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4 years ago

12

You want to estimate the number of students in a junior high school who ride the school bus. Which sample is unbiased? PLEASE HE

LP!!!

1 answer:

Tanzania [10]4 years ago

8 0

I would say C. This is because these students are the first 50 who are on the roster & has nothing to do with either side.

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A data set of 27 different numbers has a mean of 33 and a median of 33. A new data set is

Svetllana [295]

Answer:

Option D.

Step-by-step explanation:

Suppose that we have a set of N values:

{x₁, x₂, ..., xₙ}

The median is the value in the middle, and the mean is calculated as:

$M = \frac{(x_1 + x_2 + x_3 ... + x_n)}{N}$

Because here we have "the median" then we will assume that N is odd, and there is only one median, the value "k" (if N was even, the median would be the mean of the two middle values, that case is really similar to the case where N is odd, so solving only one of the cases is enough)

Now we add 7 to all the values greater than the median

We subtract 7 to all values smaller than the median.

Then the median remains unchanged (because we did not add nor subtract anything to the median).

The new mean will be:

$M' = \frac{(x_1 - 7) + (x_2 - 7) + ... + (x_{k}) + ... + (x_n + 7)}{N} = \frac{(x_1 + x_2 + ... + x_n) + (-7)*(n/2 - 0.5) + 7*(n/2 - 0.5)}{N} = \frac{(x_1 + x_2 + ... + x_n)}{N} = M$

So the mean does not change.

Because the mean is computed as the sum of the numbers divided by N, and the mean does not change, and N does not change, then the sum of the numbers does not change.

Finally, the standard deviation is computed as:

$SD = \sqrt{\frac{(x_1 - M)^2 + ... + (xn - M)^2}{N} }$

Because M does not change, if we add or subtract numbers to some of the values, the standard deviation will change (Because all the terms are squared, so the added and subtracted sevens don't cancel like in the previous cases)

Then the only value that does not have the same value in both the original and new data sets is the standard deviation.

6 0

3 years ago

In 2007, the U.S. population was 302 million people. In 2012, it was 314 million. What was the rate of population change per yea

allsm [11]

Answer:

314/302=1.039735 growth in US population in 5 years

(1.039735)^1/5=0.782% average growth in US population per year.

6 0

4 years ago

What is the value of x? <br> X + (-7) =2<br> Enter your answer in the box in simplest form.

Makovka662 [10]

move -7 to the right side of the equal sign which becomes +7

x=2+7

7 + 2 is 9

=9

5 0

3 years ago

Read 2 more answers

A new sales training program has been instituted at a rent-to-own company. Prior to the training, 10 employees were tested on th

Lunna [17]

Answer:

C. H0: µD = 0, HA: µD <0

Step-by-step explanation:

We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test.

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value for pretest , y = test value for posttest

x: 66, 94, 87, 84, 76, 88

y: 75, 100, 93, 85, 75, 90

The system of hypothesis for this case are:

Null hypothesis: $\mu_D \geq 0$

Alternative hypothesis: $\mu_D < 0$

Because the difference D is defined as Pretest-Postest. And we want to see if the postets score is higher than the pretest with th training.

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

d: -9, -6, -6, -1, 1, -2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-23}{6}=-3.833$

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

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.763$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-3.833 -0}{\frac{3.763}{\sqrt{6}}}=-2.494$

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

$df=n-1=6-1=5$

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

$p_v =P(t_{(5)}$

The p value is lower than the significance level assumed $\alpha=0.05$ , so then we can conclude that we can reject the null hypothesis. So we can say that the differences between Pretest and Posttest $\mu_{pretest}-\mu_{postest}$ are less than 0.

3 0

3 years ago

Which expression is equivalent to 5x - 5x + 3x - 3x?

Mademuasel [1]

5x - 5x + 3x - 3x = 0
****since there is one positive and one negative of both 5x and 3x everything would cancel and the equation would equal 0

7 0

3 years ago