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Elza [17]
3 years ago
9

The repetition of a previous study using a different sample or population to verify or refute the original findings is referred

to as:_______
A. Verification
B. Replication
C. Validation
Mathematics
1 answer:
zavuch27 [327]3 years ago
4 0
The answer is B. Replication
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What is the length if a right triangle with a leg that measures 16 and a hypotenuse of 20
Genrish500 [490]

Answer:

It would be 12.

Step-by-step explanation:


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3 years ago
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Azul spun a tri-color spinner twice. What is the probability that the spinner landed on the colors blue and green in any order?
guapka [62]

Answer:

The Probability that the spinner landed on the colors blue and green in any order = 2/9

Step-by-step explanation:

Given - Azul spun a tri-color spinner twice.

To find - What is the probability that the spinner landed on the colors blue and green in any order?

Solution -

Given that,

A tri-color spinner spun twice

So,

The Sample Space, S = {BB, BG, BY, GB, GG, GY, YB, YG, YY}

⇒n(S) = 9

Now,

Let A be an event that the spinner landed on the colors blue and green

So,

A = {BG, GB}

⇒n(A) = 2

Now,

Probability that the spinner landed on the colors blue and green in any order = n(A) ÷ n(S)

         = 2 ÷ 9

∴ we get

The Probability that the spinner landed on the colors blue and green in any order = 2/9

8 0
3 years ago
Reduce to simplest form.<br> -7/12 + 3/8<br> (They are fractions)
denpristay [2]

Answer:

-5/24

Step-by-step explanation:

You first change -7/12 to -14/24, and 3/8 to 9/24 so they have the same denominator. You then do -14/24 + 9/24. That will be the same as the opposite of the answer to 14/24 - 9/24.

14/24 - 9/24 = 5/24

-5/24

4 0
3 years ago
The following data lists the ages of a random selection of actresses when they won an award in the category of Best​ Actress, al
Valentin [98]

Answer:

a) p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

b) The 99% confidence interval would be given by (-21.469;2.069)

c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two

Step-by-step explanation:

Part a

Let put some notation  

x=actor's age , y = actress's age

x: 58 41 36 36 34 33 48 37 37 43

y: 26 27 34 26 35 29 23 42 30 34

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \geq 0

Alternative hypothesis: \mu_y -\mu_x

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679

The next step is calculate the degrees of freedom given by:

df=n-1=10-1=9

Now we can calculate the p value, since we have a left tailed test the p value is given by:

p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

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

\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that t_{\alpha/2}=3.25  

Now we have everything in order to replace into formula (1):  

-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469  

-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069  

So on this case the 99% confidence interval would be given by (-21.469;2.069)

Part c

We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.

8 0
3 years ago
Katie grew 26 flowers with 13 seed packets. With 55 seed packets, how many total flowers can Katie have in her garden?
Leokris [45]
110 flowers in her garden.
3 0
3 years ago
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