An environmental group wanted to estimate the proportion of fresh produce sales identified as organic in a local grocery store. In the winter, the group obtained a random sample of sales from the store and used the data to construct the 95 percent z-interval for a proportion (0.087, 0.133 ). Six months later in the summer, the group obtained a second random sample of sales from the store. The second sample was the same size as the first, and the proportion of sales identified as organic was 0.4. How does the 95 percent z-interval for a proportion constructed from the summer sample compare to the winter interval?

Winter interval is (0.087, 0.133)

Confidence interval is given by

Lower limit = 0.087 = p-moe

Upper limit = 0.133 = p+moe

Where moe = margin of error

After adding both the equations

(0.087+0.133) = (p+p) + (moe-moe)

2p = 0.22

p = 0.11

After substituting the value of p in equation 0.087 = p-moe

Moe = 0.023

Now we know that moe = z*√{p*(1-p)/n}

Where z is the critical value of 95% confidence interval which is 1.96

P = 0.11

After substituting all the values

N = (1.96/0.023)^2 * 0.11*(1-0.11)

N = 710.950170132 = 711

In summer, p = 0.4 and n = 711(as it is mentioned that sample size is same)

Here moe = 1.96*√{0.4*(1-0.4)/711}

= 0.03601031263

CI

(P-moe, p+moe)

(0.364 to 0.436)

Now by above calculation it is clear that summer interval has greater point estimate that is 0.4>0.11

Width of winter interval = 0.133-0.087 = 0.046

Width of summer interval = 0.436-0.364 = 0.072

It is clear that summer interval is wider

So,summer interval is wider and has greater point estimate.

c. 16 units

step-by-step explanation:

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13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%