Question

In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some increase the P-value by a small amount and therefore produce a slightly more "conservative" answer Is fishing better from a boat or from the shore? Pyramid Lake is located on the Palute Indian Reservation in Nevada. Presidents, movie stars, and people who just want to catch fish go to Pyramid Lake for really large cutthroat trout. Let row B represent hours per fish caught fishing from the shore, and let row A represent hours per fish caught using a boat. The following data are paired by month from October through April. Oct Nov Dec Jan Feb March April B: Shore 1.5 1.8 1.9 3.2 3.9 3.6 3.3 A: Boat 1.3 1.5 1.5 2.2 3.3 3.0 3.8 Use a 1% level of significance to test if there is a difference in the population mean hours per fish caught using a boat compared with fishing from the shore. (Let d = B - A) (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? O Mo: 40 = 0; H:M, < 0; left-tailed HoMg = 0; H: 4,#0; two-tailed HO: M*0; H, Ha-0; two-tailed O Ho: M, -0; H:40 >0; right-tailed (b) What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that d has an approximately uniform distribution. The standard normal. We assume that d has an approximately normal distribution The Student's t. We assume that d has an approximately uniform distribution. The Student's t. We assume that d has an approximately normal distribution. What is the value of the sample test statistic? (Round your answer to three decimal places.)

Answer 1

a) The level of significance is 0.01.

Option - B) $\dpi{120} H_{0}: \mu_{d} = 0; H_{1}:\mu_{d}\neq 0, two-tailed test$

b) Option - D) The student's t. We assume that d has an approximately normal distribution.

$\dpi{120} \bar d$ = (0.2 + 0.3 + 0.4 + 1 + 0.6 + 0.6 + (-0.5))/7 = 0.37

s_d = sqrt(((0.2 - 0.37) ^2 + (0.3 - 0.37)^2 + (0.4 - 0.37)^2 + (1 - 0.37)^2 + (0.6 - 0.37)^2 + (0.6 - 0.37)^2 + (-0.5 - 0.37)^2)/6) = 0.4645

The test statistic t = ($\dpi{120} \bar d$ - D)/(s_d/$\dpi{120} \sqrt n$)

= (0.37 - 0)/(0.4645/$\dpi{120} \sqrt 7$)

= 2.107

c) P-value = 2 * P(T > 2.107)

= 2 * (1 - P(T < 2.107))

= 2 * (1 - 0.9601)

= 0.0618

Option - D) 0.050 < p-value < 0.100

Option - A is correct graph.

d) Option - B) At the alpha = 0.01 level we fall to reject the null hypothesis and conclude the data are not statistically significant.

E) Fail to reject H₀, there is insufficient evidence to claim a difference in population mean hours per fish between boat fishing and shore fishing.

2) $\dpi{120} \bar d$ = (-12 + (-11) + (-9) + 4 + 10 + (-15) + 6 + 7)/8 = -2.5

s_d = sqrt(((-12 + 2.5)^2 + (-11 + 2.5)^2 + (-9 + 2.5)^2 + (4 + 2.5)^2 + (10 + 2.5)^2 + (-15 + 2.5)^2 + (6 + 2.5)^2 + (7 + 2.5)^2)/7) = 10.1559

The test statistic t = ($\dpi{120} \bar d$ - D)/(s_d/$\dpi{120} \sqrt n$)

= (-2.5 - 0)/(10.1559/$\dpi{120} \sqrt 8$)

= -0.696

critical value= +/- 2.365

Since the test statistic value is not less than the negative critical value, so we should not reject the null hypothesis.

Option - C) Fail to reject the null hypothesis, there is insufficient evidence to claim a difference in population mean percentage increases for corporate revenue and CEO salary.

P-value = 2 * P(T < -0.696)

= 2 * 0.2544 = 0.5088

Since the p-value > $\dpi{120} \alpha$ , so don't reject the null hypothesis.

Option - B) The conclusions obtained by using both methods are the same.