Algebra 1 taking stock

Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n

Oksanka [162]

Answer:

$z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097$

$p_v =P(Z>2.097)=0.018$

If we compare the p value obtained and 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 students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

$\hat p=\frac{209}{377}=0.554$ estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

$p_o=0.5$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097$

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

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

$p_v =P(Z>2.097)=0.018$

If we compare the p value obtained and 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 students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

6 0

2 years ago

PLEEEEEEEAAAAAAASSSSSSSSSSEEEEEEEEEEE help me!!!!!!!!!!

defon

Answer:

$\boxed{\sf \ \ \ \dfrac{18}{(x-7)(x+12)(x+30)} \ \ \ }$

Step-by-step explanation:

hello,

first of all we will study the quadratic expressions

we can write that, (the different answers provide good clues :-) )

$x^2+5x-84=(x-7)(x+12)$

and

$x^2+23x-210=(x-7)(x+30)$

so first of all as we cannot divide by 0 we need to take x different from 7, -12 and -30 and then we can write

$\dfrac{1}{x^2+5x-48}-\dfrac{1}{x^2+23x-210}=\dfrac{1}{(x-7)(x+12)}-\dfrac{1}{(x-7)(x+30)}\\\\=\dfrac{(x+30) -(x+12)}{(x-7)(x+12)(x+30)}=\dfrac{x+30-x-12}{(x-7)(x+12)(x+30)}\\\\=\dfrac{18}{(x-7)(x+12)(x+30)}$