Need Help On Stats Question

Need help with stats question stat... ?

1) For Xbar we know that
* E(Xbar) = µ
* Var(Xbar) = sigma²/n
Since Var(A) = E(A²) - E²(A), we have E(A²) = Var(A)+E²(A)
Hence
* E((Xbar)²) = Var(Xbar)+E²(Xbar) = sigma²/n + µ²
Since this is not equal to µ², (Xbar)² is biased

2) For s² we know that E(s²) = sigma²
We have
E((Xbar)² - ks²) = sigma²/n + µ² - k*sigma²
If we choose k = 1/n, we find ... = µ²

NEED HELP WITH STATISTICS QUESTION!!?

The deciles of any distribution are the points that mark off the lowest 10% and the highest 10%. On a density curve, these are the points with areas 0.1 and 0.9 to their left under the curve.
(a) What are the deciles of the standard Normal distribution?
and

(b) The weight of 9-ounce potato chip bags are approximately Normal with mean 9.63 ounces and standard deviation 0.18 ounces. What are the deciles of this distribution?
lowest decile
highest decile

Need help with statistics questions?

1.) From the binomial distr. equation:

P(X=x) = (nCx) * p^x * q^(n-x)

in this case, x=2, p= 0.3, n=5, q = 1 - p = 1 - 0.3

So substituting...
P(X=2) = (5C2) * 0.3^2 * (1-0.3)^3
= 5!/(2!*3!) * 0.09 * 0.7^3 .... dont have a calculator but think its...
= 10 * 0.09 * 0.49 * 0.7
= 10 * 0.09 * 0.343
= 3.43 * 0.09
= 3.087 ....should be right, just verify

2.) Many ways to do this, but ill take the shortest way, we just need to make sense of this first

If we flip a (balanced!)coin, we get either a head or a tail.
this means the probability of heads is 0.5, right?

they give us a number of times the coin is tossed, this implies we working with the binomial distribution.

For variance(binomial), var = npq
var = 72 * 0.5 * 0.5
var = 18
std dev = sqrt(var) = sqrt (18) = 4.something ...again, calculator work

3.) Now im not american (no idea what SAT is), and the question was written down a bit funny, but i think you want the who are in the top 5% ? So...

this is not a sample, since they not providing us with a number of applicants to test, so we use "population" equations...

We now try to find the test score which would put us in the top 5%. Other words, the people who beat at least 95% of the people taking the test...

We'll use the normal table for the corresping z-score for 0.95 (or 0.05 depending on the table). Cant remember but im pretty sure its 1.645, anycase i'll just call it z

(x - 1000) / 200 > z
x-1000 > 200z
x> 200z + 1000
substitute z in, and get the score for x (again look in normal table and do calculator work)
x > 200 * 1.645 + 1000 = 329 + 1000 = 1329 ... if my guess for z is right
so a minimum score of 1329 would put an applicant in the top 5%

I adivse you to work along with these answers i've given, just make sure u understand where everything is coming from, and what you want to do with them. Irrespective if your question is right or wrong, your procedure and understanding of the work is what would give you an A in statistics (hopefully a bursary and deduction of class fees as well :)

Stats Questions need help urgent please thanks?

Confidence intervals are used to find a region in which we are 100 * ( 1 - α )% confident the true value of the parameter is in the interval.

For small sample confidence intervals about the population proportion you have:

pHat ± t * sqrt(phat * (1- phat) / n)

where phat is the sample proportion
t is the t score for having α% of the data in the tails, i.e., P( |T| > t) = α
n is the sample size

The sample proportion, phat = 0.64
The sample size n = 25

The z score for a 0.99 confidence interval is the z score such that 0.005 is in each tail.
t = 2.796940

there are 24 degrees of freedom

The confidence interval is:

( phat ± t * sqrt(phat * (1 - phat) / n)
( 0.3714938 , 0.9085062 )

We are 0.99 confident the true value of the population proportion lies in this interval.


--


Confidence intervals are used to find a region in which we are 100 * ( 1 - α )% confident the true value of the parameter is in the interval.

