Can anyone help me with my statistics homework?
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How do you use statistics in your job?
Edit: The question title was "How is statistics used in real life?" when I wrote this, and that's the question I was addressing.I always struggle with these sorts of questions, because there are two different ways to interpret what's being asked, and the answers often have very little in common. But I think that here it's reasonable to spell out what those questions are, and to give the answers to both of them.The first reading is "How will the average person use statistics?", and the answer is that by and large they won't. Sure, they might have to interpret basic descriptive statistics like means and medians, but they're just not doing anything that requires tools which are more sophisticated than that. Heck, even I don't really do any serious statistics off the clock.The second reading is "How are statistics used outside of a statistics class?", and the answer is that almost every non-trivial technical problem requires some degree of statistical analysis, and the techniques you need to use are often quite complicated even if the problems are simple. Deciding which of two surgical procedures gives a better outcome for a certain type of cancer is an active research area, and the techniques that are just now coming out are showing up in practice pretty quickly.As others have mentioned, one of the key activities of statistics is prediction. What will the weather be like tomorrow? How many units of this product will we sell next month? Which of these ten links are you most likely to click on? Will there be a major earthquake in San Francisco within the next ten years? You can pretty easily see why these questions are interesting to their respective audiences, and every single one of them requires statistical methods to answer.
Statistics. Please check my answers.?
Your work is correct For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.) You can translate into standard normal units by: Z = ( X - μ ) / σ Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities. If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem. If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed with mean μ and standard deviation σ /√(n) An applet for finding the values http://www-stat.stanford.edu/~naras/jsm/... calculator http://stattrek.com/Tables/normal.aspx how to read the tables http://rlbroderson.tripod.com/statistics... In this question we have X ~ Normal( μx = 68 , σx² = 9 ) X ~ Normal( μx = 68 , σx = 3 ) Find P( 67 < X < 69 ) = P( ( 67 - 68 ) / 3 < ( X - μ ) / σ < ( 69 - 68 ) / 3 ) = P( -0.3333333 < Z < 0.3333333 ) = P( Z < 0.3333333 ) - P( Z < -0.3333333 ) = 0.6305587 - 0.3694413 = 0.2611173 == == = = = = Xbar ~ Normal( μ = 68 , σ² = 9 / 9 ) Xbar ~ Normal( μ = 68 , σ² = 1 ) Xbar ~ Normal( μ = 68 , σ = 3 / sqrt( 9 ) ) Xbar ~ Normal( μ = 68 , σ = 1 ) Find P( 67 < Xbar < 69 ) = P( ( 67 - 68 ) / 1 < ( Xbar - μ ) / σ < ( 69 - 68 ) / 1 ) = P( -1 < Z < 1 ) = P( Z < 1 ) - P( Z < -1 ) = 0.8413447 - 0.1586553 = 0.6826895 === (c) good answer
Stats help- Check my work?
Hi, This is a problem from my statistics homework. I answered a few of the questions. Mostly, I am unsure what to say for part b, and the last part, about null/alternate hypothesis. Thanks! Ashley =] **************** Chips Ahoy (from ActivStats Homework) In 1998, as an advertising campaign, the Nabisco Company announced a “1000 Chips Challenge,” claiming that every 18-ounce bag of their Chips Ahoy cookies contained at least 1000 chocolate chips. Dedicated Statistics students purchased some randomly selected bags of cookies and counted the chocolate chips. The data: mean = 1238.19 n (sample size) = 15 SD = 94.28 a) The data is independent, because we can safely assume that the number of chips in one bag does not affect the amount of chips in the next bag manufactured. The sample data is random because in the problem, it specifies that the bags purchased by students were “randomly selected.” This data demonstrates a normal distribution because, when I graph it, it is unimodal and symmetric about the peak. b) Comment on any concerns you may have about the data. c) Create a 95% confidence interval for the average number of chips in bags of Chips Ahoy Cookies. 1072.26 < μ < 1404.12 d) What does this evidence indicate about Nabisco’s Claim? Since even the lower-tail value of 1072.26 is greater than 1000, we can be more than 95% confident that each bag contains at least 1000 chips. e) Test the claim that each bag contains at least 1000 chocolate chips. Include the null and alternate hypothesis. Clearly state and support your conclusion.
