Question

Question 2 (15 marks) Police departments often try to control traffic speed by placing speed-measuring machines on roads that tell motorists how fast they are driving. Traffic safety experts must determine where machines should be placed. In one recent test, police recorded the average speed clocked by cars driving on one busy street close to an elementary school. For a sample of 16 speeds, it was determined that the average amount over the speed limit for the 16 clocked speed was 11.6 mph. Without the information of the population standard deviation, t-distribution is used to construct the 95% confidence interval for the unknown population mean. The confidence interval ranged from 7.337 mph to 15.863 mph. (a) What assumption is needed to construct the mentioned confidence interval? Why? (b) Determine the sampling error and the sample standard deviation? (c) The researchers commented that the confidence interval was too wide. With all other things remain unchanged, what sample size is needed in order to reduce the sampling error by 60%? (d) The researchers later found that one sample speed is mistakenly recorded. After the correction, both the sample mean and sample standard deviation become larger. Discuss how would this affects the mentioned confidence interval

Answer 1

(a)
Assumptions needed to construct confidence interval is,
1. The data must be sampled randomly.
2. The sample values must be independent of each other.
3. When the sample is drawn without replacement, the sample size, n, should be no more than 10% of the population.
4. The sample size must be sufficiently large (greater than 30). If the sample size is less than 30, the population distribution should by symmetric, unimodal and without outliers.

(b)
Sampling error , E = (15.863 - 7.337)/2 = 4.263
Degree of freedom = n-1 = 16-1 = 15
Critical value of t at df = 15 and 95% confidence interval is 2.13
Standard error = E / t = 4.263 / 2.13 = 2.00

Sample standard deviation = $\sqrt{n}$ * Standard error = $\sqrt{16}$ * 2 = 8.00

(c)

New Sampling error, E = 4.263 * (1 - 0.6) = 1.7052

Standard error = E / t = 1.7052 / 2.13 = 0.8

Sample size = (Sample standard deviation / Standard error)² = (8 / 0.8)² = 100

(d)

Since the sample mean becomes larger, both the limits of confidence interval will increase.

Since the sample standard deviation becomes larger, the width of confidence interval which depends on the standard error (standard deviation) will increase.