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

(-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.