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 constructed from the summer sample compare to the winter interval? (A) The summer interval is wider and has a lesser point estimate. 0.4. How does the 95 perc ent z-interval for a proportion (B) The summer interval is wider and has a greater point estimate. (C) The summer interval is narrower and has a lesser point estimate (D) The summer interval is narrower and has a greater point estimate. (E) The summer interval is the same width and has a greater point estimate

(B) The summer interval is wider and has a greater point estimate.

Step-by-step explanation:

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

Therefore 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)

So 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

Therefore it is obvious that summer interval is wider

Therefore the correct answer is ,summer interval is wider and has greater point estimate.

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation: