Use this information to answer Questions 2 trough 6: A high-school teacher in a low-income urban school in Worcester, Massachusetts, used cash incentives to encourage student learning in his AP statistics class. In 2010, 15 of the 61 students enrolled in his class scored a 5 on the AP statistics exam. Worldwide, the proportion of students who scored a 5 in 2010 was 0.15. Is this evidence that the proportion of students who would score a 5 on the AP statistics exam when taught by the teacher in Worcester using cash incentives is higher than the worldwide proportion of 0.15? Setup the appropriate null and alternative hypotesis. Continuation of Question 2. Find the p-value. tcdf(ts,999,60) = 0.017 O tcdf(-999,ts,60) = 0.983 O normalcdf(ts,999,0,1) = 0.015 O 2*normalcdf(ts,99,0,1) = 0.03

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Use this information to answer Questions 2 trough 6: A high-school teacher in a
low-income urban school in Worcester, Massachusetts, used cash incentives to
encourage student learning in his AP statistics class. In 2010, 15 of the 61 students
enrolled in his class scored a 5 on the AP statistics exam. Worldwide, the proportion
of students who scored a 5 in 2010 was 0.15. Is this evidence that the proportion of
students who would score a 5 on the AP statistics exam when taught by the teacher
in Worcester using cash incentives is higher than the worldwide proportion of 0.15?
Setup the appropriate null and alternative hypotesis.
Transcribed Image Text:Use this information to answer Questions 2 trough 6: A high-school teacher in a low-income urban school in Worcester, Massachusetts, used cash incentives to encourage student learning in his AP statistics class. In 2010, 15 of the 61 students enrolled in his class scored a 5 on the AP statistics exam. Worldwide, the proportion of students who scored a 5 in 2010 was 0.15. Is this evidence that the proportion of students who would score a 5 on the AP statistics exam when taught by the teacher in Worcester using cash incentives is higher than the worldwide proportion of 0.15? Setup the appropriate null and alternative hypotesis.
Continuation of Question 2. Find the p-value.
tcdf(ts,999,60) = 0.017
O tcdf(-999,ts,60) = 0.983
O normalcdf(ts,999,0,1) = 0.015
O 2*normalcdf(ts,99,0,1) = 0.03
Transcribed Image Text:Continuation of Question 2. Find the p-value. tcdf(ts,999,60) = 0.017 O tcdf(-999,ts,60) = 0.983 O normalcdf(ts,999,0,1) = 0.015 O 2*normalcdf(ts,99,0,1) = 0.03
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