Question

In: Statistics and Probability

1) Smart Driver Driving School has many branches across provinces. In a province different from the...

1) Smart Driver Driving School has many branches across provinces. In a province different from the one in problem 1, the teaching committee at the provincial branch randomly assigned half of their 5000 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. In another random sample of 100 students who received the new teaching method, 84 passed the road test. The committee would like to use the sample data to test if the two teaching methods have different passing rates. Let ?C and ?? be the true passing rates of the conventional and the new teaching methods, respectively.

Construct a 95% confidence interval for the difference in the true passing rates (expressed as proportions) between the conventional and new teaching methods:

95% confidence interval: (        ,          ) [Your two numbers must be rounded to 3 decimal places. Keep at least 6 significant figures in intermediate steps.]

Independent random samples, each containing 60 observations, were selected from two populations. The samples from populations 1 and 2 produced 35 and 25 successes, respectively.
Test ?0:(?1−?2)=0 against ??:(?1−?2)≠0. Use ?=0.01

(a) The test statistic is

(b) The P-value is

Solutions

Expert Solution

Here in this scenario the answer of both Questions is calculated using sampling Distribution for sample proportion.

Que 1) in this Question we have to calculate the 95% Confidence Interval for difference in population proportion. Bassd on given sample information we calculated as below,

Que 2) in this Question we have to test weather there is difference in population proportion. To test this claim we have performed this at 0.01 level of significance as below,

For question 2).

The test Statistic z cal = 1.826.

The p value is 0.0679.

The p value is calculated using Standerd normal z-table or using Excel.

Thank you.


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