the average shower uses 12 gallons of water. how many gallons of water would a person use by taking one shower a day for a year?

Jordan is five years old. his mother says to him "if you have two apples and i give you two more apples, how many apples will yo

Mars2501 [29]

Answer should be 2 and he says 4

8 0

4 years ago

a restaurant and ordered 3940 pounds of rice in 220 pound bag about how many 20 pound bags of rice to the change order

Alchen [17]

If my answer is correct,then here..
11-20 LB bags for 220
197-20 LB bags for 3940

3 0

3 years ago

Consider the box-and-whisker plot.

dedylja [7]

each part of the box and whisker is 25% (meaning each quartile) There are 4 quartiles so this makes 100%

So, the first being 25% and this is where the below 15 falls.

So, 25%

3 0

3 years ago

Last year, a comprehensive report stated that 28% of businesses in the northeast of Ohio were considered highly profitable. This

Alik [6]

Answer:

$z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575$

$p_v =2*P(z>1.575)=0.115$

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of businesses were highly profitable is not significantly different from 0.28 or 28%.

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

X=19 represent the businesses were highly profitable

$\hat p=\frac{19}{50}=0.38$ estimated proportion of businesses were highly profitable

$p_o=0.28$ 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 true proportion of businesses were highly profitable is different from 0.28 or no, the system of hypothesis is.:

Null hypothesis: $p=0.28$

Alternative hypothesis: $p \neq 0.28$

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 required we can replace in formula (1) like this:

$z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575$

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 $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>1.575)=0.115$

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of businesses were highly profitable is not significantly different from 0.28 or 28%.

4 0

4 years ago

Can you pls slove the 12 question

quester [9]

So (77 5/7)/17
5/7 = 0.7142
77.7142/17
=4.5714 rupees per metre

7 0

3 years ago