In: Statistics and Probability

Calculate the margin of error and construct the confidence interval for the population mean using the Student's t-distribution (you may assume the population data is normally distributed).

T-Distribution Table

**a.** x̄ =87.3, n=64, s=19.6, x̄ =87.3,
n=64, s=19.6, 98% confidence

E=E=

Round to two decimal places

< μ < < μ <

Round to two decimal places

**b.** x̄ =31.6, n=44, s=14.6, x̄ =31.6,
n=44, s=14.6, 90% confidence

E=E=

Round to two decimal places

< μ < < μ <

Round to two decimal places

Please provide correct answers. thanks

Solution :

a) degrees of freedom = n - 1 = 64 - 1 = 63

t/2,df = t0.01,63 = 2.387

Margin of error = E = t/2,df * (s /n)

= 2.387 * ( 19.6 / 64)

Margin of error = E = 5.85

The 98% confidence interval estimate of the population mean is,

- E < < + E

87.3 - 5.85 < < 87.3 + 5.85

( 81.45 < < 93.15 )

b) degrees of freedom = n - 1 = 44 - 1 = 43

t/2,df = t0.05,43 = 1.681

Margin of error = E = t/2,df * (s /n)

= 1.681 * ( 14.6 / 44)

Margin of error = E = 3.70

The 90% confidence interval estimate of the population mean is,

- E < < + E

31.6 - 3.70 < < 31.6 + 3.70

( 27.90 < < 35.30 )

Calculate the margin of error and construct a confidence
interval for the population proportion using the normal
approximation to the p̂ p̂ -distribution (if
it is appropriate to do so).
Standard Normal Distribution Table
a. p̂ =0.9, n=160, α =0.2
p̂ =0.9, n=160, α =0.2
E=E=
Round to four decimal places
Enter 0 if normal approximation cannot be used
< p < < p <
Round to four decimal places
Enter 0 if normal approximation cannot be used
b. p̂ =0.45, n=140, α
=0.2 p̂ =0.45, n=140, α =0.2...

Match the margin of error for an 80% confidence interval to
estimate the population mean with sigma σ equals=50
with its corresponding sample sizes.
Question 5 options:
a.) 8.07. 1.) n= 34
b.) 9.45 2.) n= 46
c.) 10.99 3.) n= 63

Determine the margin of error for a confidence interval to
estimate the population mean with nequals24 and s = 13.5 for the
confidence levels below. a) 80% b) 90% c) 99%

Determine the margin of error for a confidence interval to
estimate the population mean with n=34 and σ=48 for the following
confidence levels.
91%
95%
98%

Determine the margin of error for a confidence interval to
estimate the population mean with n=22 and s =14.4 for the
confidence levels below
a. 80%
b.90%
c. 99%
The margin of error for an 80% confidence interval
is_______?
repeat for 90% and 99%

A confidence interval
for a population mean has length 26.
a) Determine the margin of error.
b) If the sample mean is 58.9, obtain the confidence
interval.
Confidence interval: ( , ).
Lindsey thinks a
certain potato chip maker is putting fewer chips in their regular
bags of chips. From a random sample of 23 bags of potato chips she
calculated a P value of 0.029 for the sample.
(a) At a 5% level of significance, is there evidence that...

Determine the margin of error for a 99% confidence interval to
estimate the population mean when s = 43 for the sample sizes
below. a) n=12 b) n=25 c) n=46 a) The margin of error for a
99% confidence interval when n=12 is _.

Determine the margin of error for a 99% confidence interval to
estimate the population mean when s=45 for the sample sizes of
n=15, n=34, n=54. (Find the margin of error for each interval when
n=x)
Determine the margin of error for a confidence interval to
estimate the population mean with n=24 and s=12.3 for confidence
levels 80%, 90%, 99%.

Determine the margin of error for a confidence interval to
estimate the population mean with n=18 and s=12.1 for the
confidence levels below.
A) The margin of error for an 80% confidence interval
is:
B) The margin of error for an 90% confidence interval
is:
C) The margin of error for an 99% confidence interval
is:

Determine the margin of error for a confidence interval to
estimate the population proportion for the following confidence
levels with a sample proportion equal to 0.36 and n=125.
a. 90%
b. 95%
c. 98%
a. The margin of error for a confidence interval to estimate the
population proportion for the 90% confidence level is _
b. The margin of error for a confidence interval to estimate the
population proportion for the 95% confidence level is _
c. The margin of...

ADVERTISEMENT

ADVERTISEMENT

Latest Questions

- Write a fortran 90 program that sets up a 4x4 2D real array A and associate...
- Many people have very active social media lives, and it is now quite common for potential...
- Edom Company, the lessor, enters into a lease with Davis Company to lease equipment to Davis...
- Write a Powershell script to create a Windows user account that is a non-admin user. This...
- Quick Fix Ltd is a manufacturing company engaged in the production of adhesives. The company has...
- Assume that the summary is for your international marketing manager(s)/department. Consider the COVID-19 virus: In general,...
- In cryptography, a Caesar cipher, also known as Caesar's cipher, the shift cipher, Caesar's code or...

ADVERTISEMENT