In the Department of Education at UR University, student records suggest that th
ID: 3155262 • Letter: I
Question
In the Department of Education at UR University, student records suggest that the population of students spends an average of 6.00 hours per week playing organized sports. The population's standard deviation is 3.40 hours per week. Based on a sample of 49 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates. a. Compute the standard error of the sample mean. (Round your answer to 2 decimal places.) Standard error b. What is the chance HLI will find a sample mean between 5.4 and 6.6 hours? (Round z and standard error values to 2 decimal places and final answer to 4 decimal places.) Chance c. Calculate the probability that the sample mean will be between 5.6 and 6.4 hours. (Round z and standard error values to 2 decimal places and final answer to 4 decimal places.) Probability d. How strange would it be to obtain a sample mean greater than 8.40 hours?
Explanation / Answer
a. Compute the standard error of the sample mean. (Round your answer to 2 decimal places.) Standard error
SE = sigma/sqrt(n) = 3.40/sqrt(49) = 0.485714286 [ANSWER]
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b. What is the chance HLI will find a sample mean between 5.4 and 6.6 hours? (Round z and standard error values to 2 decimal places and final answer to 4 decimal places.) Chance
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 5.4
x2 = upper bound = 6.6
u = mean = 6
s = standard deviation = 0.49
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1.22
z2 = upper z score = (x2 - u) / s = 1.22
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.1112
P(z < z2) = 0.8888
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.7776 [ANSWER]
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c. Calculate the probability that the sample mean will be between 5.6 and 6.4 hours. (Round z and standard error values to 2 decimal places and final answer to 4 decimal places.) Probability
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 5.6
x2 = upper bound = 6.4
u = mean = 6
s = standard deviation = 0.49
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.82
z2 = upper z score = (x2 - u) / s = 0.82
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.2061
P(z < z2) = 0.7939
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.5878 [ANSWER]
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d. How strange would it be to obtain a sample mean greater than 8.40 hours?
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 8.4
u = mean = 6
s = standard deviation = 0.49
Thus,
z = (x - u) / s = 4.9
Thus, using a table/technology, the right tailed area of this is
P(z > 4.9 ) = 0 [ANSWER]