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

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Asked Aug 2, 2019
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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

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Expert Answer

Step 1

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.

Step 2

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.

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р- Ро Z = Р.(1- Р.) п 0.196 0.423 Г0.423 (1-0.423) 30 -2.5167

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Step 3

p-value of the test statistic:

The p-value of the test statistic is 0.0059 from the c...

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using the excel formula, "=1-NORM.S.DIST(0.563,TRUE)" | P(Z<-2.5167) = 0.005923

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