What is x for these qs<br> 2x+7=3 <br> 1/2x-4=5<br> 10-2x=5

To plan the budget for next year a college must update its estimate of the proportion of next year's freshmen class that will ne

Naily [24]

Answer:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ 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 applicants who request financial aid is significantly higher than 0.35

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

$\hat p=\frac{67}{150}=0.447$ estimated proportion of applicants who request financial aid

$p_o=0.35$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 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:

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

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 next step would be calculate the p value for this test.

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

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ 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 applicants who request financial aid is significantly higher than 0.35

5 0

3 years ago

a 4.50 g coin of copper absorbed 54 calories of heat. what was the final temperature of the copper if the initial temperature wa

I am Lyosha [343]

The final temperature of copper is 155.43°C.

Step-by-step explanation:

The formula to apply is Q=m*Δt*c where ;

Q=heat energy, m=mass, c=specific heat capacity and Δt= change in temperature

Given that Q=54 calories , m=4.50 g and c=0.092 cal/g°C

In this question find Δt then ,the final temperature.

Δt=Q/m*c

Δt=54/4.50*0.092 =130.43°C