Question

1. Use this scenario to answer questions 41-42.

   A fisherman wants to compare how many fish he catches on Lake Cumberland from a boat to on shore. The data follow with the unit of measure bing fish caught per hour.

   Sample 1 (Boat)		Sample 2 (Shore)
   3.8		1.2
   3.2		1.4
   3.6		3.2
   2.2		2.2
   4.2		1.8
   		1.6

   Construct a 99% confidence interval for . Assume .

   		(-2.958, -0.042)
   		(-0.042, 2.958)
   		(-2.958, 0.042)
   		(0.042, 2.958)

4 points   

QUESTION 42

1. Interpret the confidence Interval.

   		We are 99% confident that fishing from a boat is better than fishing from the shore by between 0.042 and 2.958 fish per hour.
   		We are 99% confident that fishing from shore is better than fishing from a boat by between 0.042 and 2.958 fish per hour.
   		It is unclear if fishing from a boat or from shore results in catching more fish. We are 99% confident that if fishing in a boat is best it is by at most 0.042 fish per hour and if fishing on shore is best it is by at most 2.958 fish per hour.
   		It is unclear if fishing from a boat or from shore results in catching more fish. We are 99% confident that if fishing in a boat is best it is by at most 2.958 fish per hour and if fishing on shore is best it is by at most 0.042 fish per hour.

Answer 1

The sample size is n = 5 . The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	X	X2
	3.8	14.44
	3.2	10.24
	3.6	12.96
	2.2	4.84
	4.2	17.64
Sum =	17	60.12

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

The sample size is n = 6n=6. The provided sample data along with the data required to compute the sample mean \bar XXˉ and sample variance s^2s2 are shown in the table below:

	X	X2
	1.2	1.44
	1.4	1.96
	3.2	10.24
	2.2	4.84
	1.8	3.24
	1.6	2.56
Sum =	11.4	24.28

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

We need to construct the 99% confidence interval for the difference between the population means μ1​−μ2​, for the case that the population standard deviations are not known. The following information has been provided about each of the samples:

Sample Mean 1	3.4
Sample Standard Deviation 1	0.762
Sample Size 1	5
Sample Mean 2	1.9
Sample Standard Deviation 2	0.724
Sample Size 2	6

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = n_1 + n_2 -2 = 5 + 6 - 2 = 9

The critical value for α=0.01 and df = 9 degrees of freedom is t_c = 3.25 The corresponding confidence interval is computed as shown below:

Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows:

Since we assume that the population variances are equal, the standard error is computed as follows:

Now, we finally compute the confidence interval:

CI =( -2.958, -0.042 )

We are 99% confident that fishing from a boat is better than fishing from the shore by between 0.042 and 2.958 fish per hour.