I have some questions that I\'m stuck on in my homework its all one question wit

Question

I have some questions that I'm stuck on in my homework its all one question with multiple parts it's basically asking how to find the CI intervals ?

I could really use the help and thank you !!!!!

Part A: You measure 37 textbooks' weights, and find they have a mean weight of 66 ounces. Assume the population standard deviation is 2.2 ounces. Based on this, construct a 95% confidence interval for the true population mean textbook weight.
  < <
Give your answers as decimals, to two places

Part B: We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 450 had kids. Based on this, construct a 95% confidence interval for the proportion of adult residents who are parents in this county.
   < <
Give your answers as decimals, to three places.

Part C:If n=25, ¯xx¯(x-bar)=31, and s=19, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population.
  < <
Give your answers to one decimal place.

Part D:Assume that a sample is used to estimate a population mean . Find the 95% confidence interval for a sample of size 940 with a mean of 47.8 and a standard deviation of 8.1. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
  < <

Part E:Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 50 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 7.3 and a standard deviation of 3.4. What is the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

< <

Part F:

In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $35 and standard deviation of $8. Find the margin of error at a 99% confidence level.

Give your answer to two decimal places.

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Part E: If n=20, ¯xx¯(x-bar)=45, and s=10, find the margin of error at a 90% confidence level

Give your answer to two decimal places.

Explanation / Answer

Q1.
given that,
sample mean, x =66
standard deviation, s =2.2
sample size, n =37
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 36 d.f is 2.03
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 66 ± t a/2 ( 2.2/ Sqrt ( 37) ]
= [ 66-(2.03 * 0.362) , 66+(2.03 * 0.362) ]
= [ 65.266 , 66.734 ]

Q2.
given that,
possibile chances (x)=450
sample size(n)=600
success rate ( p )= x/n = 0.75
CI = confidence interval
confidence interval = [ 0.75 ± 1.96 * Sqrt ( (0.75*0.25) /600) ) ]
= [0.75 - 1.96 * Sqrt ( (0.75*0.25) /600) , 0.75 + 1.96 * Sqrt ( (0.75*0.25) /600) ]
= [0.715352 , 0.784648]

Q3.
given that,
sample mean, x =31
standard deviation, s =19
sample size, n =25
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 24 d.f is 2.8
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 31 ± t a/2 ( 19/ Sqrt ( 25) ]
= [ 31-(2.8 * 3.8) , 31+(2.8 * 3.8) ]
= [ 20.36 , 41.64 ]

Q4.
given that,
sample mean, x =7.3
standard deviation, s =3.4
sample size, n =50
level of significance, = 0.2
from standard normal table, two tailed value of |t /2| with n-1 = 49 d.f is 1.3
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 7.3 ± t a/2 ( 3.4/ Sqrt ( 50) ]
= [ 7.3-(1.3 * 0.481) , 7.3+(1.3 * 0.481) ]
= [ 6.675 , 7.925 ]

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects