Question

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

a) The post estimate for the population proportion of the businesses that had an increase in sales: $\widehat{p}=\frac{66}{100}=0.66$

To find the confidence interval we will use the formula

$CI=\widehat{p}\pm Z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

n=100

for 95% confidence

$Z_{\frac{\alpha }{2}}=1.96$

$CI=0.66\pm 1.96\sqrt{\frac{0.66(1-0.66)}{100}}$

$CI=0.66\pm 0.09$

$CI=(0.57,0.75)$

b) the width of the 95% confidence interval is twice the margin of error for the confidence interval calculated above $2ME=2( 1.96\sqrt{\frac{0.66(1-0.66)}{100}})=0.186$

for 68% confidence interval

$Z_{\frac{\alpha }{2}}=0.9945$

The width of 68% confidence interval is

$2*0.9945\sqrt{\frac{0.5(1-05)}{n}}=0.186$ ( since we are not sure of the population proportion so we assume p =0.5)

solving we get n= 28.59

rounding up

the sample size required is n=29

c) If the claim made is that less than $\frac{2}{3}$ of small businesses had an increase in sales that it means that the Upper limit of the confidence interval of the population proportion with which the claim is made is less than $\frac{2}{3}$

$UL=0.66+ Z_{\frac{\alpha }{2}}\sqrt{\frac{0.66(1-0.66)}{100}}$

$0.66+ Z_{\frac{\alpha }{2}}\sqrt{\frac{0.66(1-0.66)}{100}}=\frac{2}{3}$

solving

$Z_{\frac{\alpha }{2}}=0.1407$

now $P(Z<0.1407)=0.5559$

The confidence level for this is $2*P(Z<0.1407)-1=2*0.5559-1=0.1118$

WIth 11% confidence this claim can be made

d) With repeated sampling, there is a 95% chance that a random confidence interval will cover the true population mean. However, it is not correct that ONE specific interval has a 95% probability of covering the true population mean. It is right that ‘there is 95% confidence that the specific interval covers the true population mean.