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At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

vampirchik [111]

Answer:

a) Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b) The 95% confidence interval would be given by (910.05;959.95)

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) $z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750$

Since is a bilateral test the p value is given by:

$p_v =2*P(Z>2.750)=0.0059$

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

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".

$\bar X=935$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=180$ represent the population standard deviation

n=200 represent the sample size

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

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

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=3278.222$

The sample deviation calculated $s=97.054$

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):

$935-1.96\frac{180}{\sqrt{200}}=910.05$

$935+1.96\frac{180}{\sqrt{200}}=959.95$

So on this case the 95% confidence interval would be given by (910.05;959.95)

c. Use the confidence interval to conduct a hypothesis test. Using α= .05, what is your conclusion?

Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d. What is the p-value?

The statistic is given by:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

If we replace we got:

$z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750$

Since is a bilateral test the p value is given by:

$p_v =2*P(Z>2.750)=0.0059$

So then since the p value is less than the significance we can reject the null hypothesis at 5% of significance.

8 0

3 years ago

What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC?

natulia [17]

Answer:

The endpoint coordinates for the mid-segment are: $(-2,-1)$ and $(1,0)$

Step-by-step explanation:

According to the given diagram, the coordinates of the vertices of $\triangle BCD$ are: $B(-3,1), C(3,3)$ and $D(-1,-3)$

Now, the endpoints of the mid-segment of $\triangle BCD$ which is parallel to $BC$ will be the mid-points of sides $BD$ and $CD$ .

Formula for the coordinate of mid-point : $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$ , where $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are two endpoints.