Question
Jan Villaroel
Topic: Statistics Posted 10 months ago
To determine if chocolate milk is as effective as other carbohydrate replacement drinks, 10 trained cyclists performed

To determine if chocolate milk is as effective as other carbohydrate replacement drinks, 10 trained cyclists
performed an intense workout followed by a drink and a rest period. At the end of the rest period, each
cyclist performed a 40-km timed trial in which they were instructed to try to cycle this distance as fast as
possible. The time to completion was then recorded. Each cyclist completed the entire regimen on two
different days. On one day, the drink provided was chocolate milk, and on the other day the drink provided
was a carbohydrate replacement drink. Data are given in the table below.

Time to Complete Trial (in minutes)
Cyclist 1 2 3 4 5 6 7 8 9 10
Chocolate Milk 74 75 77 79 81 82 82 84 87 87
Carbohydrate Replacement 73 75 76 82 84 89 93 96 99 101

Is there evidence that the mean time to complete the trial is shorter after chocolate milk than after a
carbohydrate replacement drink? Use a significance level of 𝛼 = 0.05. (Use 𝜇1 for chocolate milk, 𝜇2 for the
carbohydrate replacement, and 𝜇d = 𝜇1 − 𝜇2 .)

State the appropriate null and alternative hypotheses

Find the test statistic. (Round your answer to two decimal places.)

Find the test statistic. (Round your answer to two decimal places.)



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Koala Tutor
Feb 4, 2024
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The appropriate null and alternative hypotheses are:

Null hypothesis (H0): The mean time to complete the trial after chocolate milk is equal to the mean time to complete the trial after a carbohydrate replacement drink (μ1 = μ2).

Alternative hypothesis (Ha): The mean time to complete the trial after chocolate milk is shorter than the mean time to complete the trial after a carbohydrate replacement drink (μ1 < μ2).

To find the test statistic, we will use the formula for the difference in means:

t = (𝑋̄1 - 𝑋̄2) / (s / √n)

Where:
𝑋̄1 = mean time to complete the trial after chocolate milk
𝑋̄2 = mean time to complete the trial after a carbohydrate replacement drink
s = pooled standard deviation
n = number of observations (in this case, 10)

First, let's calculate the mean and standard deviation for each group:

Mean time to complete the trial after chocolate milk (𝑋̄1) = (74 + 75 + 77 + 79 + 81 + 82 + 82 + 84 + 87 + 87) / 10 = 815 / 10 = 81.5

Mean time to complete the trial after a carbohydrate replacement drink (𝑋̄2) = (73 + 75 + 76 + 82 + 84 + 89 + 93 + 96 + 99 + 101) / 10 = 888 / 10 = 88.8

Next, calculate the pooled standard deviation:

s = √(((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / (n1 + n2 - 2))

Where:
n1 = number of observations in group 1 (10)
n2 = number of observations in group 2 (10)
s1 = standard deviation of group 1
s2 = standard deviation of group 2

To find the standard deviation for each group, we can use the following formula:

s = √((∑(x - 𝑋̄)^2) / (n - 1))

For the chocolate milk group:

s1 = √(((74 - 81.5)^2 + (75 - 81.5)^2 + (77 - 81.5)^2 + (79 - 81.5)^2 + (81 - 81.5)^2 + (82 - 81.5)^2 + (82 - 81.5)^2 + (84 - 81.5)^2 + (87 - 81.5)^2 + (87 - 81.5)^2) / (10 - 1))

s1 = √((40.25 + 26.25 + 10.25 + 2.25 + 0.25 + 0.25 + 0.25 + 6.25 + 25.25 + 25.25) / 9)

s1 = √(146.5 / 9)

s1 = √16.278

s1 = 4.03 (rounded to two decimal places)

For the carbohydrate replacement group:

s2 = √(((73 - 88.8)^2 + (75 - 88.8)^2 + (76 - 88.8)^2 + (82 - 88.8)^2 + (84 - 88.8)^2 + (89 - 88.8)^2 + (93 - 88.8)^2 + (96 - 88.8)^2 + (99 - 88.8)^2 + (101 - 88.8)^2) / (10 - 1))

s2 = √((237.16 + 160.16 + 153.76 + 43.56 + 18.36 + 0.16 + 20.16 + 54.76 + 101.16 + 132.76) / 9)

s2 = √(921.72 / 9)

s2 = √102.41

s2 = 10.12 (rounded to two decimal places)

Now we can substitute the values into the t statistic formula:

t = (𝑋̄1 - 𝑋̄2) / (s / √n)

t = (81.5 - 88.8) / (sqrt(((10 - 1) * 4.03^2 + (10 - 1) * 10.12^2) / (10 + 10 - 2)) / sqrt(10))

t = -7.3 / (sqrt(((9) * (16.278^2) + (9) * (10.12^2)) / 18) / sqrt(10))

t = -7.3 / (sqrt((141.22 + 924.88144) / 18) / sqrt(10))

t = -7.3 / (sqrt(1165.10144 / 18) / sqrt(10))

t = -7.3 / (sqrt(64.727858 / 18) / sqrt(10))

t = -7.3 / (sqrt(3.596547 / 18) / sqrt(10))

t = -7.3 / (sqrt(0.199808167) / sqrt(10))

t = -7.3 / (0.4464348235 / 3.16227766)

t = -7.3 / (0.141425 / 3.16227766)

t = -7.3 / 0.044774

t = -163.227

The test statistic is approximately -163.23 (rounded to two decimal places).

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