Assuming that the population is normally distributed, construct a 9090% confidence interval for the population mean for each of the samples below. explain why these two samples produce different confidence intervals even though they have the same mean and range. sample
a.11 44 44 44 55 55 55 88 full data set sample
b.11 22 33 44 55 66 77 88

Question

13-11-2017
Mathematics

contestada

Assuming that the population is normally distributed, construct a 9090% confidence interval for the population mean for each of the samples below. explain why these two samples produce different confidence intervals even though they have the same mean and range. sample
a.11 44 44 44 55 55 55 88 full data set sample
b.11 22 33 44 55 66 77 88

Respuesta :

Otras preguntas

The table shows information about the masses of some dogs. A) Work out the minimum number of dogs that could have a mass of more than 24 kg. B) Work out the max

1. Think back to what you learned about Betty Friedan and Dolores Huerta. How did each woman's social background shape her activism? 2. What do Friedan and Huer

The engineering process defines the steps engineers take to create technology. Which of these describes these steps in the correct order? A. Identifying a need,

Determine the open intervals on which the function is increasing, decreasing, or constant. (Enter your answers using interval notation.) t + 2, t<0

the corruption scandal related to government-subsidized railroads was known as

(q1)Find the area of the region bounded by the graphs of y = x - 2 and y2 = 2x - 4.

case-control studies are particularly well suited to studying:

A small class has 10 students, 6 of whom are girls and 4 of whom are boys. The teacher is going to choose two of the students at random. What is the pr

4. parallel editing is: a. cutting between two lines of narrative action b. a form of long take c. an element of set design d. a split screen

shinmin shinmin · Answer 1 · 2017-11-23T14:05:42+01:00

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

Set B:

Mean: 4.5

Standard dev: 2.45

% = 90

Set A:

Standard Error, SE = s/ √n = 1.9/√8 = 0.67

Degrees of freedom = n - 1 = 8 -1 = 7

t- score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

Set B:

Standard Error, SE = s/ √n = 2.45/√8 = 0.87

Degrees of freedom = n - 1 = 8 -1 = 7

t- Score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.