Question

Prospective drivers who enroll in Smart Driver Driving School have always been taught by a conventional teaching method. The driving school has many branches across provinces. Last year, among all students that took driving lessons from the school in a certain province, 80% passed the provincial road test. This year, the teaching committee came up with a new teaching method. The committee randomly assigned half of its 2400 students enrolled this year to receive the conventional teaching method and the remaining half to receive the new teaching method. In a random sample of 100 students who received the conventional teaching method, 76% passed the road test.

Part i) To test if the passing rate has decreased from last year for students who received the conventional teaching method, what will be the null hypothesis?

A. The proportion of 100 students who received the conventional teaching method and subsequently passed the road test this year equals 0.80.

B. The proportion of 1200 students who received the conventional teaching method and subsequently passed the road test this year equals 0.76.

C. The proportion of 1200 students who received the conventional teaching method and subsequently passed the road test this year is lower than 0.80.

D. The proportion of 100 students who received the conventional teaching method and subsequently passed the road test this year is lower than 0.80.

E. The proportion of 100 students who received the conventional teaching method and subsequently passed the road test this year equals 0.76.

F. The proportion of 1200 students who received the conventional teaching method and subsequently passed the road test this year equals 0.80.

Part ii) For the test mentioned in the previous part, what will be the alternative hypothesis?

A. The proportion of 100 students who received the conventional teaching method and subsequently passed the road test this year equals 0.80.

B. The proportion of 100 students who received the conventional teaching method and subsequently passed the road test this year is lower than 0.80.

C. The proportion of 1200 students who received the conventional teaching method and subsequently passed the road test this year equals 0.80.

D. The proportion of 1200 students who received the conventional teaching method and subsequently passed the road test this year is lower than 0.80.

E. The proportion of 100 students who received the conventional teaching method and subsequently passed the road test this year equals 0.76.

F. The proportion of 1200 students who received the conventional teaching method and subsequently passed the road test this year equals 0.76.

Part iii) What is the approximate null model for the sample proportion of the conventional teaching group who passed the road test?

A. N(0.76,0.76⋅0.241200−−−−−−√).

B. N(0.80,0.8⋅0.21200−−−−−√).

C. N(0.80,0.76⋅0.24100−−−−−−√).

D. N(0.76,0.8⋅0.2100−−−−−√).

E. N(0.80,0.8⋅0.2100−−−−−√).

F. N(0.76,0.76⋅0.24100−−−−−−√).

Part iv) Compute the P-value: (your answer must be expressed as a proportion and rounded to 4 decimal places.)

Part v) What is an appropriate conclusion for the hypothesis test at the 2% significance level?

A. The passing rate for students taught using the conventional method this year is significantly lower than last years.

B. The passing rate for students taught using the conventional method this year is not significantly lower than last years.

C. The passing rate for students taught using the conventional method this year is the same as last years.

D. Both (b) and (c).

Answer 1

Part i) null hypothesis:

Answer: A. The proportion of 100 students who received the conventional teaching method and subsequently passed the road test this year equals 0.80.

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Part ii) Alternative hypothesis:

Answer : B. The proportion of 100 students who received the conventional teaching method and subsequently passed the road test this year is lower than 0.80.

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Part iii) Approximate null model for the sample proportion of the conventional teaching group who passed the road test:

E. N(0.80,√(0.8⋅0.2/100)).

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Part iv)

Test statistic:  

z = (p̄ -p)/√(p*(1-p)/n) = (0.76 - 0.8)/√(0.8 * 0.2/100) = -1.0

p-value = NORM.S.DIST(-1, 1) = 0.1587

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Part v)

B. The passing rate for students taught using the conventional method this year is not significantly lower than last years.