Plz help fast will mark brainliest!!!!

Can somebody please help me omg i’m so dumb

Makovka662 [10]

Answer:

A

Step-by-step explanation:

7 0

2 years ago

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In a recent year, a poll asked 2362 random adult citizens of a large country how they rated economic conditions. In the poll,

Harman [31]

Answer:

a) The 99% confidence interval is given by (0.198;0.242).

b) Based on the p value obtained and using the significance level assumed $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

c) $\alpha=0.01$

Step-by-step explanation:

<em>Data given and notation </em>

n=2362 represent the random sample taken

X represent the people who says that they would watch one of the television shows.

$\hat p=\frac{X}{n}=0.22$ estimated proportion of people rated as Excellent/Good economic conditions.

$p_o=0.24$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest) <em> </em>

<em>Concepts and formulas to use </em>

We need to conduct a hypothesis in order to test the claim that 24% of people are rated with good economic conditions:

Null hypothesis: $p=0.24$

Alternative hypothesis: $p \neq 0.24$

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 $p_o$ .

Part a: Test the hypothesis

<em>Check for the assumptions that he sample must satisfy in order to apply the test </em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np = 2362x0.22=519.64>10 and n(1-p)=2364*(1-0.22)=1843.92>10

Condition satisfied.

<em>Calculate the statistic</em>

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.22 -0.24}{\sqrt{\frac{0.24(1-0.24)}{2362}}}=-2.28$

The confidence interval would be given by:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The critical value using $\alpha=0.01$ and $\alpha/2 =0.005$ would be $z_{\alpha/2}=2.58$ . Replacing the values given we have:

$0.22 - (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.198$

$0.22 + (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.242$

So the 99% confidence interval is given by (0.198;0.242).

Part b

<em>Statistical decision </em>

P value method or p value approach . "This method consists on 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 is $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

So based on the p value obtained and using the significance level assumed $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

Part c

The confidence level assumed was 99%, so then the signficance is given by $\alpha=1-confidence=1-0.99=0.01$

6 0

2 years ago

Find the value of x. Then find the m∠S and m∠T

Sophie [7]

Answer:

1

Step-by-step explanation:

<h2>Given, y = b m x m</h2><h2>y' = -b m x (m+1)</h2><h2>At any point (x1,y1)</h2><h2 /><h2>Equation of tangent is given by </h2><h2>y - y1 </h2><h2>------- = -b m x1 -(m+1)</h2><h2>x - x1</h2><h2>Y intercept = - m</h2><h2>X intercept = (m-1) × 1</h2><h2> m </h2><h2>Area bounced = 1 (m - 1)×1</h2><h2> ---- ---------------------- x m</h2><h2> 2 m</h2><h2>For area to be constant, m = 1</h2>

4 0

3 years ago

Find the mean of the following data set: 9, 8, 11, 9, 16, 15, 13, 6, 15, 18

FromTheMoon [43]

Answer:

The mean is 12

4 0

3 years ago

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Daniel believes that people perform better in the barbell curl, on average, if they are encouraged by a coach. Be recruited 29 s

miv72 [106K]

Answer:

Step-by-step explanation:

Hello!

a)

One of the advantages of assigning the units randomly to the different treatments is to avoid selection bias. An example of selection bias would be selecting the best subjects at the barbell curl to the "encouraged" group and the worst ones to the "not encouraged" group.

Randomization also guarantees independence between the experimental units and treatment groups.

b) (see boxplots in attachment)

Encouraged group:

Box:

Q₁≅ 10

Q₂≅ 30

Q₃= 50

IQD= 50 - 10 = 40

The box seems symmetric, the distance between Q₁-Q₂ and Q₂-Q₃ is the same and the median is exactly in the middle.

Whiskers:

The left whisker is shorter than the right whisker, which is longer.

If you were to look only at the box, the distribution seems symmetrical but adding the whiskers there is a pronounced right asymmetry.

Min value ≅ 5

Max value ≅ 75

Range= 75 - 5 = 70

Not encouraged group:

Box:

Q₁≅ 10

Q₂≅ 20

Q₃= 35

IQD= 35 - 10 = 25

The box shows a little skewness to the right (the 2nd Quartile is closer to the first one than the 3rd) and is overall shorter compared to the first group.

Whiskers:

The left whisker is shorter than the right whisker, so in general, this distribution is also right-skewed.

Min value ≅ 0

Max value ≅ 60

Range= 60 - 0 = 60

Compared with the "encouraged" group, the IQD and the Range are shorter, which means that the "not encouraged" group shows less dispersion.

c) and d)

Daniel claims that people perform better in the barbell curl, on average, when being encouraged.

He divided the 29 subjects into two groups and measured their performance, then the study variables are:

X₁: Performance in the barber curl of a subject that was encouraged by his coach.

X₂: Performance in the barber curl of a subject that did not receive encouragement.

Logically, since Daniel wants to study the performance of both groups on average, you'd think that the parameters of interest will be the population means. But to study the average you need the population to have a normal distribution and looking at those boxplots, none of them is near the normal distribution and the size of both samples isn't big enough for an approximation.

The best option is to conduct a non-parametric analysis, for example, the Mann Whitney U-test and instead of comparing the population means, you'll compare the population medians: θ₁ vs. θ₂ ⇒ In the hypothesis of this test, you'll state that both samples come from the same population and if both have the same median, then each observation of the first sample x₁i will have an equal probability of being smaller or greater than each observation of the second sample x₂i (probability 0.5)

Depending of the statistics course, the hypotheses may change, I'm used at working directly with the medians, if they are equal, it will mean that the top 50% and bottom 50% of each population will be the same, which is the same as saying that P(x₁i > x₂i)= 0.5 vs. P(x₁i > x₂i)≠ 0.5

The conditions for the Mann-Whitney U-test are:

1) All observations on both groups should be independent. Check

2) The variables should be continuous or at least ordinal. Check

3) Both variables have the same distribution under the null hypothesis. Check (Looking at the boxplots, both are right-skewed)

4) Under the alternative hypothesis, the values of one of the populations exceed the other. In this case: H₁: θ₁ > θ₂

e)

H₀: θ₁ ≤ θ₂ vs. H₁: θ₁ > θ₂

Using the p-value approach, the decision rule is always the same:

If p-value ≤ α ⇒ Reject the null hypothesis.

If p-value > α ⇒ Do not reject the null hypothesis.

p-value: 0.107 > α: 0.05 ⇒ The decision is to not reject the null hypothesis.

At a 5% significance level, there is no significant evidence to reject the null hypothesis. Both samples seem to be from the same population. He can conclude that encouraging the subjects doesn't change significantly their performance in the barbell curl.

I hope you have a SUPER day!

4 0

3 years ago