use the given information to find the number of degrees of freedom the critical

Question

use the given information to find the number of degrees of freedom the critical values x 2 L. and x2r and the conference interval estimate of o it is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. nicotine and mental cigarettes 99% confidence n equal 29 s equal 0.29 mg

DF equals type a whole number

x 2 L equals round to three decimal places

X2 R are equals round to asthma place as needed.

the confidence interval estimate of ? mg<o ? mg

round to two decimal places as neede

you are the operations manager for an airline and you are considering the higher field-level for passengers in aisle seats how many randomly select the air passengers must you serve a assume that you want to be 9% confident that the sample percentage is within 1.5 percentage points of the true population percentage

assume that nothing is known about the percentage of passengers who prefer aisle seats.   n=? round up to the nearest integer

assume that a prior service suggests that about 32% of air passengers prefer and I'll seat n= we're on up to the nearest integer

Explanation / Answer

Part 1

Solution:

Here, we are given

Sample size = n = 29

Sample standard deviation = 0.29

Confidence level = 99%

We have to find confidence interval for population standard deviation.

Formula for confidence interval for population standard deviation is given as below:

Sqrt[(n – 1)*S2 / 2/2, n– 1 ] < < sqrt[(n – 1)*S2 / 21 -/2, n– 1 ]

Degrees of freedom = n – 1 = 29 – 1 = 28

Lower Chi square value = 21 -/2 = 2L = 12.4613

Upper Chi square value = 2/2, n– 1 = 2R = 50.9934

Sqrt[(29 – 1)*0.292 / 20.01/2,29– 1 ] < < sqrt[(29 – 1)*0.292 / 21 – 0.01/2,29– 1 ]

Sqrt[(28)*0.292 /50.9934 ] < < sqrt[(29 – 1)*0.292 / 12.4613 ]

0.2149< <0.4347

0.21 mg < < 0.43 mg

Part 2

We are given

Confidence level = 99%

E = 1.5% = 0.015

Percentage of passengers is not given, so default p-hat = 0.5

Sample size = n = (Z/E)^2*p-hat-(1 – p-hat)

Critical Z value = 2.5758

Sample size = n = (2.5758/0.015)^2*0.5*(1 – 0.5) = 7372.1073

Required Sample size = n = 7373

Now, we have to find number of air passengers when p-hat = 32% = 0.32

Sample size = n = (2.5758/0.015)^2*0.32*(1 – 0.32) = 6416.6822

Required Sample size = n = 6417

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