HI SOMEONE PLEASE HELP!!!!!!!!!!

in a program designed to help patients stop smoking 232 patients were given sustained care and 84.9% of them were no longer smok

grandymaker [24]

Answer:

$z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869$

$p_v =2*P(Z>1.869)=0.0616$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 adults were no longer smoking after one month is not significantly different from 0.8 or 80% .

Step-by-step explanation:

1) Data given and notation

n=232 represent the random sample taken

X represent the adults were no longer smoking after one month

$\hat p=0.849$ estimated proportion of adults were no longer smoking after one month

$p_o=0.80$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.8.:

Null hypothesis: $p=0.8$

Alternative hypothesis: $p \neq 0.8$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869$

4) 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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(Z>1.869)=0.0616$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 adults were no longer smoking after one month is not significantly different from 0.8 or 80% .

6 0

4 years ago

Suppose you have a 6-face unfair dice with numbers 1,2,3,4,5,6 on each of its faces. If the probability distribution of throwing

kotegsom [21]

Answer:

D is correct

Step-by-step explanation:

Here, we want to select which of the options is correct.

The correct option is the option D

Since the die is unfair, we expect that the probability of each of the numbers turning up

will not be equal.

However, we should also expect that if we add the chances of all the numbers occurring together, then the total probability should be equal to 1. But this does not work in this case;

In this case, adding all the probabilities together, we have;

1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/2

= 5(1/12) + 1/2 = 5/12 + 1/2 = 11/12

11/12 is not equal to 1 and thus the probability distribution cannot be correct

4 0

4 years ago

35 POINTS HELP ASAP!!!

Allisa [31]

Answer: 3

Step-by-step explanation:

$3:1$