1st person to answer gets brainliest!!!

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

Ugh geez can ANYONE help with 2 questions????????????

just olya [345]

What's the 2 question

5 0

3 years ago

I need the answers to both. Will give brainiest!

miskamm [114]

Answer:

1. Number 1 and 2 and 4 is a function, 2. number 1 is a function

Step-by-step explanation:

1)To know if it's a function or not run vertical lines through multiple places of the graph. If it is a function every single time you do the vertical line test it should only go over the line once. If you do the vertical line test on 3 you will see that it went over the line on the graph, so we know not a function. Graphs 1, 2, and 4m however, are different, when you do the vertical line test on those graphs it only goes over them once.

2) Choice (1) is a function because when drawing vertical lines through the graph it only goes over one.

Choice (2) is not a function because when drawing vertical lines through the graph it covers two points on the graph.

Choice (3) is not a function because when drawing vertical lines through the graph it goes over multiple points.

Choice (4) is not a function because when a vertical line is drawn, it goes over more than one point on the graph.

The vertical test is a way to determine if it is a function.

When looking at a table functions are one-to-one and many-to-one

Non-functions are one-to-many and many-to-many

Hoped this helped you : )

5 0

3 years ago

What is 4 1/3= 1/3 x _/_ = 4 _/_

Oxana [17]

4 1/3 = 13/3 = 1/3 x 13/1 = 4 1/3

3 0

3 years ago

What is 10 and 2/3 divided by 8?

postnew [5]

10 and 2/3 = 32/3
32/3 x 8 = 256/3
256/3 = 85.33333333333 repeating

4 0

4 years ago