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.

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keonis201 keonis201 · Answer 1 · 2021-05-21T20:22:27+02:00

I’m just tryna get up my bag yurt

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jtellezd jtellezd · Answer 2 · 2021-05-22T15:29:38+02:00

Answer:

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