Respuesta :
P-value is0.1558 and the z- value is 1.012 when a random sample of 40 drivers was questioned to determine whether the commute time differs in the winter.
Given that,
The Census Bureau informs us that Cleveland, Ohio residents commute on average for 33 minutes. A random sample of 40 drivers was questioned to determine whether the commute time differs in the winter.
The standard deviation of the population is estimated to be 7.5 minutes, and the average commute time for the month of January was calculated to be 34.2 minutes.
Let = μ average commute time in winter.
So, Null Hypothesis, H₀ : μ = 33 minutes {means that the commuting times are same in the winter}
Alternate Hypothesis, H₁ : μ≠ 33 minutes {means that the commuting times are different in the winter}
The test results that would be utilised in this z test statistics for one sample given our understanding of population standard deviation;
T.S. = ((X bar-mean)/ standard deviation)/root n ~ N(0,1)
Here, X bar = sample mean commute time for the month of January = 34.2
Standard deviation = population standard deviation = 7.5 minutes
n = sample of drivers = 40
The value of z test statistics is 1.012.
Also,
P-value of the test statistics is given by;
P-value = P(Z > 1.012) = 1 - P(Z 1.012)
= 1 - 0.84423
= 0.1558
Therefore, P-value is0.1558 and the z- value is 1.012 when a random sample of 40 drivers was questioned to determine whether the commute time differs in the winter.
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