Fill This Form To Receive Instant Help

Help in Homework
trustpilot ratings
google ratings


Homework answers / question archive / Researchers designed a survey to compare the proportions of children who come to school without eating breakfast in two low-income elementary schools

Researchers designed a survey to compare the proportions of children who come to school without eating breakfast in two low-income elementary schools

Math

Researchers designed a survey to compare the proportions of children who come to school without eating breakfast in two low-income elementary schools. An SRS of 80 students from School 1 found that 19 had not eaten breakfast. At School 2, an SRS of 150 students included 26 who had not had breakfast. More than 2500 students attend each school. Using PANIC, construct and interpret a 90% confidence interval for the true difference between the proportion of all students in School 1 and School 2 who come to school without eating breakfast (School 1 - School 2). 

P - ?

A - ?

N - ?

I - ?

C - ?

pur-new-sol

Purchase A New Answer

Custom new solution created by our subject matter experts

GET A QUOTE

Answer Preview

(-0.0292,0.1575)

 

Parameters

Assumptions

Name

Interval caluclations

Conclusion

Step-by-step explanation

Parameter: difference between the proportion of all students in School 1 and School 2 who come to school without eating breakfast

Assumptions: The samples are randomly selected, independent, and a normal approximation is able to be used.

Name of the interval: As it is a two sample proportion confidence interval we will use a Z distribution (One sample proportion Z confidence interval)

Interval caluclations:

?Group 1: x1=19.00n1=80p1=0.24Group 2: x2=26.00n2=150p2=0.17Confidence level= 0.900We find the confidence interval for the population mean using the following formula: p1?−p2?±Margin of errorp1?−p2?±Za/2?∗n1p1?∗(1−p1?)?+n2p2?∗(1−p2?)??Finding Za/2Where a/2 is (1-confidence)/2  In this problem confidence=0.900 then a/2=(1-0.900)/2=0.0500 The z score for 0.0500 is 1.645 we can find the Za/2 using the excel formula: "abs(Norm.inv(a/2,0,1))" Calculating the Confidence interval p1?−p2?±Za/2?∗n1p1?∗(1−p1?)?+n2p2?∗(1−p2?)??0.2375−0.1733±1.64∗800.2375∗(1−0.2375)?+1500.1733∗(1−0.1733)??0.0642±0.0933(0.0642-0.0933,0.0642+0.0933)(-0.0292,0.1575)?

Conclusion: At a 90% confidence level he true difference between the proportion of all students in School 1 and School 2 who come to school without eating breakfast is between -0.0292 and 0.1575.