Refer to the technology output given to the right that results from measured hemoglobin levels (g/dL) in 100 randomly selected adult females. The Tntrval confidence level of 90% was used. 13.032,13.418) x= 13.225 a. Express the confidence interval in the format that uses the "less than" symbol. Assume that the original listed data use two decimal places, andSx-1.164 round the confidence interval limits accordingly. b. Identify the best point estimate of μ and the margin of error. c. In constructing the confidence interval estimate of μ, why is it not necessary to confirm that the sample data appear to be from a population with a normal distribution?

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Answer:

Step-by-step explanation:

a)

Confidence interval in less than symbol expressed as

[tex]\bar{x} - E < \mu < \bar{x} + E [/tex]

Where [tex]\bar{x}[/tex] is sample mean and  [tex]E[/tex] is margin of error.

[tex]13.03 < \mu 13.42 [/tex]

b)

The given t interval is [tex](13.032 , 13.418 ) [/tex]

That is [tex]\bar{x} - E = 13.032[/tex] and [tex]\bar{x} + E = 13.418[/tex]

Solve these two equation by adding together.

[tex]2 \bar{x} = 13.032 + 13.418 \\\\\bar{x} = 13.225[/tex]

Solve this value of \bar{x} in equation [tex]\bar{x} - E = 13.032[/tex] and solve for [tex]E[/tex]

[tex]13.225 - E = 13.032 \\\\E = 0.193 [/tex]

Best point estimate of [tex]\mu = \bar{x} = 13.225 [/tex]

Best point estimate of margin of error = 0.193

c)

Since sample size = 100 which is sufficiently large (Greater than 30) , it is no need to confirm that

sample data appear to be form a population with normal distribution.

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