Mathematics, 02.03.2020 21:27 MuntazerMehdi9746

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.)

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Answer from: viodsenpai

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (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".

$\bar X$ represent the sample mean for the sample

$\mu$ 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (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,

Answer from: Quest

dont think so

step-by-step explanation:

Answer from: Quest

The answer to your question is 1/9

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