I NEED HELP WITH THIS QUESTION/PROBLEM

Let n1equals50, Upper X 1equals30, n2equals50, and Upper X 2equals10. Complete parts (a) and (b) below. a. At the 0.05 leve

Viefleur [7K]

Answer:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

$z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082$

$p_v =2*P(Z>4.082)=4.46x10^{-5}$

So the p value is a very low value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.

Step-by-step explanation:

1) Data given and notation

$X_{1}=30$ represent the number of people with a characteristic in 1

$X_{2}=10$ represent the number of people with a characteristic in 2

$n_{1}=50$ sample of 1 selected

$n_{2}=50$ sample of 2 selected

$p_{1}=\frac{30}{50}=0.6$ represent the proportion of people with a characteristic in 1

$p_{2}=\frac{10}{50}=0.2$ represent the proportion of people with a characteristic in 2

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion 1 is different from proportion 2 , the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and 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 $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{30+10}{50+50}=0.4$

3) Calculate the statistic

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

$z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082$

4) Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test.

Since is a two sided test the p value would be:

$p_v =2*P(Z>4.082)=4.46x10^{-5}$

So the p value is a very low value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.

3 0

3 years ago

A system of inequalities is graphed below. The variable x represents the number of cans someone is buying, and y represents the

8090 [49]

We have that
We will analyze each case to verify the answer
see the attached figure

Point A) 7 cans and 140 bottles------------> (x,y)=(7,140)----> is not solution
Point B) 12 cans and 40 bottles------------> (x,y)=(12,40)----> is not solution
Point C) 8 cans and 120 bottles------------> (x,y)=(8,120)----> is a solution
Point D) 9 cans and 70 bottles------------> (x,y)=(9,70)-------> is a solution

the answer is
the solutions are
8 cans and 120 bottles
9 cans and 70 bottles

6 0

3 years ago

Janet completed this entire column of math homework in 10 minutes. How many seconds per question did Janet spend?

puteri [66]

Answer:

Assuming that Janet spent an equal amount of time on each question, he would have spent 75 seconds on each question.

Step-by-step explanation:

10 minutes = 600 seconds

600 seconds / 8 questions = 75 seconds.

5 0

3 years ago

Which is the quotient for 28/8? A. 0.25 B. 0.35 C. 3.25 D. 3.5

Marysya12 [62]

Answer:

D. 3.5

Step-by-step explanation:

28 divided by 8 = 3.5

4 0

2 years ago

The list below shows the Highlight in inches of the player on a man college basketball team

boyakko [2]

We have the data set:

{79, 83, 82, 78, 79, 77, 80, 73, 82, 72}

This data set has 10 items.

To complete the data summary is helpful to sort the data:

{72, 73, 77, 78, 79, 79, 80, 82, 82, 83}

The minimum value is 72.

The quartile divides the data set in 4 parts.

The first quartile is the value for which 25% of the data is below.

The second quartile is the value for which 50% of the data is below. The second quartile is equivalent to the median.

The third quartile is the value for which 75% of the data is below.

In this data set, as its size is a even number, the quartile fall between two positions, so we will calculate the average of this numbers.

The first quartile will fall in the position 0.25*10=2.5. Then we can calculate the 1st quartile as the average between the 2nd and 3rd number of the data set.

This values are 73 (2nd position) and 77 (3rd position), so the average is:

$Q_1=\frac{73+77}{2}=75$

We can repeat this with the other quartiles:

$\begin{gathered} Q_2=M=\frac{79+79}{2}=79 \\ Q_3=\frac{82+82}{2}=82 \end{gathered}$

Then:

The first quartile is 75.

The median (second quartile) is 79.

The third quartile is 82.

The maximum value is 83.

7 0

1 year ago