Question

1. Language Survey

About 42.3% of Californians speak a language other than English at home. Using your class as the sample, conduct a hypothesis test to determine if the percent of students at De Anza Collegethat speak a language other than English at home is different from 42.3%.

DATA TO USE: 16 out of 26 students in the sample speak a language other than English at home

1. Language Survey

a.  Ho: _______________  b.  Ha: ___________________

c.  In words, CLEARLY state what your random variable  or P'represents.

     

d.  State the distribution to use for the test. If t, include the degrees of freedom. If normal, include the mean and standard deviation.

e.  Test Statistic:  t or z (state which) = ______________ (numerical value)

f.  p-value = ______________    In 1 – 2 complete sentences, explain what the p-value means for this problem.

g.  Use the previous information to sketch a picture of this situation.  CLEARLY, label and scalethe horizontal axis and shade the region(s) corresponding to the  p-value.

                  

h.  Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using COMPLETE SENTENCES.

     alpha                    decision                    reason for decision

     

     Conclusion:  

i.  Construct a 95% Confidence Interval for the true mean or proportion.   Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the Confidence Interval.

                  

Confidence Interval:

( ___________________ , ____________________ )

j. Interpret the confidence interval. (This should be a complete sentence.)

Answer 1

NULL HYPOTHESIS H0 : p= 42.3% or p= 0.423

ALTERNATIVE HYPOTHESIS Ha:

c) Here random variable is the proportion of students speak a language other than English at home.

P= 16/26= 0.615

d) Normal distribution used for the test.

mean= n*P= 26*0.615= 15.99

standard deviation=

=

=

=

=

Test statistic= Z= P-p/sqrt(p*q/n)

Z= 0.615-0.423/sqrt(0.423*0.577/26)

Z= 0.192/sqrt(0.244/26)

Z= 0.192/sqrt(0.00938)

Z= 0.192/0.0968

Z= 1.983

The P-Value is 0.047367.

alpha= 0.05

Decision = Reject NULL HYPOTHESIS

REASON: SInce p value is significant means smaller than 0.05 level of significance.

Conclusion: Since p value is significant we therefore reject NULL HYPOTHESIS and conclude that the percent of students at De Anza College that speak a language other than English at home is different from 42.3%.

95% confidence interval is

where Zc is critical value of Z at 0.05 level of significance.

is 95% confidence interval.

The confidence interval does not contain the null hypothesis value, the results are statistically significant.