A _______ is a variable, an integer or a product of variables and<br> integers.

dimaraw [331]

78 gallons is the same as 312 quarts

8 0

3 years ago

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

Need help asap 15mins best

devlian [24]

Answer: