Mathematics, 21.04.2020 01:24, krystalScott17

A research center claims that at least 4545% of adults in a certain country think the government is not aggressive enough in pursuing people who cheat on their taxes. In a random sample of 700700 adults from that country, 3939% say that the government is not aggressive enough in pursuing people who cheat on their taxes. At alphaαequals=0.010.01, is there enough evidence to reject the center's claim? Complete parts (a) through (e) below.

Answers: 2

Show answers

Answers

answered: hjamya17

$z=\frac{0.39 -0.45}{\sqrt{\frac{0.45(1-0.45)}{700}}}=-3.19$

p_v =P(z

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interst is significantly lower than 0.45 and the claim is not appropiate

Step-by-step explanation:

Data given and notation

n=700 represent the random sample taken

$\hat p=0.39$ estimated proportion of interest

p_o=0.7 is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v 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 the true proportion is at least 0.45.:

Null hypothesis: $p\geq 0.45$

Alternative hypothesis: p < 0.45

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value p_o .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.39 -0.45}{\sqrt{\frac{0.45(1-0.45)}{700}}}=-3.19$

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 $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a lef tailed test the p value would be:

p_v =P(z

answered: Guest

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5