Which describes the relationship between the number of tickets sold and the profit?

Statements about regression

galben [10]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

It is true because the square of the standard error of its estimate was its total square error divided only by the degree of freedom.
It is true because Its coefficient with Standardized Regression, beta, will have the same value as r, the approximate similarity.
It is false because Its slope b, of its equation of regression, will have the same value as r, the projected correlation.

3 0

4 years ago

Asphere has a radius of 27 inches. A horizontal plane passes through the center of the sphere.

Westkost [7]

9514 1404 393

Answer:

circle of radius 27 inches

Step-by-step explanation:

Anywhere a plane cuts a sphere, the cross section is a circle. When the plane includes the center of the sphere, the circle has the same radius the sphere has.

The cross section is a circle of radius 27 inches.

7 0

3 years ago

Explain why algebra problems involve letters

qaws [65]

To challenge you to see if you can do it

6 0

3 years ago

Meg colored 3/8 of the tiles for a class art project. Max colored 4/15 of the tiles. What fraction of the tiles did Meg and Max

borishaifa [10]

Answer:

I need help

Step-by-step explanation:

I need help I need help

4 0

3 years ago

Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test

slamgirl [31]

Answer:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$

The 95% confidence interval would be given by (96.625;100.915)

a) $df=n-1= 19-1= 18$

b) $p_v =2*P(t_{18}$

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

c) The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and data given

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

$\bar X=98.77$ represent the sample mean

$s=4.77$ represent the sample standard deviation

n=19 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if we have significant difference on the mean of 100, the system of hypothesis would be:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can calculate the Standard error for the mean like this:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

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

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

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

$98.77-1.96\frac{4.77}{\sqrt{19}}=96.625$

$98.77+1.96\frac{4.77}{\sqrt{19}}=100.915$

So on this case the 95% confidence interval would be given by (96.625;100.915)

Part a

The degree of freedom are given by:

$df=n-1= 19-1= 18$

Part b

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$

Then since is a two sided test the p value would be:

$p_v =2*P(t_{18}$

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Part c

If the system of hypothesis on this case are:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu > 100$

The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

7 0

3 years ago