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3 years ago

Please help me answer this question

Mathematics

1 answer:

prisoha [69]3 years ago

5 0

Answers:

The sign is 4 ft on a side

The poster is 2 ft on a side

================================================

Explanation:

s = side length of the sign

p = side length of the poster

The poster "has sides measuring 2 ft less than the sides of the square sign", so p = s-2. Whatever the value of 's' is, we subtract off 2 to get the value of p. For example, if the sign has side lengths of s = 10 ft, then p = s-2 = 10-2 = 8 ft is the side length of the poster

We don't know s or p right now, so they are simply placeholders for some numbers.

If s is the side length of the sign, then s*s = s^2 is the area

If p is the side length of the poster, then p*p = p^2 is the area of the poster. We can replace p with s-2 since p = s-2. So we end up with p^2 = (s-2)^2 = s^2 - 4s + 4 after using the FOIL rule

Now subtract the two areas and set that difference equal to 12. Solve for s

(area of sign) - (area of poster) = 12

(s^2) - (s^2 - 4s + 4) = 12

s^2 - s^2 + 4s - 4 = 12

4s - 4 = 12

4s - 4+4 = 12+4

4s = 16

4s/4 = 16/4

s = 4

The sign's side length is 4, so its area is 4^2 = 16

If s = 4, then p is

p = s-2

p = 4-2

p = 2

The poster's side length is 2, so its area is 2^2 = 4

Subtract the areas: 16 - 4 = 12

The answer is confirmed

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Answer:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$

$p_v =P(z>1.82)=0.034$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

$\hat p=\frac{120}{300}=0.4$ estimated proportion of college students that have a smart phone

$p_o=0.35$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

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 the proportion is >0.35.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

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:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$

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 $\alpha=0.1$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>1.82)=0.034$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

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