×

### Let's log you in.

or

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

×

or

## Week 1 Electronic Text MicroSoft Word Solutions (8.48+8.64) - All work shown + megastat output

by: smartwriter Notetaker

12

0

7

# Week 1 Electronic Text MicroSoft Word Solutions (8.48+8.64) - All work shown + megastat output

Marketplace > Week 1 Electronic Text MicroSoft Word Solutions 8 48 8 64 All work shown megastat output
smartwriter Notetaker
CSU - Dominguez hills
GPA 3.0

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

×
Unlock Preview

Week 1 Electronic Text MicroSoft Word Solutions (8.48+8.64) - All work shown + megastat output
COURSE
PROF.
No professor available
TYPE
Study Guide
PAGES
7
WORDS
KARMA
50 ?

## Popular in Department

This 7 page Study Guide was uploaded by smartwriter Notetaker on Sunday November 15, 2015. The Study Guide belongs to a course at a university taught by a professor in Fall. Since its upload, it has received 12 views.

×

## Reviews for Week 1 Electronic Text MicroSoft Word Solutions (8.48+8.64) - All work shown + megastat output

×

×

### 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: 11/15/15
Week 1 Electronic Text  Name RES/342 Instructor  Date A. ¿maketheconfidenceinterval ,weneedthestandard deviation ,sample¿¿meanobserved Sample¿20 Mean=∑ of allva=u0+260+356….+13= 6930=346.5 n 20 20 2 Standard deviation=−x) ≈170.38 √[ n ] 0 Sample Sample Sample Standard Size Mean Deviation 260 20 346.5 170.3783715 356 403 536 0 268 369 428 536 268 396 469 536 162 338 403 536 536 130 We have to decide to use the normal interval or the t­interval. The rule is: Sample¿30−uset distributioninterval Sample¿30−use normal distributioninterval Our sample size is 20. 20 is less than 30, and not greater than 30, so we use the t­interval.  20<30 The formula for a t distribution confidence interval is: ´±t σ [ 2c√n To find t1−c , we need to find out the degrees of freedom. [ 2 Degrees of freedom = n – 1 = 20 – 1 = 19 Confidence level = 95% = 0.95 t =t =t =2.093 [ ]c,df [1−0.]19 0.025,19 2 2 σ = 170.38 √n √20 x−−μ x−−μ ¯ = ¯ σ 170.38 √n √ 20 Thisistheteststatistic ,¿herewecanderivetheconfidenceinterval Wewillmakea95 confidenceinterval −2.093≤ ´−μ ≤2.093 170.38 √20 ¿´x− 2.093 170.38≤μ≤x ´ + 2.093170.38 ( )( √20 ) ( )( √20 ) 170.38 170.38 ¿ x− 2.093)( ),´+(2.09)( ) ( √20 √20 ) x=346.5 170.38 170.38 ¿(346.5−(2.09)( √20 ),346.52.093)( √20 ) ¿(346.5−79.73937814,346.5+79.7393781) ¿(266.7606219,426.239378) ¿(266.76,426.2±79.74 One is 95% confident that the true mean area devoted to advertisements on a randomly chosen  page is between 266.76 and 426.24 millimeters squared. Confidence interval ­ mean 95% confidence level 346.5 mean 170.38 std. dev. 20 n 79.740 half­width) 426.240 upper confidence limit 266.760 lower confidence limit B. Normality might be an issue since we don’t know if this sample of 20 was chosen  randomly. Also, the data is not continuous, and due to the two zero values, normality is  an issue. C. σ=170.38 Confidencelevel=99 =0.99 z1−c=z 1−0.99z 0.01z 0.005.576 [2] [ 2 ] 2 Marginof error=10 n 2   Sample¿(¿)= z   (E) 2 ¿ 170.38∗2.576 ( 10 ) ¿1926.322269 ≈1927 Sample size ­ mean 10 E, error tolerance 170.38 standard deviation 99% confidence level 2.576 z 1926.067 sample size 1927 rounded up D. 99% confidence level is way too narrow for a sample size of 20, and one which is unknown to be normal or even chosen randomly. A better level would be 90%, or lower. A. Sample¿ 773 Unpopped kernelscount=86 p= 86 =0.1112548512≈0.1113 773 86 687 1−p=q=1− 773 773=0.8887451488≈0.8888 Confidenceinterval level=90 =0.90 α=1−0.9=0.1 zα=z0.1 0.05.645 2 2 p(1−p) Confidence interval2√   n    0.1113∗0.8888 ¿0.1113±1.6√[ 773 ] ¿ 0.1113−1.6450.1113∗0.88,0.1113+1.645.1113∗0.8888 ( √[ 773 ] √[ 773 ) ¿ 0.1113−1.6450.0989234,0.1113+1.645.09892344 ( √[ 773 ] √[ 773 ]) −4 −4 ¿(0.1113−1.6√5 1.279734023×10 ,0.1113√1.645 1.27973)023×10 ¿(0.1113−0.0186091167,0.1113+0.018)091167 ¿(0.0926908833,0.12990)1167 ≈ 0.093,0.1±0.019 One can be 90% confident that the true proportion of unpopped kernels is between 0.093 and 0.130. Confidence interval ­  proportion 90% confidence level 0.1113 proportion 773 n 1.645 z 0.019 half­width 0.130 upper confidence limit 0.093 lower confidence limit (b) The normality assumption says: Normality is confirmed if: n∗p>5 ¿ n∗(1−p )>5 773∗0.1113>5−−−86.0349>5  – TRUE! 773∗0.8888>5−−−687.0424>5  –TRUE! Normality is confirmed. (c) The “Very Quick Rule” works when p should be close to 0.5. Out proportion is 0.1113, which  isn’t close to 0.5, so the “Very Quick Rule” can’t be applied here. (d) The confidence interval of [0.093, 0.130] applies to the population proportion of unpopped  kernels. A sample of 86 which didn’t pop is unrepresentative.

×

×

### 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!"

Janice Dongeun University of Washington

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

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!"

Parker Thompson 500 Startups

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

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