Mathematics, 19.06.2020 04:57 areynaguzman100

The mean and standard deviation of a random sample of n measurements are equal to 34.8 and 3.9, respectively. a. Find a 90% confidence interval for mu if nequals81. b. Find a 90% confidence interval for mu if nequals324. c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?

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Answer from: chrissycherry

a) The 90% confidence interval for the mean is (34.08, 35.52).

b) The 90% confidence interval for the mean is (34.44, 35.16).

c) Width for n=81: 1.44

Width for n=324: 0.72

d) The width of the interval is reduced by a factor of √4=2.

Step-by-step explanation:

a) We have to calculate a 90% 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=34.8.

The sample size is N=81.

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

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.9}{\sqrt{81}}=\dfrac{3.9}{9}=0.43$

The degrees of freedom for this sample size are:

df=n-1=81-1=80

The t-value for a 90% confidence interval and 80 degrees of freedom is t=1.66.

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

$MOE=t\cdot s_M=1.66 \cdot 0.43=0.72$

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

$LL=M-t \cdot s_M = 34.8-0.72=34.08\\\\UL=M+t \cdot s_M = 34.8+0.72=35.52$

The 90% confidence interval for the mean is (34.08, 35.52).

b) As the sample size has changed, we recalculate the standard error:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.9}{\sqrt{324}}=\dfrac{3.9}{18}=0.22$

The degrees of freedom for this sample size are:

df=n-1=324-1=323

The t-value for a 90% confidence interval and 323 degrees of freedom is t=1.65.

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

$MOE=t\cdot s_M=1.65 \cdot 0.22=0.36$

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

$LL=M-t \cdot s_M = 34.8-0.36=34.44\\\\UL=M+t \cdot s_M = 34.8+0.36=35.16$

The 90% confidence interval for the mean is (34.44, 35.16).

c. The widths are:

For n=81

w_a=UL-LL=35.52-34.08=1.44

For n=324

w_b=UL-LL=35.16-34.44=0.72

d. The effect of quadrupling the sample size, with all the other parameters constant, is that the width of the interval is reduced by a factor of 2.

This is because the standard error, and therefore the margin of error, is reduced by a factor of √4=2, when the sample size is quadrupled.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1

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