(20.60 S-AQ) To study the metabolism of insects, researchers fed cockroaches mea
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(20.60 S-AQ) To study the metabolism of insects, researchers fed cockroaches measured amounts of a sugar solution. After 2, 5, and 10 hours, they dissected some of the cockroaches and measured the amount of sugar in various tissues. Five roaches fed the sugar D-glucose and dissected after 10 hours had the following amounts (in micrograms) of D-glucose in their hindguts:
The researchers gave a 98% confidence interval for the mean amount of D-glucose in cockroach hindguts under these conditions. The insects are a random sample from a uniform population grown in the laboratory. We therefore expect responses to be Normal.
The mean (±0.01) and the standard deviation (±0.0001) of the SRS are:
x¯ = s = .
The critical value (±0.001) from the distribution for 98% confidence interval is:
t* = .
What confidence interval (±0.01) did the researchers give? to .
(20.60 S-AQ) To study the metabolism of insects, researchers fed cockroaches measured amounts of a sugar solution. After 2, 5, and 10 hours, they dissected some of the cockroaches and measured the amount of sugar in various tissues. Five roaches fed the sugar D-glucose and dissected after 10 hours had the following amounts (in micrograms) of D-glucose in their hindguts:
Amounts of D-glucose 55.95 68.24 52.73 21.5 23.78The researchers gave a 98% confidence interval for the mean amount of D-glucose in cockroach hindguts under these conditions. The insects are a random sample from a uniform population grown in the laboratory. We therefore expect responses to be Normal.
The mean (±0.01) and the standard deviation (±0.0001) of the SRS are:
x¯ = s = .
The critical value (±0.001) from the distribution for 98% confidence interval is:
t* = .
What confidence interval (±0.01) did the researchers give? to .
Explanation / Answer
By technology,
X = sample mean = 44.44 [ANSWER]
s = sample standard deviation = 20.74080158 [ANSWER]
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As
alpha/2 = (1 - confidence level)/2 = 0.01
df = n - 1 = 4, by table/technology for alpha/2 = 0.01,
t(alpha/2) = critical t for the confidence interval = 3.747 [ANSWER]
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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 = 0.01
X = sample mean = 44.44
t(alpha/2) = critical t for the confidence interval = 3.746947388
s = sample standard deviation = 20.74080158
n = sample size = 5
df = n - 1 = 4
Thus,
Lower bound = 9.684933039
Upper bound = 79.19506696
Thus, the confidence interval is
( 9.684933039 , 79.19506696 )
or
( 9.68 , 79.20 ) [ANSWER]
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