Question

A simple random sample of size n is drawn. The sample​ mean,

x overbarx​,

is found to be

18.618.6​,

and the sample standard​ deviation, s, is found to be

4.84.8.

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Click the icon to view the table of areas under the​ t-distribution.

​(a) Construct a

9595​%

confidence interval about

muμ

if the sample​ size, n, is

3434.

Lower​ bound:

nothing​;

Upper​ bound: nothing

​(Use ascending order. Round to two decimal places as​ needed.)

​(b) Construct a

9595​%

confidence interval about

muμ

if the sample​ size, n, is

5151.

Lower​ bound:

nothing​;

Upper​ bound: nothing

​(Use ascending order. Round to two decimal places as​ needed.)

How does increasing the sample size affect the margin of​ error, E?

A.

The margin of error increases.

B.

The margin of error decreases.

C.

The margin of error does not change.

​(c) Construct a

9999​%

confidence interval about

muμ

if the sample​ size, n, is

3434.

Lower​ bound:

nothing​;

Upper​ bound: nothing

​(Use ascending order. Round to two decimal places as​ needed.)

Compare the results to those obtained in part​ (a). How does increasing the level of confidence affect the size of the margin of​ error, E?

A.

The margin of error decreases.

B.

The margin of error increases.

C.

The margin of error does not change.

​(d) If the sample size is

1212​,

what conditions must be satisfied to compute the confidence​ interval?

A.

The sample data must come from a population that is normally distributed with no outliers.

B.

The sample must come from a population that is normally distributed and the sample size must be large.

C.

The sample size must be large and the sample should not have any outliers.

Answer 1

We have given :

A simple random sample of size n is drawn.

x̄ = sample mean  = 18.6, and the

s = sample standard​ deviation = 4.8.

Click the icon to view the table of areas under the​ t-distribution.

## ​(a) Construct a 95​% confidence interval about mμ

if the sample​ size, n = 34

Answer : we know that

mμ = [ x̄ ± Critical value * standard error ]

Critical value = t ( α /2 , df ) here

df = degree of freedom = n-1 = 34 -1 = 33

therefore ,

Critical value = t ( 0.05 /2 , 33 ) = ± 2.0345 ( used statistical t table )

standard error = s / SQRT( n) = 4.8 / SQRT ( 34)

= 4.8 / SQRT (34) = 0.82319

# margin of error = Critical value * standard error

= 2.0345 * 0.82319 = 1.674799

mμ = [ 18.6 ± ( 2.0345 * 0.82319) ]   

mμ = [ 18.6   ± 1.674799 ]

Lower​ bound: = [ x̄ - margin or error ] =  [ 18.6 -   1.674799 ]

= 16 .9252 ie 16.93

Upper ​ bound: = [ x̄ +  margin or error ] =   [ 18.6 +   1.674799 ]

= 20 .2748 ie 20.27

# Interpretation : we can say 95 % confident that population mean mμ is lies within 16.93 to 20.27

## ​(b) Construct a 95​% confidence interval about mμ

if the sample​ size, n, is = 51.

Answer : we know that

mμ = [ x̄ ± Critical value * standard error ]

Critical value = t ( α /2 , df ) here df = degree of freedom = n  - 1

= 51  -1 = 50   therefore ,

Critical value = t ( 0.05 /2 , 50 ) = ± 2. 008559 ( used statistical t table )

standard error = s / SQRT( n)

= 4.8 / SQRT ( 51 ) = 0.672134

# margin of error = Critical value * standard error

= 2. 008559   * 0.672134 =  1.350022

mμ = [ 18.6 ± ( 2. 008559   * 0.672134) ]   

mμ = [ 18.6   ± 1.350022]

Lower​ bound: = [ x̄ - margin or error ] [ 18.6 -   1.350022 ]

= 17.24998 ie 17. 25

Upper ​ bound: = [ x̄ +  margin or error ] = [ 18.6 +   1.350022 ]

= 19.95002 ie 19.95  

# Interpretation : we can say 95 % confident that population mean mμ is lies within 17.25 to 19.95 .

## How does increasing the sample size affect the margin of​ error, E?

Answer :

B. The margin of error decreases.

​## (c) Construct a 99​% confidence interval about mμ

if the sample​ size, n, is = 34.

Answer : we know that

mμ = [ x̄ ± Critical value * standard error ]

Critical value = t ( α /2 , df ) here df = degree of freedom = n-1

= 34 -1 = 33 therefore ,

Critical value = t ( 0.01 /2 , 33 ) = ± 2.733277 ( used statistical t table )

standard error = s /  SQRT( n) = 4.8 / SQRT ( 34)

= 4.8 / SQRT (34) = 0.82319

# margin of error = Critical value * standard error

= 2.733277 * 0.82319 = 2.250015

mμ = [ 18.6 ± ( 2.733277 * 0.82319 ) ]   

mμ = [ 18.6   ± 2.250015 ]

Lower​ bound: = [ x̄ - margin or error ] =  [ 18.6 -   2.250015 ]

= 16 .34999 ie 16.35

Upper ​ bound: = [ x̄ +  margin or error ] =   [ 18.6 +   2.250015 ]

= 20. 85001 ie 20.85  

# Interpretation : we can say 99 % confident that population mean mμ is lies within 16.35 to 20.85

## Compare the results to those obtained in part​ (a). How does increasing the level of confidence affect the size of the margin of​ error, E?

Answer :

B. The margin of error increases.

## ​(d) If the sample size is 12​, what conditions must be satisfied to compute the confidence​ interval?

Answer :

A. The sample data must come from a population that is normally distributed with no outliers.