+1 678 648 4277 

Central Limit Theorem Assignment Solution


Sampling Distribution of Poll Data

1. According to a Quinnipiac poll released 4/15/21, 91% of adults who identify as Democrats support stronger gun laws in the US.

a. Determine whether the Central Limit Theorem of Proportions applies to the sampling distribution of proportions of samples of 108 adults who identify as Democrat. Provide proof.

b. Show that the Central Limit Theorem of Proportions applies to the sampling distribution of proportions of samples of 180 adults who identify as Democrat. Provide proof.

c. Describe the sampling distribution of proportions of samples of 180 adults who identify as Democrats.

d. Find the probability that the proportion that supports stronger gun laws in the US of a sample of 180 adults who identify as Democrat is between 89% and 95%.

e. Find the probability that the proportion that supports stronger gun laws in the US of a sample of 180 adults who identify as Democrat is above 92%.

f. Above what proportion that supports stronger gun laws in the US of a sample of 180 adults who identify as Democrat does the largest 70% of all sample proportions that support stronger gun laws in the US of 180 adults who identify as Democrat fall?

g. Between what two proportions that support stronger gun laws in the US of a sample of 180 adults who identify as Democrats do the middle 75% of all sample proportions that support stronger gun laws in the US of 180 adults who identify as Democrat fall?

2. According to a Quinnipiac poll released 4/15/21, 54% of adults support stronger gun laws in the US.

a. Show that the Central Limit Theorem of Proportions applies to the sampling distribution of proportions of samples of 1100 adults. Provide proof.

b. Describe the sampling distribution of proportions of samples of 1100 adults.

c. A member of Governor Cuomo’s administration wants to know the proportion of New York adults who support stronger gun laws. Her staff takes a sample of 1100 adults and 593 say they do support stronger gun laws.

    i. Create a 96% confidence interval estimate for the proportion of all New York adults who support stronger gun laws.

    ii. Interpret the 96% confidence interval estimate.

3. A particular Dunkin store wishes to determine the proportion of customers served within two minutes of ordering.

a. What size sample is necessary to create a 98% confidence interval estimate for the proportion of customers served within two minutes of ordering with a maximum error of 1.5%?

b. They believe 78.7% of customers are served within two minutes of ordering. Assuming this is correct, what size sample is now necessary to calculate a 98% confidence interval estimate for the proportion of customers served within two minutes of ordering with a maximum error of 1.5%?

Solutions:

Q1:

We have to check if np(1-p) is greater than 108*0.91*0.09 = 8.8452, which is not greater than 10. Thus, the Central Limit theorem of proportions can’t be applied here.

 Now here n=108, then np(1-p) = 14.742 which is greater than 10. Thus, the Central Limit theorem of proportions can be applied now.

 The sampling distribution of proportion follows a normal distribution with a mean of 0.91 and variance np(1-p)=14.742.

 P(0.85

 P(p>0.92) = 0.3583

 The 70th percentile is 0.924

 The middle 75% covers 37.5% of data on both sides of the mean, i.e. 12.5th percentile and 87.5th percentile. Thus, the proportion between which 75% of all sample proportions are 0.878 and 0.942.

Q2:

 Here n=1100 and p=0.54. Hence, np(1-p) = 273.2 which is greater than 10. Thus, the Central Limit theorem of proportions can be applied now.

 The sampling distribution of proportion follows a normal distribution with a mean of 0.54 and the standard deviation is the square root of np(1-p) which is 0.015

 i) The 96% confidence interval is given by p±z_0.98*Std.Dev.=0.54±2.054*0.015=(0.509,0.571)

ii) Interpretation: We have 96% confidence that the true proportion of all new York adults who support stronger laws are within the above interval.

Q3:

 The Margin of error for 98% confidence interval with no information of proportion is given by z_0.99*0.5/√n, as we suppose the proportion to be 0.5. Now we need to equate it to 0.015 margins of error. Thus, we get n=(z_0.99*0.5/0.015)^2=6013.2. Now rounding up to the next integer we get the required number to be 6014.

 Here we have a prior knowledge of p which is 0.787. Thus the margin of error now is z_0.99*√(0.787*213)/√n. Now we need to equate it to a 0.015 margin of error. Thus, we get n=(z_0.99*√(0.787*213)/0.015)^2=4032.01. Now rounding up to the next integer we get the required number to be 4033.