Mathematics, 07.07.2020 19:01 Squara

The mean and standard deviation of a random sample of n measurements are equal to 34.5 and 3.4, respectively. A. Find a 95 % confidence interval for μ if n=49.B. Find a 95% confidence interval for μ if n=196.C. Find the widths of the confidence intervals found in parts a and b. D. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?1. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 4.2. Quadrupling the sample size while holding the confidence coefficient fixed increases the width of the confidence interval by a factor of 2.3. Quadrupling the sample size while holding the confidence coefficient fixed increases the width of the on confidence interval by a factor of 4.4. Quadrupling the sample size while holding the confidence coefficient fixed does not affect the width of the confidence interval.5. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 2.

Answers: 2

Show answers

Answers

Answer from: hillisaiah734

a. The 95% confidence interval for the mean is (33.52, 35.48).

b. The 95% confidence interval for the mean is (34.02, 34.98).

c. n=49 ⇒ Width = 1.95

n=196 ⇒ Width = 0.96

Note: it should be a factor of 2 between the widths, but the different degrees of freedom affects the critical value for each interval, as the sample size is different. It the population standard deviation had been used, the factor would have been exactly 2.

d. 5. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 2.

Step-by-step explanation:

a. We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=34.5.

The sample size is N=49.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{49}}=\dfrac{3.4}{7}=0.486$

The degrees of freedom for this sample size are:

df=n-1=49-1=48

The t-value for a 95% confidence interval and 48 degrees of freedom is t=2.011.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.011 \cdot 0.486=0.98$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 34.5-0.98=33.52\\\\UL=M+t \cdot s_M = 34.5+0.98=35.48$

The 95% confidence interval for the mean is (33.52, 35.48).

b. We have to calculate a 95% confidence interval for the mean.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{196}}=\dfrac{3.4}{14}=0.243$

The degrees of freedom for this sample size are:

df=n-1=196-1=195

The t-value for a 95% confidence interval and 195 degrees of freedom is t=1.972.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.972 \cdot 0.243=0.48$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 34.5-0.48=34.02\\\\UL=M+t \cdot s_M = 34.5+0.48=34.98$

The 95% confidence interval for the mean is (34.02, 34.98).

c. The width of the intervals is:

$n=49\rightarrow UL-LL=33.52-35.48=1.95\\\\n=196\rightarrow UL-LL=34.02-34.98=0.96$

d. The width of the intervals is decreased by a factor of √4=2 when the sample size is quadrupled, while the others factors are fixed.

Answer from: Quest

true

step-by-step explanation:

Answer from: Quest

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post
Can someone plz me 1 to 37 . i will earn 10 points

Can someone plz me 1 to 37 . i will earn 10 points

Another question on Mathematics

Mathematics, 21.06.2019 21:30

Ten members of balin's soccer team ran warm ups for practice. each member ran the same distance. their combined distance was 5/6 of a mile. to find the distance that each member ran, balin wrote the expression below.

Answers: 3

Answer

Mathematics, 22.06.2019 01:30

If 5x - 2 = -12, then x = a) -50 b) -4.4 c) -2.8 d) -2

Answers: 1

Answer

Mathematics, 22.06.2019 05:00

Atreasure chest sinks at a rate no less than 20 feet er second. the floor of the icean is at most at -4,200 feet. will the tresure chest reach the ocean floor in less than 3 minutes? explain.

Answers: 3

Answer

Mathematics, 22.06.2019 06:00

Angle a and angle b are both supplementary angles. what must be true about the angles?

Answers: 1

Answer

You know the right answer?

The mean and standard deviation of a random sample of n measurements are equal to 34.5 and 3.4, resp...

Questions