4.14 Thanksgiving spending, Part I

The 2009 holiday retail season, which kicked off on November 27, 2009 (the day after Thanksgiving), had been marked by somewhat lower self-reported consumer spending than was seen during the comparable period in 2008. To get an estimate of consumer spending, 436 randomly sampled American adults were surveyed. Daily consumer spending for the six-day period after Thanksgiving, spanning the Black Friday weekend and Cyber Monday, averaged $84.71. A 95% confidence interval based on this sample is ($80.31, $89.11). Determine whether the following statements are true or false, and explain your reasoning.

tgSpending <- read.csv("https://raw.githubusercontent.com/ahmshahparan/Homework_4.14/master/thanksgiving_spend.csv")
hist(tgSpending$spending)

summary(tgSpending$spending)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   5.719  49.177  75.792  84.707 112.255 282.803
sd(tgSpending$spending)
## [1] 46.92851
  1. We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.

FALSE. We know 100% for certain that the average spending costs of these 436 American adults is between $80.31 and $89.11. The point estimate is always in the confidence interval.

  1. This confidence interval is not valid since the distribution of spending in the sample is right skewed.

FALSE. The skew is acceptable for the sample size of 436, so the skew can be overlooked.

  1. 95% of random samples have a sample mean between $80.31 and $89.11.

FALSE. The confidence interval is not about a sample mean.

  1. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.

TRUE. This is the definition of a 95% Confidence Interval.

  1. A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.

TRUE. If we do not need to be as sure, then we can use a lower number for confidence interval. The range would also be smaller vs. a 95% confidence interval.

  1. In order to decrease the margin of error of a 95% confidence interval to a third of what it is now, we would need to use a sample 3 times larger.

FALSE. The sample size needs to be 9x larger (3^2) to decrease the error by 1/3 (In the calculation of the standard error, we divide the standard deviation by the square root of the sample size.).

  1. The margin of error is 4.4.

TRUE. It is the product of (1.96)∗(sd/sqrt(436))