A researcher claims that 45% of U. S. adults think that the IRS is not aggressive enough in pursuing people who cheat on their taxes. You believe the true percentage is more than 45%. To test this, you take a simple random sample of 500 U. S. adults and find that 245 of them say the IRS is not aggressive enough. Required:
a. What's the minimum population size required?
c. How many successes were there?
c. Test at 0.01 significance.

I’m just tryna get up my bag yurt

a) Then minimum sample size is  n = 12

b) We got  245 successes  

c)p-value is greater than 0,01 Therefore we accept H₀.

We can´t support our claim

Step-by-step explanation:

a) The minimum population size required to use the approximation of binomial distribution to normal distribution can be obtained from:

n*p > 5      p  = 0,45     0,45 * n  >  5     n > 11  

Then minimum sample size is  n = 12

b)  We got  245 successes  out of  n = 500

x =  245       p  =  245/500    p  = 0,49

q  =  1 - p     q  =  1  - 0,49    q  =  0,51

Test Hypothesis:

Null Hypothesis                        H₀      p  =  0,45

Alternative Hypothesis            Hₐ      p  > 0,45

Alternative Hypothesis indicates that the test is a one tail test to the right

z( s)  =   (  p  -  0,45  )  / √ p*q/n

z( s)  =   (  0,49 - 0,45 ) / √ 0,49*0,51 / 500

z( s)  =  0,04 /  0,0223

z(s)  =  1,793

p- value for  z = 1.79  is from z-table    p-value = 0,0367

Comparing p-value with the significance level  0,01 we see that

p-value is greater than 0,01 Therefore we accept H₀.

We can´t support our claim

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!

let's fu(!

step-by-step explanation: