The 2010 General Social Survey asked the question: "After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?" to a random sample of 1,155 Americans.^41 A 95% confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). Interpret this interval in context of the data. Suppose another set of researchers reported a confidence interval with a larger margin of error based on the same sample of 1,155 Americans. How does their confidence level compare to the confidence level of the interval stated above? Suppose next year a new survey asking the same question is conducted, and this time the sample size is 2,500. Assuming that the population characteristics, with respect to how much time people spend relaxing after work, have not changed much within a year. How will the margin of error of the 95% confidence interval constructed based on data from the new survey compare to the margin of error of the interval stated above?

- How does their confidence level compare to the confidence level of the interval stated above?

Provided all the other parameters involved in the determination of margin of error remains constant, the confidence level of the interval obtained by the second set of researchers (with a larger margin of error) is definitely greater than the confidence of the first stated interval.

- How will the margin of error of the 95% confidence interval constructed based on data from the new survey compare to the margin of error of the interval stated above?

Provided all the other parameters involved in the determination of margin of error remains constant, the margin of error of the confidence interval of data from the new survey will be less than the margin of error of the initially stated interval because the margin of error is inversely proportional to the sample size and the sample size of the new survey (2500) is more than the sample size of the old survey (1155).

Step-by-step explanation:

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Since the sample sizes are so large, the critical value for this would be obtained from the z-distribution tables.

And critical value increases with higher confidence level.

Which directly translates to margin of error increasing with higher confidence level.

A) Hence, if all other parameters stay the same, the margin of error can only be larger if the confidence level of the second researchers is more than than the confidence level at which the initial research was carried out.

B) In the determination of the margin of error

Margin of Error = (Critical value) × (standard Error of the mean)

Standard error of the mean = σₓ = (σ/√n)

where σ = population standard deviation or sample standard deviation for a sample size that's very large

n = sample size

If the critical value stays the same (the tests in the two years being considered are performed at the same level) and the standard deviation too remain almost the same (the question asks us to assume that the population characteristics, with respect to how much time people spend relaxing after work, have not changed much within a year), the margin of error is obviously inversely proportional to the sample size.

Margin of error = (zσ/√n)

zσ = constant = k

Margin of error = (k/√n)

So, if the sample size increases, the new margin of error that will be obtained will be lesser.

Hope this Helps!!!

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

answer is 12345678900987654321234567890987654321

step-by-step explanation: