A marketing researcher wants to estimate the mean savings ($) realized by shoppers who showroom. Showrooming is the practice of inspecting p

Question

A marketing researcher wants to estimate the mean savings ($) realized by shoppers who showroom. Showrooming is the practice of inspecting products in retail stores and then purchasing the products online at a lower price. A random sample of 100 shoppers who recently purchased a consumer electronics item online after making a visit to a retail store yielded a mean savings of $58 and a standard deviation of $55. a) Construct a 95% confidence interval for the mean savings for all showroomers who purchased a consumer electronics item. b) Suppose the owners of a consumer electronics retailer wants to predict the average total value of lost sales attributed to the next 1000 showroomers that enter their retail store. Please use the results in part (a) to help him do the estimation.

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Adeline 2 weeks 2021-10-08T08:31:21+00:00 1 Answer 0

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    2021-10-08T08:32:29+00:00

    Answer:

    a) 58-1.984\frac{55}{\sqrt{100}}=47.087    

    58+1.984\frac{55}{\sqrt{100}}=68.913    

    So on this case the 95% confidence interval would be given by (47.087;68.913)    

    b) The confience interval calculated from part a is in order to estimate the true mean of savings. And we can extrapolate the results obtained for this new sample size:

     47.087 < \mu < 68.913

    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)

    In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

    df=n-1=100-1=9

    Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: “=-T.INV(0.025,99)”.And we see that t_{\alpha/2}=1.984

    Now we have everything in order to replace into formula (1):

    58-1.984\frac{55}{\sqrt{100}}=47.087    

    58+1.984\frac{55}{\sqrt{100}}=68.913    

    So on this case the 95% confidence interval would be given by (47.087;68.913)    

    Part b

    The confience interval calculated from part a is in order to estimate the true mean of savings . And we can extrapolate the results obtained for this new sample size:

     47.087 < \mu < 68.913

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