Thanksgiving spending, Part II. Exercise 4.14 provides a 95% confidence interval for the average spending by American adults during the six-day period after Thanksgiving 2009: ($80.31, $89.11). (a) A local news anchor claims that the average spending during this period in 2009 was $100. What do you think of her claim? (b) Would the news anchor’s claim be considered reasonable based on a 90% confidence interval? Why or why not? (Do not actually calculate the interval.)
Thanksgiving spending, Part II. Exercise 4.14 provides a 95% confidence interval for the
average spending by American adults during the six-day period after Thanksgiving 2009: ($80.31,
$89.11).
(a) A local news anchor claims that the average spending during this period in 2009 was $100.
What do you think of her claim?
(b) Would the news anchor’s claim be considered reasonable based on a 90% confidence interval?
Why or why not? (Do not actually calculate the interval.)
Please type clearly so I can follow the steps
Given :
For the average spending by American adults during the six day period after Thanksgiving 2009.
95% confidence interval = ($80.31 , $89.11)
Formula to calculate confidence interval :
CI = Mean ± Zalpha/2( Standard deviation/√n )
Where Z = critical value
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