Question

Chapter 6, Section 5, Exercise 256 Do Hands Adapt to Water? Researchers in the UK designed a study to determine if skin wrinkled from submersion in water performed better at handling wet objects. They gathered 20 participants and had each move wet objects and a set of dry objects before and after submerging their hands in water for 30 minutes (order of trials was randomized). The response set of objects with wrinkled hands minus the time with unwrinkled hands. The mean difference for moving dry objects was 0.85 seconds with a standard deviation of 11.5 seconds. The mean difference for moving wet objects was -15.1 seconds with a standard deviation of 13.4 seconds. set of the time (seconds) it took to move the specific Kareklas, K., Nettle, D., and Smulders, T.V., "2013 Waterinduced finger wrinkles improve handling of wet objects." Biol. Lett vol 9 (2013). Your answer is partially correct. Try again. (a) Perform the appropriate test to determine if the wrinkled hands were significantly faster than unwrinkled hands at moving dry objects. State the null and alternative hypotheses. Use for in the response area. > Pi P2 0 0 0 >

(a) Perform the appropriate test to determine if the wrinkled hands were significantly faster than unwrinkled hands at moving dry objects. State the null and alternative hypotheses. Use μ1 for μd in the response area. H₀: __________ H_a: __________ (This is the question from the screenshot above)

(b) Find the p-value =

(c) Perform the appropriate test to determine if the wrinkled hands were significantly faster than unwrinkled hands at moving wet objects.State the null and alternative hypotheses.
Use μ1 for μd in the response area.  H₀: __________ H_a: __________

Answer 1

a) Sample mean, x̅d = 0.85

Sample standard deviation, sd = 11.5

Sample size, n = 20  

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd < 0

Test statistic:

t = (x̅d)/(sd/√n) = (0.85)/(11.5/√20) = 0.3305

df = n-1 = 19

b) p-value :

Left tailed p-value = T.DIST(0.3305, 19, 1) = 0.6277

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c) x̅d = -15.1, sd = 13.4, n = 20

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd < 0

Test statistic:

t = (x̅d)/(sd/√n) = (-15.1)/(13.4/√20) = -5.0395

df = n-1 = 19

p-value :

Left tailed p-value = T.DIST(-5.0395, 19, 1) = 0.0000