Question

Suppose, for a random sample selected from a normally distributed population, X^- =51.70 and s= 8.8

Round your answers to 2 decimal places.

a. Construct a 95% confidence interval for μ assuming n = 16.

.... to ....

b. Construct a 90% confidence interval for μ assuming n = 16.

.... to ....

Is the width of the 90% confidence interval smaller than the width of the 95% confidence interval calculated in part a?

Yes or No?

c. Find a 95% confidence interval for μ assuming n = 25.

to Is the width of the 95% confidence interval for μ with n = 25 smaller than the width of the 95% confidence interval for μ with n = 16 calculated in part a?

Yes or No?

Answer 1

Solution:

a)

t value at 95% = 2.1314

CI = mean +/- t *(s/sqrt(n))
= 51.70 +/- 2.1314*(8.8/sqrt(16))
= (47.0109 , 56.3890 )

The 95% cI is (47.0109 , 56.3890 )

b)

t value at 90% = 1.7531

CI = mean +/- t *(s/sqrt(n))
= 51.70 +/- 1.7531*(8.8/sqrt(16))
= (47.8431 , 55.5568 )

The 95% cI is (47.8431 , 55.5568 )

yes

the width of the 90% confidence interval smaller than the width of the 95% confidence interval calculated in part a

c)

for n = 16

t value at 95% = 2.0639

CI = mean +/- t *(s/sqrt(n))
= 51.70 +/- 2.0639*(8.8/sqrt(16))
= (47.1594 , 56.2405 )

The 95% cI is (47.1594 , 56.2405 )

for n = 25

The 95% cI is (48.0675 , 55.3324 )

yes, the width of the 95% confidence interval for ? with n = 25 smaller than the width of the 95% confidence interval for ? with n = 16 calculated in part a

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