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Analysis of the venom of seven eight-day-old worker bees yielded the following observations on histamine content in nanograms: 649, 832, 418, 530, 384, 899, 755.
(a) Construct by band a 90% CI for the true mean histamine content for all worker bees of this age. What assumptions, if any, are needed for the validity of the CI?
(b) The true mean histamine content will be in the CI you constructed in part (a) with probability 90%. True or false?
(c) Find the confidence level of the CI (418, 832) for the population median histamine content.

Sagot :

Answer:

a.

• The population is normally distributed

• The 7 subjects represent a random sample from this population

b. True

c. 74.16%

Step-by-step explanation:

on using data [649, 832, 41 8, 530, 384, 899, 755] (please correct if wrong)

sample mean: 638.143

s: 201.72

Sample Mean = 638.143

SD = 201.72

Sample Size (n) = 7

Standard Error (SE) = SD/root(n) = 76.243

alpha (a) = 1-0.9 = 0.1

we use t-distribution as population standard deviation is unknown t(a/2, n-l ) = 1.9432

Margin of Error (ME) = SE = 148.1554

90% confidence interval is given by: Sample Mean +/- (Margin of Error) 638.143 +/- 148.1554 = (489.9876 , 786.2984)

• The population is normally distributed

• The 7 subjects represent a random sample from this populatio

(b) true, this is the definition of Cl

(c) (41 8, 432) has width = 14

hence, Margin of error = 7

sorted data is 384, 418, 530, 649, 755, 832, 899

Here 418 and 432 are 2nd and 6th entry of sample size of 7

Since median is given by (1 + n/2 + z(alpha/2)*sqrt(n) /2 )th entry on the right, we conclude that 1 + (3.5 + this gives z = 1.1339 ==> alpha = 0.2584

Hence this Cl is of 74.16% (approximately)

Note that your course might have given a different formula for calculating Cl medians from sample.