PLEASE HELP!!!! FIND THE MEASURE OF <2

grin007 [14]

Lol u will not believe if i tell u tht my sister have banana phobia... if u show banana to her she will shout loud and if u take it near to her she will cry loud

3 0

3 years ago

A study is planned to compare the proportion of men who dislike anchovies with the proportion of women who dislike anchovies. Th

andrey2020 [161]

Answer:

The requirements are satisifed since we have the proportion estimated, the sample sizes provided, we assume random sampling for the selection of the data and the distribution for the difference of proportions can be considered as normal

$z=\frac{0.67-0.84}{\sqrt{0.755(1-0.755)(\frac{1}{41}+\frac{1}{56})}}=-1.923$

Step-by-step explanation:

The requirements are satisifed since we have the proportion estimated, the sample sizes provided, we assume random sampling for the selection of the data and the distribution for the difference of proportions can be considered as normal

Data given and notation

$n_{M}=41$ sample of male selected

$n_{W}=56$ sample of demale selected

$p_{M}=0.57$ represent the proportion of men who dislike anchovies

$p_{WCB}=0.84$ represent the proportion of women who dislike anchovies

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

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

Null hypothesis: $p_{M} = p_{W}$

Alternative hypothesis: $p_{M} \neq p_{W}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$ (1)

Where $\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{0.67+0.84}{2}=0.755$

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

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

$z=\frac{0.67-0.84}{\sqrt{0.755(1-0.755)(\frac{1}{41}+\frac{1}{56})}}=-1.923$

4 0

4 years ago

Im about to cry <br> please help

enot [183]

Answer:

Try D

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Plzz help ... I'm thinking it's either c or d

iren [92.7K]

Looking at the slope of the "best-fit" line, you can see that for every two hours of homework, 15 more points are scored. (compare t=0 to t=2).

For t=5, the line seems to go through 47.5. So for t=7, which is two hours more, you'd expect a score of 47.5+15 = 62.5.

That answer is in the list, so I'd choose that.

In reality, you would never be able to have any score you like just by putting more hours in, so this can never be a straight line forever...

5 0

3 years ago

Read 2 more answers

The mean one-way commute to work in Chowchilla is 7 minutes. The standard deviation is 2.4 minutes, and the population is normal

adell [148]

Answer:

The answer is below

Step-by-step explanation:

Given that:

The mean (μ) one-way commute to work in Chowchilla is 7 minutes. The standard deviation (σ) is 2.4 minutes.

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

$z=\frac{x-\mu}{\sigma}$

a) For x < 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

From normal distribution table, P(x < 2) = P(z < -2.08) = 0.0188 = 1.88%

b) For x = 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

For x = 11:

$z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67$

From normal distribution table, P(2 < x < 11) = P(-2.08 < z < 1.67 ) = P(z < 1.67) - P(z < -2.08) = 0.9525 - 0.0188 = 0.9337

c) For x = 11:

$z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67$

From normal distribution table, P(x < 11) = P(z < 1.67) = 0.9525

d) For x = 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

For x = 5:

$z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83$

From normal distribution table, P(2 < x < 5) = P(-2.08 < z < -0.83 ) = P(z < -0.83) - P(z < -2.08) = 0.2033- 0.0188 = 0.1845

e) For x = 5:

$z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83$

From normal distribution table, P(x < 5) = P(z < -0.83) = 0.2033

8 0

4 years ago