Stats hypothesis questions - can you check my work?
As you increase the number of subjects in your sample, the calculated value of a t-test will b.) increase *This is because as "n" increases, the denominator of the t-test decreases, increasing the value of T(observed). As you decrease the true distance between the null and alternative hypotheses, the likelihood of rejecting the null hypothesis b.) increases *As H(0) and H(A) gets closer, it becomes more difficult to distinguish them apart so it's more likely to reject H(0). Keeping everything else the same, if you were to decrease your alpha level from .05 to .01, the likelihood of rejecting the null hypothesis a.) decreases *alpha = likelihood of rejecting null when it is true. If it's smaller, then the likelihood of rejecting the null on the whole is also smaller.
Statistics. Please check my answers (a- how do I determine s)?
the data set provided and the questions that follow are different. you went from the number of wolf pups to the weight of mountain lions. the work looks like you're on the right track. Let X1, X2, X3, ... , Xn be a random sample. The mean, xbar is found by: xbar = n ∑ Xi / n i = 1 the sample variance is: n ∑ (Xi - xbar)² / (n - 1) i = 1 the variance is divided by n -1, not n. this is done because if you divide by n you have a biased estimator for the population variance. using n - 1 yields an unbiased estimator. the sample standard deviation is found by taking the square root of the variance. for the number of wolf pups: the mean is 5.625 std dev = 1.78419 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. In order for the Confidence Interval to be valid you must have data from a normal distribution, at least if you are using the method here. If you do not have normal data then this type of confidence interval is not valid. To clear up the notation I will use here. "t" is the test statistic and "t_(n-1)" is a Student t random variable with n - 1 degrees of freedom, e.g. a Student t random variable with 18 degrees of freedom is denoted as t_18. For small sample confidence intervals about the mean you have: xBar ± t * sx / sqrt(n) where xBar is the sample mean t is the t - score with n - 1 degrees of freedom such that α% of the data in the tails, i.e., P( |t_(n-1)| > t) = α sx is the sample standard deviation n is the sample size For small sample confidence bounds we have the same estimators save the t-score. A 100 * (1 - α )% Confidence Lower Bound is: xBar - t * sx / sqrt(n) where t is such that P( t_(n-1) < t ) = α A 100 * (1 - α )% Confidence Upper Bound is: xBar + t * sx / sqrt(n) where t is such that P( t_(n-1) > t ) = α The sample mean xbar = 5.625 The sample standard deviation sx = 1.78419 The sample size n = 16 The t score for a 0.75 confidence interval is the t score such that 0.125 is in each tail. t = 1.196689 The confidence interval is: ( xbar - t * sx / sqrt( n ) , xbar + t * sx / sqrt( n ) ) ( 5.09122 , 6.15878 )
Could someone please check my AP statistic answers?
Abby, Barbara, Carla, Dan, and Ennis work in a firm’s public relations office. Their employer must choose two of them to attend a conference in Chicago. To avoid unfairness, the choice will be made by drawing two names from a hat. (This is a sample size of 2.) Write down all possible choices of two of the five names. For convenience, you can simply use the first letter of their names (for example Abby and Barbara would be AB) A.) AB, AC, AD, AE, BC, BD, BE, CD, CE, DE B.) AB, AC, AD, AE, CD, CE, DE * C. ) AB, AC, AD, AE, BC, BD, BE D.) AB, AC, AD, AE, BC, BD, BE, CD, CE What is the probability that neither of the two men (Dan and Ennis) is chosen? a)7/10 = 0.7 b)9/10 = 0.9* c)1/2 = 0.5 d)3/10 = 0.3