Exercise 4.12 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 this 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.)

a) The confidence interval for the mean is given by the following formula:

  (1)

After calculate the 95% confidence interval they got (80.31; 89.11)

As we can see the confience interval not contains the value of 100 so then the claim not makes sense since the value reportes $100 is above the right limit of the confidence interval for this case.

b) If the confidence decrease the width of the confidence interval decrease because we will have a lower standard of error. On this case the conclusion will not change since the new interval not contains the vaue of $ 100 purposed,

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Part a

The confidence interval for the mean is given by the following formula:

  (1)

After calculate the 95% confidence interval they got (80.31; 89.11)

As we can see the confience interval not contains the value of 100 so then the claim not makes sense since the value reportes $100 is above the right limit of the confidence interval for this case.

Part b

If the confidence decrease the width of the confidence interval decrease because we will have a lower standard of error. On this case the conclusion will not change since the new interval not contains the vaue of $ 100 purposed,

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

77

step-by-step explanation: