Plz help...................

Devaughn is 13 years younger than Sydney. The sum of their ages is77 What is Sydney's age?

IrinaK [193]

S is sydney's age. devaughn's age is d. d=s-13. The total sum is 77=s+d OR 77=s+s-13. 77=2s-13. add 13 to both sides. 90=2s. sydney's age is 45. to check, substitute sydney's age in. d=45-13=32. 77=45+32=77.

3 0

3 years ago

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Help please !!!!!!!!!!!!

zhenek [66]

I'm pretty sure it is 20 hundreths

8 0

4 years ago

Solve for x -5/4(4/5+16)=x-6

Gala2k [10]

Answer:

x=1991/331

Step-by-step explanation:

6 0

2 years ago

which is not a step in the rock cycle? sediments melt deep beneath earth’s surface. rocks at earth’s surface are broken down int

Oksanka [162]

<span>sediments melt deep beneath earth’s surface.</span>

5 0

3 years ago

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In a survey conducted by a website, employers were asked if they had ever sent an employee home because they were dressed inappr

castortr0y [4]

Answer:

$z=\frac{0.353 -0.333}{\sqrt{\frac{0.333(1-0.333)}{2759}}}=2.23$

$p_v =P(z>2.23)=0.0129$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of more than one-third of employers have sent an employee home to change clothes is higher than 1/3 or 0.333.

Step-by-step explanation:

Data given and notation

n=2759 represent the random sample taken

X=974 represent the people saying that they had sent an employee home for inappropriate attire

$\hat p=\frac{974}{2759}=0.353$ estimated proportion of people saying that they had sent an employee home for inappropriate attire

$p_o=0.333$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that more than one-third of employers have sent an employee home to change clothes.:

Null hypothesis: $p \leq 0.33$

Alternative hypothesis: $p > 0.33$

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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.353 -0.333}{\sqrt{\frac{0.333(1-0.333)}{2759}}}=2.23$

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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed the p value would be:

$p_v =P(z>2.23)=0.0129$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of more than one-third of employers have sent an employee home to change clothes is higher than 1/3 or 0.333.

4 0

3 years ago