PLEASE HELP MEEE!!!!!!!!!

In an experiment, colored letters flash briefly on a computer screen. On each trial, participants are asked to report the letter

katen-ka-za [31]

The alternatives or options aren't given.

Answer:

Kindly check explanation

Step-by-step explanation:

The dependent variable in an experiment refers to the va ibale which we intend to measure based on the outcome or values of one or more predictor variables, this variable which we intend to measure is the dependent variable while the the variables which causes a change or upon which the dependent variable relies are the independent variable. Hence, from the scenario described above, the possible dependent variables are :

The letters in their correct colors and the confidence in response Given.

8 0

3 years ago

In a survey of 1000 women age 22 to 35 who work full time, 540 indicated that they would be willing to give up some personal in

morpeh [17]

Answer:

$z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.530$

$p_v =P(Z>2.530)=0.0057$

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 women indicated that they would be willing to give up some personal in order to make more money is significantly higher than 0.5.

The 95% confidence interval would be given by (0.509;0.571)

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

X=540 represent the women indicated that they would be willing to give up some personal in order to make more money

$\hat p=\frac{540}{1000}=0.54$ estimated proportion of women indicated that they would be willing to give up some personal in order to make more money

$p_o=0.5$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the majority of woman age 22 to 35 who work full-time would be willing to give up some personal time for more money.:

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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.530$

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

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

$p_v =P(Z>2.530)=0.0057$

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 women indicated that they would be willing to give up some personal in order to make more money is significantly higher than 0.5.

Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.54 - 1.96\sqrt{\frac{0.54(1-0.54)}{1000}}=0.509$

$0.54 + 1.96\sqrt{\frac{0.54(1-0.54)}{1000}}=0.571$

The 95% confidence interval would be given by (0.509;0.571)

5 0

4 years ago

Alice purchased 4 1⁄2 kilograms of olive oil for $27. What is the price per kilogram

Bas_tet [7]

Do 27 divided by 4 1/2 or 4.5. That will be your answer. Let me know if this helps!

5 0

3 years ago

Read 2 more answers

Which is the most convenient placement of an isosceles triangle in the coordinate plane for finding the side lengths?

Usimov [2.4K]

Answer:

The bottom left

Step-by-step explanation:

5 0

3 years ago

How do you start this

lidiya [134]

You have two equations.
since the second is already isolated, sub in x-4 for every y in equation 1 so that
$x^{2} - 4 [(x-4)^{2}] =16
$
expand, collect like terms, factor to find x, then plug x value back into original equation to find y

3 0

3 years ago

Read 2 more answers