×

### Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

## Stat200 quiz 3 october 2015

by: Topseller Notetaker

23

0

9

# Stat200 quiz 3 october 2015 PRG211

Marketplace > Ashford University > Computer Programming > PRG211 > Stat200 quiz 3 october 2015
Topseller Notetaker
AU
GPA 3.9

Get a free preview of these Notes, just enter your email below.

×
Unlock Preview

Stat200 quiz 3 october 2015
COURSE
PROF.
TYPE
Study Guide
PAGES
9
WORDS
KARMA
50 ?

## Popular in Computer Programming

This 9 page Study Guide was uploaded by Topseller Notetaker on Monday November 9, 2015. The Study Guide belongs to PRG211 at Ashford University taught by in Fall 2015. Since its upload, it has received 23 views. For similar materials see in Computer Programming at Ashford University.

×

## Reviews for Stat200 quiz 3 october 2015

×

×

### What is Karma?

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 11/09/15
Quiz 3  Statistics 200 – October 2015  There are again 5 problems on this quiz. Again, it is not meant to be tricky or hard, but to ensure  you are comfortable with the concepts in the assigned chapters.  Please submit your quiz to your assignment folder. If you have issues submitting, please do not  use email. Submit with your homework assignment and I will pick it up there. It has to be  submitted into LEO for credit.  Answer key will be posted on Tuesday. ______________________________________________________ Problem 1 Below are five (5) separate distributions. a­ Calculate the Range b­ Use the range to calculate the standard deviation c­ Calculate the standard deviation (and show any difference) a) Range = Max – Min b) Using the range rule of thumb, the relation between range and standard deviation,  Range   4 c) The actual formula for standard deviation,   1  x x  n1    The following table gives the necessary values for each distribution as follows: 1 2 3 4 5 6 6 6 6 6 5 5 6 6 6 4 5 5 6 6 4 4 5 5 6 4 4 4 4 4 4 4 3 3 2 4 3 3 2 2 3 3 2 2 2 2 2 2 2 2  x 37 38 39 40 41 x2  154 156 164 170 176 max 6 6 6 6 6 min 2 2 2 2 2 range 4 4 4 4 4 standard deviatio 1 1 1 1 1 1.1180 1.2247 1.5811 1.8027 stdev 34 45 39 76 2 Problem 2  Two (2) students took the same statistics test.  Convert their scores to percentiles and plot on a chart.  You will need the z­table in the back of the text (or internet search).  Math Test Results  X = 73.0  s = 8.0  John scores 79 on the test  Mary scored 68 on the test  X    The Z­score is defined as         8 Given that 7973 6 ZJ   0.75 8 8 Then, the Z-score for John score, P   0.75  0.7734 77.34% The Z score is 0.75 and the corresponding percentile, The corresponding chart is shown as follows:       8 Given that 6873 5 Z M    0.625 8 8 Then, the Z-score for John score, The Z score is –0.625 and the corresponding percentile, P   0.624  0.266  26.6% The corresponding chart is shown as follows: Problem 3 I have finished analyzing the data from a recent study. There are 25 subjects in the sample The mean of the sample is 75 The standard deviation is 4.7 Please answer the following: a­ What is the standard error o the mean for these data? b­ What are the 95% and 99% confidence intervals c­ What is the probability of obtaining by chance the following: i ­ a mean less than 72 ii­ a mean less than 74 iii­ a mean greater than 76 iv­ a mean greater than 78 Answer:  n  25,x  75,s  4.7 Given that  s 4.7  x   0.94 n 25 a)The standard error of the mean, b) 1  The confidence interval for the mean is defined by s x t n The 95% critical value of ‘t’ at 25–1=24DF is 2.064. Then, the 95% confidence interval for the mean is given by s 4.7 x t 75 2.064 n 25  751.94  3.06, 26.94  The 99% critical value of ‘t’ at 25–1=24DF is 2.064. Then, the 95% confidence interval for the mean is given by s 4.7 x t n 752.7969 25  752.629  2.3709, 27.6291  c) P x 72    x   7275   s  n 4.7/ 25   P   3.1915  24  0.00196 Uing excel function =tdist 3.191,24,1   x  7475  P x 74       s  n 4.7/ 25   P  241.0638   0.14899 Uing excel function =tdist 1.063,24,1    x  7675  P x 76       s  n 4.7/ 25   P  24.0638  1 P t 1.0638  24  10.14899 Usig excel function =tdist 1.0638,4,1   0.8563  x  7875  P x 78    s  n  4.7/ 25   P  243.1915  1 P t24.1915  10.00196 Usig excel function =tdist 1.0638,4,1   0.998 Problem 4 Use the following information to write the correspondingnull and alternative hypothesis: a­ A sample mean of 23 is statistically different from a population mean of 30    Null hypothesis: The mean is same as population mean     Alternative hypothesis: The mean is not the same as population mean  b­ A sample mean of 56 is less than the population mean of 70   70 Null hypothesis: The mean is not less than population mean     Alternative hypothesis: The mean is less than population mean  c­ A sample mean of 75 is greater than the population mean of 70   70 Null hypothesis: The mean is not greater than population mean     Alternative hypothesis: The mean is greater than population mean  Problem 5 I am back in the lab, collecting samples. The mean of the sample is 75 My associate wants to determine whether the sample meanstatistically greater than the  population mean of 70.  My associate set α = 0.05 Please provide responses to: a­ The probability associated with a mean of 75 is p = 0.05Can my associate reject the null  hypothesis? b­ If the null hypothesis is a correct statement, what is the probability that the researcher  will make the correct decision not to reject the null hypothesis? Answer: Given that x  75 The sample mean, a)The probability associated with mean of 75 (P-value is 0.05) Since the P-value is equal to the level of significance, do not reject the null hypothesis. b) Ifthe null hypothesis is a correct statement, the probability that the researcher will make the  correct decision not to reject the null hypothesis is nothing but  1–P(type I error) = 1 – 0.05 = 0.95

×

×

### BOOM! Enjoy Your Free Notes!

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Kyle Maynard Purdue

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made \$280 on my first study guide!"

Bentley McCaw University of Florida

Forbes

#### "Their 'Elite Notetakers' are making over \$1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!
×

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com