An agricultural researcher is interested in estimating the mean length of the gr
ID: 3125336 • Letter: A
Question
An agricultural researcher is interested in estimating the mean length of the growing season in a region. Treating the last 10 years as a simple random sample, he obtains the following data, which represent the number of days of the growing season.
155 164 150 141 173 186 191 185 161 156
Construct a 95% confidence interval for the mean lenght of the growing season in the region
___, ____ (use ascending order, round to two decimal places)
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A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1033 people age 15 or older, the mean amount of time spend eating or drinking per day is 1.01 hours with a standard deviation of .74 hour.
(a) A histogram of time spent eating and drinking each day is skewed right. Use this result to explain why a large sample size is needed to construct a confidence interval for the mean time spent eating and drinking each day.
A.The distribution of the sample mean will always be approximately normal.
B. Since the distribution of time spent eating and drinking each day is not normally distributed the sample must be large so that the distribution of the sample mean will be approximately normal.
C. Since the distribution of time spent eating and drinking each day is normally distributed the sample must be large so that the distribution of the sample mean will be approximately normal.
D. The distribution of the sample mean will never be approximately normal.
(b). In 2010, there were over 200 million people nationally age 15 or older. Explain why this, along with the fact that the data were obtained using random sample, satisifed the requirements for constructing a confidence interval.
A. Sample size less than 10% of population
B. Sample size greater than 5% of population.
C. Sample size less than 5% of population
D Sample size greater than 10% population
(c) Determine and interpret a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking a day.
A. There is a 95% probability that the mean amount of time spent eating or drinking per day is between ___ and ___ hours.
B. The nutritionist is 95% confident that hte amount of time spent eating or drinking per day is between __ and __ hours.
C. The nutritionist is 95% confident that mean amount of time spent eating and drinking per day is between __ and __ hours.
D. The requirements for constructin a confidence interval are not satisified.
(d) Could the interval be used to estimate the mean amout of time a 9 year old spends eating and drinking each day? Explain.
A. No. The interval is about individual time spent eating or drinking per day and cannot be used to find the mean time spent eating or drinking per day for specific age.
B. Yes, the interval is about the mean amount of time spent eating or drinking per day for people age 15 or older and can be used to find the mean amount of time spent eating or drinking per day for 9 year olds.
C. No the interval is about people age 15 or older. The mean amount of time spent eating or drinking per day for 9 year olds may differ.
D . Yes, the interval is about individual time spent eating or drinking per day and can be used to find the mean amount of time a 9 year old spends eating or drinking each day.
E. A confidence interval could not be constructed in part (c).
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265000 143000 279900
459900 211900 205800
191500 147800 219900
276500 187500 264900
The following data represent the asking price of a simple random sample of homes for sale.
(a). Construct a 99% confidence interval with the outlier included
____,____ (round the nearest integer)
(b). Construct a 99% confidence interval with the outlier removed
___,___ (round to the nearest integer)
(c) Comment on the effect the outlier has on the confidence interval.
A. the outlier has no effect on the width of confidence interval
B. the outlier caused the width of confidence interval to increase
C. the outlier caused the width of confidence interval to decrease
265000 143000 279900
459900 211900 205800
191500 147800 219900
276500 187500 264900
Explanation / Answer
An agricultural researcher is interested in estimating the mean length of the growing season in a region. Treating the last 10 years as a simple random sample, he obtains the following data, which represent the number of days of the growing season.
155 164 150 141 173 186 191 185 161 156
Construct a 95% confidence interval for the mean lenght of the growing season in the region.
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 166.2
t(alpha/2) = critical t for the confidence interval = 2.262157163
s = sample standard deviation = 16.8839174
n = sample size = 10
df = n - 1 = 9
Thus,
Lower bound = 154.1219731
Upper bound = 178.2780269
Thus, the confidence interval is
( 154.12 , 178.28 ) [ANSWER]
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