For large sample confidence intervals about the population proportion you have:

pHat ± z * sqrt(phat * (1- phat) / n)

where phat is the sample proportion
z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α
n is the sample size

The sample proportion, phat = 0.46
The sample size n = 12500

The z score for a 0.99 confidence interval is the z score such that 0.005 is in each tail.
z = 2.575829
The confidence interval is:

( phat ± z * sqrt(phat * (1 - phat) / n)
( 0.4485175 , 0.4714825 )

We are 0.99 confident the true value of the population proportion lies in this interval.


--

Need help on statistics question. Can anyone help thx. 10 points :)?

for any discrete random variable the expectation is:

E(X) = μ = ∑x * P(X = x)

the variance of a discrete random variable is:

Var(X) = σ² = ∑(x - μ)² * P(X = x) = {∑x² * P(X = x) }- μ²

the standard deviation is just the square root of the variance.




1)

the mean is
0 * 0.1296 + 1 * 0.3456 + 2 * 0.3456 + 3 * 0.1536 + 4 * 0.0256 = 1.6

the variance is:

0^2 * 0.1296 + 1^2 * 0.3456 + 2^2 * 0.3456 + 3^2 * 0.1536 + 4^2 * 0.0256 - 1.6^2
= 0.96

the standard deviation is sqrt(0.96) = 0.9797959






2)

mean
0 * 0.16 + 1 * 0.29 + 2 * 0.22 + 3 * 0.09 + 4 * 0.24
= 1.96

variance =
0^2 * 0.16 + 1^2 * 0.29 + 2^2 * 0.22 + 3^2 * 0.09 + 4^2 * 0.24 - 1.96^2
= 1.9784

standard deviation = sqrt(1.9784) = 1.406556






3)

X has the binomial distribution with n = 10 trials and success probability p = 0.3333333

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n = 10 , p = 0.3333333 )

the mean of the binomial distribution is n * p = 3.333333
the variance of the binomial distribution is n * p * (1 - p) = 2.222222
the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.490712

The Probability Mass Function, PMF,
f(X) = P(X = x) is:

P( X = 0 ) = 0.01734153
P( X = 1 ) = 0.08670765
P( X = 2 ) = 0.1950922 <<<< ANSWER
P( X = 3 ) = 0.2601229
P( X = 4 ) = 0.2276076
P( X = 5 ) = 0.1365645
P( X = 6 ) = 0.0569019
P( X = 7 ) = 0.01625768
P( X = 8 ) = 0.003048316
P( X = 9 ) = 0.0003387018
P( X = 10 ) = 1.693509e-05

Need help on statistics question.. Can anyone help? thanks.. 10 points if u do?

In a group of 12 scores, the largest score is increased by 36 points. What effect will this have on the mean of the scores?
a. it will be increased by 12 points.
b. it will remain unchanged.
c. it will be increased by 3 points.
d. it will increase by 36 points.
e. there is no way of knowing exactly how many points the mean will be increased.

If a teacher computes the mean for a set of test scores and then subtracts this mean from each score, the SUM of the resulting set of difference scores will equal
a. zero.
b. unity.
c. n, the number of scores.
d. the mean.
e. n times the mean.

Let us define a new statistic as the distance between 70th sample percentile and 30th sample percentile. This new statistic would give us information concerning
a. central tendency.
b. variability
c. relative position.
d. skewness.
e. symmetry.

Need help on statistics question. can anyone help me. thanks 10 points if u do :)?

Hi,

2. Make a scatter diagram for the data. Use the scatter diagram to describe how, if at all, the variables are related.
X 10 8 3 7 6 12 5
Y 13 15 11 14 17 13 20
A. The variables appear to be positively, linearly related.
B. The variables do not appear to be related. <==ANSWER
C. The variables appear to be negatively, linearly related.

If you graph the points, there is no strong linear pattern in any direction. If you work out the best fit linear equation, its correlation coefficient is -.21. Since this is relatively close to 0, there is a very weak, non-existent correlation. So the answer is B.

