Could someone please answer this And explain I would really appreciate it thank you so much :)

According to a study by the American Pet Food Dealers association, 63 percent of U.S. households own pets. A report is being pre

Nastasia [14]

Answer

$z=\frac{0.7 -0.63}{\sqrt{\frac{0.63(1-0.63)}{300}}}=2.511$

$p_v =2*P(z>2.511)=0.012$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of households with pets is different from 0.63. This data disagree deom the Pet food association

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

X=210 represent the households with pets

$\hat p=\frac{210}{300}=0.7$ estimated proportion of households with pets

$p_o=0.63$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is 0.63:

Null hypothesis: $p=0.63$

Alternative hypothesis: $p \neq 0.63$

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:

$z=\frac{0.7 -0.63}{\sqrt{\frac{0.63(1-0.63)}{300}}}=2.511$

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:

$p_v =2*P(z>2.511)=0.012$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of households with pets is different from 0.63. This data disagree deom the Pet food association

7 0

3 years ago

Given that 4^-n=32 Find the value of n

ch4aika [34]

Step-by-step explanation:

${4}^{ - n} = 32$

${2}^{ -2 n} = {2}^{5}$

$- 2n = 5$

$n = \frac{ - 5}{2}$

8 0

2 years ago

6z less than or equal to 18

yan [13]

Answer:

z<=3

Step-by-step explanation:

6z<=18

z<=18/6

z<=3

8 0

3 years ago

The width is 7 and the diagonal is 25 what is the height

Korvikt [17]

The height is 24.

We can use Pythagorean’s Theorem: a^2+b^2=c^2

substituting values we get:

a^2+7^2=25^2

Solving we get:
a^2+49=625
a^2=576
a=24.

So the height is 24.

Hope this helps!

8 0

3 years ago

Question 1

xxTIMURxx [149]

Answer:

The teacher randomly divides the class into two groups. One of the groups receives computer-based instruction and the other group receives traditional instruction without computers. After instruction, each student takes a test and the teacher wants to compare the test scores of the two groups.

Step-by-step explanation:

A two sample t-test is used to compared the means of two samples who were exposed to different scenarios and therefore, it helps us to see the difference between these two samples.

In this problem, the teacher is experimenting with a new computer-based instruction. To be able to use the two-sample t-test she would have to have two randomly assigned samples and expose each of these two samples to different scenarios (one would receive a traditional instruction and the other one would receive a computer instruction), and finally she would need to compare the scores in both samples.

Therefore, the correct answer would be b) The teacher randomly divides the class into two groups. One of the groups receives computer-based instruction and the other group receives traditional instruction without computers. After instruction, each student takes a test and the teacher wants to compare the test scores of the two groups.

<u>*Option A is incorrect</u> because the students are only exposed to the lesson using a computer program and therefore they aren't exposed to different scenarios.

<u>*Option C is incorrect</u> because although in this case the teacher uses both traditional and computer methods, there are not two different samples and she only gets the opinion of the students which is not an exact method of getting information on the effectiveness of the computer based method.

3 0

4 years ago