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We wish to test an effect of a drug that claimed to improves the patients with 90% of chance. The drug is given to 20 patients and we observed the number Y of people who improved. The hypothesis is Ho: p=0.9 vs II: p<0.9. Use the rejection region RR= {y≤17) and answer question a-e. a) State type I error using Y and p. Find a. State type II error using Y and p. Find the rejection region {y

We wish to test an effect of a drug that claimed to improves the patients with 90% of chance. The drug is given to 20 patients and we observed the number Y of people who improved. The hypothesis is Ho: p=0.9 vs II: p<0.9. Use the rejection region RR= {y≤17) and answer question a-e. a) State type I error using Y and p. Find a. State type II error using Y and p. Find the rejection region {y

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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We wish to test an effect of a drug that claimed to improves the patients with 90% of chance. The drug is given to 20 patients and we observed the number Y of people who improved. The hypothesis is Ho: p=0.9 vs II: p<0.9. Use the rejection region RR= {y <17) and answer question a-e. a) b) d) State type I error using Y and p. Find a. State type II error using Y and p. Find the rejection region {y <c) with a 0.01. For the rejection region in e), find ß B when p=0.6.
Transcribed Image Text:We wish to test an effect of a drug that claimed to improves the patients with 90% of chance. The drug is given to 20 patients and we observed the number Y of people who improved. The hypothesis is Ho: p=0.9 vs II: p<0.9. Use the rejection region RR= {y <17) and answer question a-e. a) b) d) State type I error using Y and p. Find a. State type II error using Y and p. Find the rejection region {y <c) with a 0.01. For the rejection region in e), find ß B when p=0.6.
Credit Card Balance Data A data frame with 400 observations on a number of variables. • Income: Income in $1,000's . Limit: Credit limit . Rating: Credit rating . Cards: Number of credit cards Age: Age in years . Education: Education in years • Own: A factor with levels No and Yes indicating whether the individual owns a home . Student: A factor with levels No and Yes indicating whether the individual is a student • Married: A factor with levels No and Yes indicating whether the individual is married • Region: A factor with levels East, South, and West indicating the individual's geographical location Balance: Average credit card balance in $. library(tidyr) library (ISLR2) dat <- drop_na (Credit) head(dat) ** Income Limit Rating Cards Age Education Own Student Married Region Balance ##1 14.891 3606 283 2 34 11 No Yes South Yes Weat ## 2 106.025 6645 ##3 104.593 7075 ## 4 148.924 9504 ##5 55.882 4897 483 514 681 357 ## 6 80.180 8047 569 x1 <- x1[x1>0] x2 <- x2[x2>0] x1 <- dat $Balance [dat$Student=="Yes"] x2 <- dat $Balance [dat$Student=="No"] Frequency 03 par (nfrow-c (2,1)) hist(x1, main="Student", breaks-30) hist(x2, main="Non-student", breaks-30) Frequency 3 82 4 71 3 36 0 15 2 68 4 77 0 Use the following R code separate Balance for Student and Non-student and create histogram of Balance for Student and Non-student. We wish to compare mean Balance of Student and Non-student group. podo 0 500 15 Yea 11 No 11 Yes 16 No 10 No 500 Student X1 No Yes No No No No 1000 x2 ubddp. Non-student durupp...cm 1000 No West No West Yes South 331 No South 1151 333 903 580 964 1500 1500 2000
Transcribed Image Text:Credit Card Balance Data A data frame with 400 observations on a number of variables. • Income: Income in $1,000's . Limit: Credit limit . Rating: Credit rating . Cards: Number of credit cards Age: Age in years . Education: Education in years • Own: A factor with levels No and Yes indicating whether the individual owns a home . Student: A factor with levels No and Yes indicating whether the individual is a student • Married: A factor with levels No and Yes indicating whether the individual is married • Region: A factor with levels East, South, and West indicating the individual's geographical location Balance: Average credit card balance in $. library(tidyr) library (ISLR2) dat <- drop_na (Credit) head(dat) ** Income Limit Rating Cards Age Education Own Student Married Region Balance ##1 14.891 3606 283 2 34 11 No Yes South Yes Weat ## 2 106.025 6645 ##3 104.593 7075 ## 4 148.924 9504 ##5 55.882 4897 483 514 681 357 ## 6 80.180 8047 569 x1 <- x1[x1>0] x2 <- x2[x2>0] x1 <- dat $Balance [dat$Student=="Yes"] x2 <- dat $Balance [dat$Student=="No"] Frequency 03 par (nfrow-c (2,1)) hist(x1, main="Student", breaks-30) hist(x2, main="Non-student", breaks-30) Frequency 3 82 4 71 3 36 0 15 2 68 4 77 0 Use the following R code separate Balance for Student and Non-student and create histogram of Balance for Student and Non-student. We wish to compare mean Balance of Student and Non-student group. podo 0 500 15 Yea 11 No 11 Yes 16 No 10 No 500 Student X1 No Yes No No No No 1000 x2 ubddp. Non-student durupp...cm 1000 No West No West Yes South 331 No South 1151 333 903 580 964 1500 1500 2000
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