the following expression :
1 answer:
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Answer:
a)
If we compare the p value obtained and using the significance level given 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 rejected items is less than 0.03.
b) We can use a 95 percent upper confidence interval.
On this case we want a interval on this form :
So the critical value would be on this case and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"
We found the upper limit like this:
And the interval would be:
And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03
Step-by-step explanation:
Part a
Data given and notation
n=500 represent the random sample taken
X=10 represent the number of objects rejected
estimated proportion of objects rejected
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:
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
.
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
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 . The next step would be calculate the p value for this test.
Since is a one tailed left test the p value would be:
If we compare the p value obtained and using the significance level given 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 rejected items is less than 0.03.
Part b
We can use a 95 percent upper confidence interval.
On this case we want a interval on this form :
So the critical value would be on this case and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"
We found the upper limit like this:
And the interval would be:
And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03
The limit does not exist. There are infinitely many infinite discontinuities at , where
. The function oscillates wildly between negative and positive infinity.