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SVETLANKA909090 [29]

3 years ago

12

Suppose that x has a binomial distribution with n = 201 and p = 0.45. (Round np and n(1-p) answers to 2 decimal places. Round yo

ur answers to 4 decimal places. Round z values to 2 decimal places. Round the intermediate value (σ) to 4 decimal places.) (a) Show that the normal approximation to the binomial can appropriately be used to calculate probabilities about x.

1 answer:

ycow [4]3 years ago

3 0

Answer:

a) It can be used because np and n(1-p) are both greater than 5.

Step-by-step explanation:

Binomial distribution and approximation to the normal:

The binomial distribution has two parameters:

n, which is the number of trials.

p, which is the probability of a success on a single trial.

If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.

In this question:

$n = 201, p = 0.45$

So, lets verify the conditions:

np = 201*0.45 = 90.45 > 5

n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5

Since both np and n(1-p) are greater than 5, the approximation can be used.

You might be interested in

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

Just the answer please.

Bond [772]

Answer:

No, they are not similar.

5 0

3 years ago

A screening test for a disease shows a positive result in 91% of all cases when the disease is actually present and in 7% of all

Jobisdone [24]

Answer:

0.0713 = 7.13% probability that the result is positive.

Step-by-step explanation:

Probability of a positive test:

91% of 1/650(has the disease).

7% of 1 - 1/650 = 649/650(does not have the disease). So

$p = 0.91\frac{1}{650} + 0.07\frac{649}{650} = \frac{0.91 + 0.07*649}{650} = 0.0713$

0.0713 = 7.13% probability that the result is positive.

4 0

2 years ago

Find the product.<br> (x+2)(x−2)=

kaheart [24]

Answer:

x² - 4

Step-by-step explanation:

Step 1: Write expression

(x + 2)(x - 2)

Step 2: FOIL

x² + 2x - 2x - 4

Step 3: Combine like terms

x² - 4

8 0

3 years ago

Read 2 more answers

The table lists how poverty level income cutoffs (in dollars) for a family of four have changed over time. Use the midpoint for

frez [133]

The poverty level cutoff in 1987 to the nearest dollar was $10787.

<h3>How to find a midpoint?</h3>

The midpoint as the point that divides the line segment exactly in half having two equal segments. Therefore, the midpoint presents the same distance between the endpoints for the line segment. The midpoint formula is: $\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)$ .

For solving this exercise, first you need plot the points in a chart. See the image.

Your question asks to approximate the poverty level cutoff in 1987 to the nearest dollar using the midpoint formula. Note that the year 1987 is between 1980 and 1990, thus you should apply the midpoint formula from data for this year (1987).

$\mathrm{Midpoint\:of\:}\left(1980,8429),\:\left(1990,13145):\quad \left(\frac{1990+1980}{2},\:\:\frac{13145+8429}{2}\right)\\\\ \\$

$\mathrm{Midpoint\:of\:}\left(1980,8429)\:=\left(\frac{1990+1987}{2},\:\frac{13145+3843}{2}\right)\\ \\ =\left(\frac{3977}{2},\:\frac{21574}{2}\right)$

$\mathrm{Midpoint\:of\:}\left(1980,8429)\:=\=(\frac{3977}{2},10787)$

The answer for your question will be the value that you calculated for the y-coordinate. Then, the poverty level cutoff in 1987 to the nearest dollar was $10787.

Read more about the midpoint segment here:

brainly.com/question/11408596

7 0

2 years ago