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Social ScienceAbout 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English at home. Using your class as a sample, conduct a hypothesis test to determine if the percent of the students at your school who speak a language other than English at home is different from 42.3% Ho=____Ha=____In words, define the random variable ___ =___ The distribution to use for the test is ____ Determine the test statistic using your data Draw a graph and label it appropriately. Shade the actual level of significance
About 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English at home. Using your class as a sample, conduct a hypothesis test to determine if the percent of the students at your school who speak a language other than English at home is different from 42.3%
Ho=____
Ha=____
In words, define the random variable ___ =___
The distribution to use for the test is ____
Determine the test statistic using your data
Draw a graph and label it appropriately. Shade the actual level of significance
Expert Answer
Solution:
Let us consider that your class has 30 students.
Null and alternative hypotheses:
Here, to check whether the percent of the students at your school who speak a language other than English at home is different from 0.423. The appropriate test is z-test for single sample proportion.
Here, the sample proportion p = 0.196.
The hypotheses for the test is given below.
Null hypothesis:
H0: p0 = 0.423
Alternative hypothesis:
H1: p0 ≠ 0.423
The random variable, x = the percent of the students who speak a language other than English at home.
Computation of test statistic:
Here, n = 30.
The sample proportion, p = 0.196.
The population proportion, p0 = 0.423.
The value of test statistic is −2.5167 from the calculations given below.
Image Transcriptionclose
р- Ро Z = Р.(1- Р.) п 0.196 0.423 Г0.423 (1-0.423) 30 -2.5167
p-value of the test statistic:
The p-value of the test statistic is 0.0059 from the c...
Image Transcriptionclose
using the excel formula, "=1-NORM.S.DIST(0.563,TRUE)" | P(Z<-2.5167) = 0.005923
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