In a certain large city, the mean birth weight of babies is 7.1 pounds with a standard deviation of 1.2 pounds. You believe that in one neighborhood in the city the mean birth weight is different from 7.1 pounds. You sample 95 births and find the mean weight of the sample to be 7.4 pounds. Can the claim be supported to a level of significance of α = .05, test the hypothesis?

No, the claim that the mean birth weight is 7.1 pounds cannot be supported at 0.05 significance level.

Step-by-step explanation:

A z-test is used to test the hypothesis because the population standard deviation is known.

Null hypothesis: The mean birth weight is 7.1 pounds.

Alternate hypothesis: The mean birth weight is not equal to 7.1 pounds.

The test is a two-tailed test because the alternate hypothesis is expressed with the inequality not equal to.

Test statistic (z) = (sample mean - population mean) ÷ (population sd/√n)

sample mean = 7.4 pounds

population mean = 7.1 pounds

population sd = 1.2 pounds

n = 95

z = (7.4 - 7.1) ÷ (1.2/√95) = 0.3 ÷ 0.123 = 2.44

At 0.05 significance level, the critical values from the standard normal distribution table are -1.96 and 1.96.

Conclusion:

Reject the null hypothesis because the test statistic 2.44 falls outside the region bounded by the critical values -1.96 and 1.96.

The claim cannot be supported at a 0.05 significance level.

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).