What is the median of the data set?

Ect the correct answer.<br> Which of the following graphs represents the equation below?

Masja [62]

Seems like none of them +4 unit up yaxis

7 0

3 years ago

Please help me with this homework

klio [65]

Answer:

The answer is B

Step-by-step explanation:

8 0

3 years ago

Gummy bears: red or yellow? Chance (Winter 2010) presented a lesson in hypothesis testing carried out by medical students in a b

xxMikexx [17]

Answer:

$z=\frac{0.802 -0.5}{\sqrt{\frac{0.5(1-0.5)}{121}}}=6.64$

$p_v =2*P(z>6.64)=3.14x10^{-11}$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of students who will correctly identify the color is different than 0.5

Step-by-step explanation:

Data given and notation

n=121 represent the random sample taken

X=97 represent the people who identify the color of the gummy bear

$\hat p=\frac{97}{121}=0.802$ estimated proportion of people who identify the color of the gummy bear

$p_o=0.5$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Part a

For this case the parameter that we want to test is 0.5 for the proportion of the population of students will correctly identify the color

Part b: Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that there is no relationship between color and gummy bear flavor (population proportion different from 0.5).:

Null hypothesis: $p=0.5$

Alternative hypothesis: $p \neq 0.5$

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 $p_o$ .

Part c: Calculate the statistic

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

$z=\frac{0.802 -0.5}{\sqrt{\frac{0.5(1-0.5)}{121}}}=6.64$

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 right tailed test the p value would be:

$p_v =2*P(z>6.64)=3.14x10^{-11}$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of students who will correctly identify the color is different than 0.5

8 0

4 years ago

At the farmers market a farmer has bunches of radishes for sale. The

stealth61 [152]

Answer:

The probability a random selected radish bunch weighs between 5 and 6.5 ounces is 0.8185

Step-by-step explanation:

The weight of the radish bunches is normally distributed with a mean of 6 ounces and a standard deviation of 0.5 ounces

Mean = $\mu = 6$