What is the economically efficient output level? 600 units 800 units 940 units 1160 units?

An Austrian study was completed to determine if untrained sea lions and sea lionesses could follow various experimenter-given cu

Sholpan [36]

Using the z-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, this result shows that Zwerg can correctly follow this type of direction by an experimenter more than 50% of the time.

<h3>What are the hypothesis?</h3>

At the null hypothesis, it is tested if Zwerg cannot correctly follow this type of direction by an experimenter more than 50% of the time, that is:

$H_0: p \leq 0.5$

At the alternative hypothesis, it is tested if Zwerg can correctly follow this type of direction by an experimenter more than 50% of the time, that is:

$H_1: p > 0.5$

<h3>Test statistic</h3>

The <em>test statistic</em> is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

$\overline{p}$ is the sample proportion.

p is the proportion tested at the null hypothesis.

n is the sample size.

For this problem, the parameters are:

$n = 48, \overline{p} = \frac{37}{48} = 0.7708, p = 0.5$

The value of the <em>test statistic</em> is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.7708 - 0.5}{\sqrt{\frac{0.5(0.5)}{48}}}$

$z = 3.75$

Considering a <u>right-tailed test</u>, as we are testing if the proportion is greater than a value, with a <u>significance level of 0.05</u>, the critical value for the z-distribution is $z^{\ast} = 1.645$ .

Since the test statistic is greater than the critical value for the right-tailed test, this result shows that Zwerg can correctly follow this type of direction by an experimenter more than 50% of the time.

To learn more about the z-distribution, you can take a look at brainly.com/question/16313918

3 0

2 years ago

Kevin is an auto mechanic. He spends 2 hours when he replaces the shocks on a car and 2 hours when he replaces the brakes. He wo

nata0808 [166]

Answer:

Kevin should do 15 shock replacements and 6 brake replacements every week to maximize his weekly income.

Step-by-step explanation:

Both jobs take the same amount of time to complete (2 hours), therefore, in order to maximize his income, Kevin should prioritize the job that gives him the greatest income, which is replacing shocks.

If each job takes 2 hours and he works at most 42 hours per week, Kevin can complete 21 jobs per week. He should aim to do as little break replacements as possible and as many shock replacements as possible to maximize income.

If he completes at least 6 break replacements, the remaining jobs should all be shock replacements:

$S=21-6\\S=15$

Kevin should do 15 shock replacements and 6 brake replacements every week to maximize his weekly income.

4 0

3 years ago

Suppose you want to have $400,000 for retirement in 30 years. Your account earns 4% interest.

Ne4ueva [31]

Answer:

if he wants to have $400000 on gis bank in 30 years at 4 %rate then he needs to deposit $15151.5 per month

5 0

4 years ago

The Girl Scouts were selling cookies outside the local grocery store. Each box sold for $4.50.

dusya [7]

If your needing to find the amount just multiply the number of boxes by it’s price. So the answers C.

3 0

3 years ago

8th grade math (will give brainliest to best answer)

algol [13]

Das is disguised pythagorean theorme

the height is one leg
the distance from house is other leg
length of ladder=hyptoonuse

so
remmeber
if legs of right triangle are a and b and hyptonuse is c then
a²+b²=c²

so
legs are 15 and 20
15²+20²=c²
225+400=c²
625=c²
sqrt both sides
25=c

25ft is length of laddder

4 0

3 years ago