Question

Traffic police monitor the speed of vehicles as they travel over a new bridge. The average speed for a sample of 31 vehicles was 82.72 km/h, with the sample standard deviation being 4.61 km/h. We will assume that the speeds are Normally distributed, and the police are interested in the mean speed.

Part a) Since the variance of the underlying Normal distribution is not known, inference here would involve the t distribution. How many degrees of freedom would the relevant t distribution have?

Part b) Create a 95 % confidence interval for the mean speed of vehicles crossing the bridge. Give the upper and lower bounds to your interval, each to 2 decimal places. (

,)

Part c) The police hypothesized that the mean speed of vehicles over the bridge would be the speed limit, 80 km/h. Taking a significance level of 5 %, what should infer about this hypothesis?

A. We should reject the hypothesis since 80 km/h is not in the interval found in (b).
B. We should not reject the hypothesis since 80 km/h is in the interval found in (b).
C. We should not reject the hypothesis since the sample mean is in the interval found in (b).
D. We should reject the hypothesis since the sample mean was not 80 km/h.
E. We should reject the hypothesis since 80 km/h is in the interval found in (b).

Part d) Decreasing the significance level of the hypothesis test above would (select all that apply)

A. either increase or decrease the Type I error probability.
B. not change the Type I error probability.
C. increase the Type I error probability.
D. decrease the Type I error probability.
E. not change the Type II error probability.

Answer 1

a. The degrees of freedom, df=n-1, where, n is sample size. Therefore, df=31-1=30.

b. The 95% c.i=xbar+-t_{alpha/2, df=n-1}(s/sqrt n), where, xbar is sample mean, s is sample standard deviation, n is sample size, and t denotes t critical at alpha/2, and df=n-1.

=82.72+-2.0423(4.61/sqrt 31)

=81.03, 84.41

c. Rule of thumb says that reject H0, if population mean is not in the interval. Option A is correct.

d. The significance level, alpha is the Type I error. Therefore, decreasing the significance level, decrease the Type I error probability. Option D.