How does translation affect the properties of a two-dimensional figure? please i need help

A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a populat

eimsori [14]

Answer:

$t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$

$p_v =P(t_{15}>2)=0.0320$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.

Step-by-step explanation:

1) Data given and notation

$\bar X=25$ represent the sample mean

$s=2$ represent the standard deviation for the sample

$n=16$ sample size

$\mu_o =24$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

Confidence =0.95 or 95%

$\alpha=0.05$

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is higher than 24, the system of hypothesis would be:

Null hypothesis: $\mu \leq 24$

Alternative hypothesis: $\mu > 24$

We don't know the population deviation, so for this case 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".

Calculate the statistic

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

$t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>2)=0.0320$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.

3 0

3 years ago

Please Help Me!

Rainbow [258]

For the first option, the range is a measure of variability which measures the spread of the data set from the least value to the greatest value, but it does not take into account the variability of the other data values of the data set. The range is easily affected by the presence of outliers (data points that are away from other data points). Thus the range is regarded as a weak measure of variability and is not used when other measures of variability are available. Thus, that the range of the two data sets are equal does not mean that the data sets have the same variability. Therefore, the first option is not the correct answer.

For the second option, the median is not a measure of variability. Thus, that a data set has a greater median than another data set does not mean that the data set would have a greater variability. Therefore, the second option is not the correct answer.
For the third option, the inter-quartile range (IQR) is a better measure of variability than the range because it takes into account more data points than the range. Now, because, the the IQR of Team 2 is less than the IQR of Team 1, this shows that Team 1 have greater variability than Team 2 and thus the conclusion of the coaches are inaccurate. Therefore, the third option is the correct answer.

For the fourth option, the mean absolute deviation, MAD, is a better measure of variability than the IQR because it takes into account all the points of the data set. While IQR measures variability with respect to the median, MAD measures variability with respect to the mean. Because we are told that the data sets are not symmetrical, the median will be a better measure of the center than the mean, thus the IQR will present a better measure of the variability of the data sets. Thus, though the MAD for Team 2 was calculated to be a larger number than the MAD for Team 1, the information can be misleading in arriving at a conclusion on which data set has more variability because the data sets are not symmetrical. Therefore, the fourth option is not the correct answer.

4 0

3 years ago

Read 2 more answers

Can someone plsss help me with this one problem plsss I’m trying to get a 90 and also can you explain how you got your answer

Karo-lina-s [1.5K]

From question,
lacy learned 34 recipes in 17 weeks,
so in 1 week she learns = 34/17 =2 recipes,
hence she learns 40 recipes in
= 40*2
=80 recipes.

7 0

3 years ago

Which place value would you look to compare 225998 and 225988

Jobisdone [24]

225998 225988
I believe it is the tenths place
(the bold digits)

6 0