Please help with my math I'm begging you

In a recent poll, 778 adults were asked to identify their favorite seat when they fly, and 492 of them chose a window seat. Use

seropon [69]

Answer:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

$p_v =P(Z>7.36)=9.19x10^{-14}$

Since the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .

Step-by-step explanation:

1) Data given and notation

n=778 represent the random sample taken

X=492 represent the people that chose a window seat.

$\hat p=\frac{492}{778}=0.632$ estimated proportion of people that chose a window seat.

$p_o=0.5$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the majority of adults prefer window seats when they fly:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statisitc, 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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.632 -0.5}{\sqrt{\frac{0.5(1-0.5)}{778}}}=7.36$

4) 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 $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(Z>7.36)=9.19x10^{-14}$

5) Conclusion

Since the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .

6 0

3 years ago

Pls answer fast I will give you hundred points

iris [78.8K]

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Question number 2
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3 0

3 years ago

Find two consecutive even integers such that twice the smaller is 16 more than the larger.

qwelly [4]

Answer: smaller: 18 ; larger: 20

Step-by-step explanation:

Let n be the smaller integer. Then the larger integer is n+2. So:

2n=n+2+16

n=18

The smaller integer is 18; the larger one is 20.

6 0

3 years ago

Por la seguridad de su personal y sus clientes una agencia bancaria se instalara una camara de video en un soporte de pared de m

yulyashka [42]

Answer:

16.24°

Step-by-step explanation:

Are you trying to find the angle of depression from the camera to the cashier?

Assume that the camera is 2.24 m high and the cashier is 7.69 m away.

1. Angle of elevation from cashier

tanθ = 2.24/7.69 = 0.2913

θ = arctan(0.2913) = 16.24°

2. Angle of depression from camera

∠ of depression = ∠ of elevation = 16.24°

6 0

3 years ago

The teachers of the two classes want to compare the heights of their students. Which statements about the data sets are accurate

aleksandr82 [10.1K]

This question is incomplete.

Complete question

The sets of data below show the heights, in inches, of students in two different preschool classes.

A) a box plot titled class 1.

The number line goes from 38 to 49. the whiskers range from 39 to 48, and the box ranges from 40 to 43. a line divides the box at 41.

class 1

B) A box plot titled class 2.

The number line goes from 38 to 49. the whiskers range from 38 to 49, and the box ranges from 39 to 42. a line divides the box at 41.

class 2

The teachers of the two classes want to compare the heights of their students. Which statements about the data sets are accurate? Select three options.

• Because the sets are symmetrical, the mean should be used to compare the data sets.

• Because the sets do not contain outliers, the MAD should be used to compare the data sets.

•Because the sets are not symmetrical, the IQR should be used to compare the data sets.

•Because the sets contain outliers, the median should be used to compare the data sets.

•The mean and mode cannot be accurately determined based on the type of data display.

Answer:

• Because the sets are not symmetrical, the IQR should be used to compare the data sets.

• Because the sets contain outliers, the median should be used to compare the data sets.

• The mean and mode cannot be accurately determined based on the type of data display.

Step-by-step explanation:

In the above question, we can see that a teacher intends to compare the heights of students from two different classes.

Because the data sets to be obtained would contain heights from different classes the results obtained would be non symmetrical.

Hence the best statistical analysis to use to solve for or to compare a set of non symmetrical data is known as either the Median or an Interquartile range.

It is important to note that a non symmetrical data would most likely contain or include the presence of outliers.

8 0

3 years ago