Suppose that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%. Use a 0.10 significance level to test that claim. A. What is the test statistic for the hypothesis test?
B. What is the p value?
C. Reject/fail to reject sufficient evidence.

We conclude that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

We are given that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%.

Let p = population percentage of blue candies

So, Null Hypothesis, : p = 29%     {means that the percentage of blue candies is equal to 29%}

Alternate Hypothesis, : p 29%     {means that the percentage of blue candies is different from 29%}

The test statistics that will be used here is One-sample z-test for proportions;

                         T.S.  =    ~ N(0,1)

where, = sample proportion of blue coloured candies = 28%

           n = sample of colored candies = 100

So, the test statistics =  

                                    =  -0.22

The value of the z-test statistics is -0.22.

Also, the P-value of the test statistics is given by;

               P-value = P(Z < -0.22) = 1 - P(Z 0.22)

                            = 1 - 0.5871 = 0.4129

Now, at a 0.10 level of significance, the z table gives a critical value of -1.645 and 1.645 for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of z, so we insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the percentage of blue candies is equal to 29%.

umumumumum

im not sure

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).