Complete the hypothesis test with alternative hypothesis ƒÊd �‚ 0 based on the paired data that follows and d = O - Y. Use ƒ¿ = 0.01. Assume normality.

Oldest 174 173 169 180 51
Youngest 168 169 175 187 53

(a) Find t. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer correct to four decimal places.)

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What does mean?

ƒÊd �‚ 0

Oldest Youngest Difference

174 168 6
173 169 4
169 175 -6
180 187 -7
51 53 -2

Sum of d = -5

d bar = -5/5 = -1

S = 5.83
n = 5

t = -1/[ (5.83/sqrt((5))]
t = -.38

P-value = 0.7209

To complete the hypothesis test with an alternative hypothesis based on the paired data, we will follow these steps:

Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): µd = 0 (No difference between the two sets of data)
- Alternative hypothesis (Ha): µd ≠ 0 (There is a difference between the two sets of data)

Step 2: Calculate the difference (d) between the two matched pairs:
d = O - Y
d = (174 - 168), (173 - 169), (169 - 175), (180 - 187), (51 - 53)
d = 6, 4, -6, -7, -2

Step 3: Calculate the sample mean (x̄) and the sample standard deviation (s) of the differences:
x̄ = (Σd) / n
s = √((Σ(d - x̄)^2) / (n - 1))

x̄ = (6 + 4 + (-6) + (-7) + (-2)) / 5
x̄ = -5 / 5
x̄ = -1

s = √(((6 - (-1))^2 + (4 - (-1))^2 + (-6 - (-1))^2 + (-7 - (-1))^2 + (-2 - (-1))^2) / (5 - 1))
s = √((49 + 25 + 25 + 36 + 1) / 4)
s = √(136 / 4)
s = √34

Step 4: Calculate the test statistic (t):
t = (x̄ - µd) / (s / √n)
t = (-1 - 0) / (√34 / √5)
t = -1 / (√(34/5))
t ≈ -1 / 0.606
t ≈ -1.65

Step 5: Find the p-value:
To find the p-value, we need to consult the t-distribution table or use statistical software. For a two-tailed test, the p-value is the probability of observing a more extreme test statistic than the one calculated. In this case, the p-value is the probability of observing a t-value less than -1.65 or greater than 1.65.

From the t-distribution table, with degrees of freedom (df) = n - 1 = 5 - 1 = 4 and a significance level (α) of 0.01, we find that the p-value is approximately 0.10.

Therefore, the p-value is approximately 0.10.