Question

1. A study aims to understand the year-to-year changes in small business sales during the winter holiday period in the US. Researchers randomly sampled 100 small businesses in December 2014 and found that 66 of them had an increase in sales relative to December 2013. a. Find a 95% confidence interval for the proportion of businesses that had an increase in sales Find the sample size needed for the 68% confidence interval to be of the same width as the 95% confidence interval calculated above (round up the sample size to the nearest integer if necessary). Note that the 95% confidence interval calculated above was based on a sample of size 100. Another research paper (funded by an agency known to be critical of the government's policies) cited this study claiming that less than 2/3rd of the small businesses had an increase in sales. With what level of confidence can this statement be made? [10 pt] Similar to the study done for December 2014, if a 95% confidence interval was computed for each of the months starting from January 2011 to December 2014 (that is 4x12-48 months), what is the probability that at least 95% of the confidence intervals cover the true proportions? Feel free to assume (and state clearly), if required, that the sales for each small business for each month are independent of that for any other combination of small business and month. Clearly specify if, where, and how you have used this assumption. Hint: As per our definition in Lecture 6, a 95% confidence interval covers the true proportion 95% of the times. b. c. d.

Answer 1

Solution

Let X = number of small businesses out of a sample 100, registering an increase in sales of December 2014 relative to December 2013.

Then X ~ B(100, p), where p =population proportion of small businesses registering an increase in sales of December 2014 relative to December 2013.

Back-up Theory

Given X ~ B(n, p), 100(1 - α) % Confidence Interval for p is:

p_cap ± Z_α/2[sq.rt{p_cap(1 –p_cap)/n}] ……………………………………………………………… (1)

where

Z_α/2 is the upper (α/2)% point of N(0, 1),

pcap = sample proportion, and

n = sample size.

Now to work out the solution,

Given X = 66 and n = 100, p_cap = 0.66

Part (a)

Vide (1), 95% Confidence Interval for proportion of small businesses registering an increase in sales

= 0.66 ± 1.96[sq.rt(0.66 x 0.34/100)]

= 0.66 ± 0.0928

= [0.567, 0.753] Answer

Part (b)

Vide (1), 68% Confidence Interval for proportion of small businesses registering an increase in sales

= 0.66 ± 0.9945[sq.rt(0.66 x 0.34/n)]

Now, width of the above interval = 2 x 0.9945[sq.rt(0.66 x 0.34/n)].

Width of the interval obtained in Part (a), i.e., 95% CI, = 2 x 0.0928.

We want, 2 x 0.9945[sq.rt(0.66 x 0.34/n)] = 2 x 0.0928

Or, [sq.rt(0.66 x 0.34/n)] = 0.0928/0.9945 = 0.0933

Or, n = (0.66 x 0.34)/0.0933²

= 25.20

Thus, the required sample size = 25 Answer

Part (c)

Here we want to find the confidence level of an interval with upper bound as 2/3 or 0.6667

i.e., 0.66 + Z_α/2[sq.rt(0.66 x 0.34/100)] = 0.6667

Or, Z_α/2 = (0.6667 – 0.66)/sq.rt(0.66 x 0.34/100)

= 0.0067/0.0474

= 0.1413

=> α/2 = 0.444 [Using Excel Function: Statistical NORMSDIST]

=> α = 0.888 or 88.8%

=> confidence level = 100 – 88.8

= 11.2%   Answer

Part (d)

As already indicated under ‘Hint’, 95% confidence interval => probability, p that the true proportion would lie in the interval is 0.95.

Now, let Y = Number of confidence intervals out of 48 intervals that would hold the true population proportion. Then, Y ~ B(48, 0.95).

Again 95% of 48 = 45.6.

So, the required probability is: P(Y ≥ 45.6)

= P(Y > 45)

= 0.567 [Using Excel Function: Statistical BINOMDIST] Answer

DONE