Question

A new disease has annihilating amphibian populations at unprecedented levels. Researchers has been trying to develop some treatment that can counteract the disease and save some of the amphibian populations. To test which of two potential solutions is more effective, they take a sample of frogs and randomly divide them into two groups. After being treated, 256 of the 400 frogs in group 1 ended up surviving exposure to the pathogen; 300 of the 482 frogs in the second group survived. For the following questions, calculate your difference in the order: group 1 - group 2. The standard error of the difference was 0.033.

Calculate the Test statistic:

Calculate the p-value:

Answer 1

Answer:

$z=\frac{0.64-0.622}{\sqrt{0.630(1-0.630)(\frac{1}{400}+\frac{1}{482})}}=0.551$  

$p_v =P(Z>0.551)=0.291$  

Step-by-step explanation:

Data given

$X_{1}=256$ represent the number of frogs in group 1 ended up surviving exposure to the pathogen

$X_{2}=300$ represent the number of frogs in the second group survived

$n_{1}=400$ sample of frogs in group 1 ended up surviving exposure to the pathogen

$n_{2}=482$ sample of frogs in the second group survived

$p_{1}=\frac{256}{400}=0.64$ represent the proportion of frogs in group 1 ended up surviving exposure to the pathogen

$p_{2}=\frac{300}{482}=0.622$ represent the proportion of frogs in the second group survived

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the proportion of survived frogs is higher with the treatment , so the system of hypothesis would be:  

Null hypothesis:$p_{1} \leq p_{2}$  

Alternative hypothesis:$p_{1} > p_{2}$  

The statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{256+300}{400+482}=0.630$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.64-0.622}{\sqrt{0.630(1-0.630)(\frac{1}{400}+\frac{1}{482})}}=0.551$  

Statistical decision

We are assuming a a one right tailed test the p value would be:  

$p_v =P(Z>0.551)=0.291$  

Answer 2

From the hypothesis test, we have that:

• The test statistic is z = 0.53.
• The p-value is 0.5962.

At the null hypothesis, it is tested if the proportions are the same, that is:

$H_0: p_1 - p_2 = 0$

At the alternative hypothesis, it is tested if they are different, that is:

$H_1: p_1 - p_2 \neq 0$

The test statistic is:

$z = \frac{\overline{p} - p}{s}$

In which $p = 0$ is the value tested at the null hypothesis.

The estimate of the difference is:

$\overline{p} = p_1 - p_2$

$\overline{p} = \frac{256}{400} - \frac{300}{482}$

$\overline{p} = 0.0176$

The standard error is $s = 0.033$, and hence, the test statistic is:

$z = \frac{\overline{p} - p}{s}$

$z = \frac{0.0176 - 0}{0.033}$

$z = 0.53$

We are testing if the mean is different of a value, hence it is a two-tailed test, and the p-value is P(|z| > 0.53), which is 2 multiplied by the p-value of z = -0.53.

Looking at the z-table, z = -0.53 has a p-value of 0.2981

2 x 0.2981 = 0.5962.

The p-value of the test is of 0.5962.

A similar problem is given at https://brainly.com/question/16162795