3. The data below are the number of absences and the final grades of 9 randomly selected students from a statistics class. Calculate the correlation coefficient, r.
# of Absences, X 0 5 6 4 9 2 15 8 5
Final grade, Y 95 85 79 81 70 91 54 75 81
a .-0.888
b. -0.918
c. -0.899
d. -0.988 <==ANSWER

The best fit linear equation for these points is y = -2.74x + 95.42. The correlation coefficient "r" is -.988. Since this is close to -1, there is a strong negative correlation, that as absences go up, grades go down.


4. The data below are the number of absences and the final grades of 9 randomly selected students from a statistics class. Calculate the correlation coefficient, r.
# of Absences, X 0 5 6 4 9 2 15 8 5
Final grade, Y 95 85 79 81 70 91 54 75 81
A. Yhat= -2.74X+95.45 <== ANSWER see explanation above
B. Yhat= 96.14X-2.75
C. Yhat=-2.75X-96.14
D. Yhat= -95.45+2.74X

Thank goodness for graphing calculators!!

I hope that helps!! :-)

Need help with Statistics problem...?

You have to test the significance of the difference between the means of two different populations. A percentage or proportion of individuals with a certain attribute *is* a special case of the mean: the mean of 0-1 valued random variables, where 0 = i-th pupil stays, 1 = i-th pupil drops out.
The statistics you have available are the sample proportion, call it p, for the last year (sample size m, unknown true population proportion pi) and the sample proportion r for this year (sample size n, unknown true population proportion rho). Your null hypothesis is that pi = rho (alternative pi unequal rho), and you want to test it using p and r. Assuming m and n are not too small, the proportion difference p - r is normally distributed (otherwise use Student's distribution), with (true) mean pi - rho and (true) variance pi(1-pi)/m + rho(1-rho)/n. This is because the true variance of each 0-1 vble is pi(1-pi) for last year, and rho(1-rho) for this year.
You have to replace these unknown true variances by sample variances; the sample-based variance of p - r is

S.E.^2 = p(1-p)/(m-1) + r(1-r)/(n-1).

(The -1 is the correction to make sample variance unbiased, but it doesn't matter much for larger m, n. Still, may as well be exact.)

Now take the Z-score of p - r under the null hypothesis that pi = rho, i.e.,

Z = [(p - r)-0]/SE
= (p - r) / sqrt[p(1-p)/(m-1) + r(1-r)/(n-1)]

and compare its value with the two-sided critical point (1.96 for 5% significance). If Z > 1.96 or Z < -1.96, reject the null h.


PS: Under the null hypothesis (that pi - rho = 0), it is probably better to pool the two sample proportions p and r to estimate the common true proportion pi = rho as

q = (mp + nr) / (m+n)

and use *this* to estimate the true variance. Then

SE = sqrt{q(1-q)[1/(m-1) + 1/(n-1)]}

(as in Matthew's answer, with the -1s added).

Need help with statistics question about distribution for sample mean.?

This is an application of the Central Limit Theorem (also known as the Fundamental Theorem of Statistics). It states that when taking a larg enough sample from a population of any distribution, the means of those samples will form a normal model with a mean the same as the parent distribution and standard deviation = (population standard deviation) / (square root of sample size).

Emphasis on "large enough." If the population is normal, any sample size will do. Typically, once you get a sample size around 40, that will work for a population of any shape. So n=50 will do just fine.

The standard deviation of a sampling distribution of means for this situation, then, equals 2.3/sqrt(50) which, to 4 decimal places, is 0.3253.

The z-score of a mean of 15 = (15-12)/0.3253 = 9.222

The corresponding probability is really small, on the order of 1.5x10^-20. In other words pretty much impossible.



if you take a step back to think, that should make sense. The probability that exactly one corn husk will be 15 inches is P(z>(15-12)/2.3)=0.096. Less than 10%. In order to get FIFTY corn husks to average more than 15 inches long, you need to get a less than 1 in 10 chance outcome to happen 50 times in a row. That's some small odds.