Y6.2.35-T Assigned MediaQuestion HelpUse the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why.Interpret the results.In a random sample of 48 people, the mean body mass index (BMI) was 28.3 and the standard deviation was 6.01.Which distribution should be used to construct the confidence interval? Choose the correct answer below.A. Use a t-distribution because the sample is random, the population is normal, and o is unknown.B. Use a normal distribution because the sample is random, the population is normal, and o is known.C. Use a t-distribution because the sample is random, n2 30, and o is unknown.D. Use a normal distribution because the sample is random, n2 30, and o is known.er a normal distribution nor a t-distribution can be used because either the sample is not random, or n< 30, and the population is not known to be normal.rfSelect the correct choice below and, if necessary, fill in any answer boxes to complete your choice.O A. The 95% confidence interval is ( ).(Round to two decimal places as needed.)B. Neither distribution can be used to construct the confidence interval.greeClick to select and enter your answer(s) and then click Check Answer.へ@e Type here to searchPrtScn17HomeFBEndPgUpFSF64.58.to

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Description: Y6.2.35-T Assigned MediaQuestion HelpUse the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why.Interpret the res

Given that Random sample size(n) =48 Mean body mass index (x¯)=28.3 Sample standard deviation (s)…

Answered: In a random sample of 48 people, the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Step 1 - Writing given data

Given that

Random sample size(n) =48

Mean body mass index ( $\bar{x}$ )=28.3

Sample standard deviation (s) =6.01

Confidence level (C)= 95% = 0.95

Level of significance ( $α$ ) = $(1 - 0.95) = 0.05$

We use t-distribution because the sample is random and population standard deviation is unknown.

Critical value ( $t_{(\frac{α}{2)}}$ )

Degrees of freedom (df) =n-1

df = 48-1 =47

From t-table

At $(\frac{α}{2}) = 0.025$ ; df=47 ; One-tailed test

$t_{(\frac{α}{2)}}$ = 2.01 =t_c (t-critical value)

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