One research study of illegal drug use among 12-to 17-year-olds reported a decrease in use (from 11.4 % in 1997) to 9.8 % now. Suppose a survey in a large high school reveals that, in a random sample of 1,045 students, 95 report using illegal drugs. Use a 0.05significance level to test the principal's claim that illegal drug use in her school is below the current national averageFormulate the null and alternative hypotheses. Choose the correct answer belowOA. H p 0.098Ha p>0.098O B. H p

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Asked Oct 13, 2019
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One research study of illegal drug use among 12-to 17-year-olds reported a decrease in use (from 11.4 % in 1997) to 9.8 % now. Suppose a survey in a large high school reveals that, in a random sample of 1,045 students, 95 report using illegal drugs. Use a 0.05
significance level to test the principal's claim that illegal drug use in her school is below the current national average
Formulate the null and alternative hypotheses. Choose the correct answer below
OA. H p 0.098
Ha p>0.098
O B. H p<0.098
Ha p 0.098
O C. H p 0 098
Ha p 0098
O D. H p 0.098
Ha p 0.098
Find the test statistic
(Round to two decimal places as needed)
Find the P-value for the found test statistic
(Round to four decimal places as needed.)
P-value
Now make a conclusion Choose the correct answer below
OA Reject H There is insufficient evidence to support the claim that illegal drug use in this principal's school is below the current national average
O B. Reject H
There is sufficient evidence to support the claim that illegal drug use in this principal's school is below the current national average
There is sufficient evidence to support the claim that illegal drug use in this principal's school is below the current national average.
O C. Do not reject H
There is insufficient evidence to support the claim that illegal drug use in this principal's school is below the current national average.
OD. Do not reject H
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One research study of illegal drug use among 12-to 17-year-olds reported a decrease in use (from 11.4 % in 1997) to 9.8 % now. Suppose a survey in a large high school reveals that, in a random sample of 1,045 students, 95 report using illegal drugs. Use a 0.05 significance level to test the principal's claim that illegal drug use in her school is below the current national average Formulate the null and alternative hypotheses. Choose the correct answer below OA. H p 0.098 Ha p>0.098 O B. H p<0.098 Ha p 0.098 O C. H p 0 098 Ha p 0098 O D. H p 0.098 Ha p 0.098 Find the test statistic (Round to two decimal places as needed) Find the P-value for the found test statistic (Round to four decimal places as needed.) P-value Now make a conclusion Choose the correct answer below OA Reject H There is insufficient evidence to support the claim that illegal drug use in this principal's school is below the current national average O B. Reject H There is sufficient evidence to support the claim that illegal drug use in this principal's school is below the current national average There is sufficient evidence to support the claim that illegal drug use in this principal's school is below the current national average. O C. Do not reject H There is insufficient evidence to support the claim that illegal drug use in this principal's school is below the current national average. OD. Do not reject H

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Expert Answer

Step 1

The appropriate Null and Alternative Hypotheses are given below:

 

From the given information, the claim of the problem is testing whether illegal drug in the school is below the current national average or not.

 

Null hypothesis:

 

H0p = 0.098.

 

That is, the illegal drug in the school is equal to the current national average, 9.8%.

 

Alternative hypothesis:

 

Ha< 0.098.

 

That is, the illegal drug in the school is below the current national average, 9.8%.

 

Hence, the correct option is D. H0p = 0.098 and Ha< 0.098.

Step 2

The test statistic value is ­-0.76 and it is obtained below:

 

Here, the sample size is, n = 1045; in those 95 using illegal drugs. The sample proportion is   =0.091(=95/1045)

 

р-р
test
|P(1-р)
0.091-0.098
J0.098(1-0.098)
1045
0.007
0.0092
-0.76
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р-р test |P(1-р) 0.091-0.098 J0.098(1-0.098) 1045 0.007 0.0092 -0.76

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Step 3

The P-value is 0.2236 and it is calcula...

P-value P(z <
[Since, Given Hypothesis is left tail test]
test
-P(z <-0.76)
From standard normal table,
-0.2236
P(z<-0.76) 0.223 6
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P-value P(z < [Since, Given Hypothesis is left tail test] test -P(z <-0.76) From standard normal table, -0.2236 P(z<-0.76) 0.223 6

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