A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driv
er’s license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A’s students and 22 of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A’s students and 22 of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?
1 answer:
Answer:
Yes, there is evidence to support that claim that instructor 1 is more effective than instructor 2
Step-by-step explanation:
We can conduct a hypothesis test for the difference of 2 proportions. If there is no difference in instructor quality, then the difference in proportions will be zero. That makes the null hypothesis
H0: p1 - p2 = 0
The question is asking whether instructor 1 is more effective, so if he is, his proportion will be larger than instructor 2, so the difference would result in a positive number. This makes the alternate hypothesis
Ha: p1 - p2 > 0
This is a right tailed test (the > or < sign always point to the critical region like an arrowhead)
We will use a significance level of 95% to conduct our test. This makes the critical values for our test statistic: z > 1.645.
If our test statistic falls in this region, we will reject the null hypothesis.
<u>See the attached photo for the hypothesis test and conclusion</u>
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Answer:
a) 0.259
b) 0.297
c) 0.497
Step-by-step explanation:
To solve this problem, it is important to know the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
By the Central Limit Theorem
has a pvalue of 0.6293.
X = 10.8
has a pvalue of 0.3707.
0.6293 - 0.3707 = 0.2586
0.259 probability, rounded to three decimal places.
b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?
Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So
X = 11
has a pvalue of 0.5.
X = 10.5
has a pvalue of 0.2033.
0.5 - 0.2033 = 0.2967
0.297, rounded to three decimal places.
c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
By the Central Limit Theorem
has a pvalue of 0.7486.
X = 10.8
has a pvalue of 0.2514.
0.7486 - 0.2514 = 0.4972
0.497, rounded to three decimal places.
Let 
be the identity matrix. Then pick whatever matrix you like for 
.