Question

To investigate the fluid mechanics of swimming, twenty swimmers each swam a specified distance in a water-filled pool and in a pool where the water was thickened with food grade guar gum to create a syrup-like consistency. Velocity, in meters per second, was recorded and the results are given in a table below. The researchers concluded that swimming in guar syrup does not change swimming speed. (Use a statistical computer package to calculate the P-value. Use ?_water ? ?_{guar syrup}. Round your test statistic to two decimal places and the P-value to three decimal places.)

Swimmer	Velocity (m/s)
	Water	Guar Syrup
1	1.52	1.68
2	1.95	1.03
3	1.65	1.23
4	1.20	1.94
5	1.54	1.90
6	1.65	1.24
7	1.44	1.28
8	1.87	0.97
9	1.93	1.50
10	1.16	1.75
11	1.06	1.99
12	1.89	1.70
13	1.65	1.90
14	1.85	1.81
15	1.20	1.86
16	1.18	1.14
17	1.70	1.26
18	1.75	0.91
19	1.07	0.96
20	1.97	1.83

t	=
df	=
P	=

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ? 0

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 0.53824

SE = s / sqrt(n)

S.E = 0.12035

DF = n - 1 = 20 -1

D.F = 19

t = [ (x₁ - x₂) - D ] / SE

t = 0.56

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 19 degrees of freedom is more extreme than 0.56; that is, less than -0.56 or greater than 0.56.

Thus, the P-value = 0.582

Interpret results. Since the P-value (0.582) is greater than the significance level (0.05), we have to accept the null hypothesis.

Do not reject H0. The mean difference appears to be zero.