Question

Friday the $13th$Refer to Exercise $88$.

a. Construct and interpret a $90\%$confidence interval for the true mean difference. If you already defined parameters and checked conditions in Exercise $88$, you don’t need to do them again here.

b. Explain how the confidence interval provides more information than the test in Exercise $88$.

Short answer

The genuine difference in the number of shoppers at each retailer on these two days is between $1.8156$and$91.1844$ shoppers lower on Friday the $6th$ than on Friday the $13th$, according to our $90$ percent confidence level.

Step-by-step solution

Part a: Step 1: Given Information

We have been given that

$$\begin{gathered} n=Samplesize=45, \\ \overline{x_d}=Meandifferences=-46.5, \\ {s}_d=S\tan darddeviationofdifferences=178.0, \\ c=Confidencelevel=90\%=0.90 \end{gathered}$$

Part a: Step 2: Simplification

Random, Independent (ten percent condition), and Normal/Large sample are the three criteria for determining a confidence interval for the population mean difference.

Because the sample is a random sample, I'm satisfied.

Independent: Satisfied, because the sample of $45$grocery stores represents less than ten percent of the total number of grocery stores (assuming that there are more than $450$grocery stores).

Satisfied with the sample size of $45$because it is at least $30$and hence the sample is substantial.

Because all of the prerequisites have been met, it is time to calculate the confidence interval for the population mean difference.

We now determine the t-value by looking in the row starting with degrees of freedom $df=n-1=45–1=44$(the table does not contain $df=44$, so we will use the nearest smaller degrees of freedom $df=40$instead) and in the column with $c=90\%$in the table of the Student's T distribution:

$t_{\alpha /2}=1.684$

The margin of error is then:

$E=44.6844$

The confidence interval's boundaries then become:

$\overline{x}–E=-46.5–44.6844=-91.1844$

$\overline{x}+E=-46.5+44.6844=-1.8156$

We are $90\%$confident that the true difference in the number of shoppers at each store on these two days is between $1.8156$and $91.1844$, lower on Friday the$6th$than on Friday the $13th.$

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Part b: Step 1: Given Information

We have to find how the confidence interval provides more information than the test .

Part b: Step 2: Explanation

Because the confidence interval provides a range of possible values for the true mean difference, whereas the significance test only tests one, the confidence interval provides more information than the significance test.

The confidence interval for the true mean difference offers us a range of possible values, but the significance test simply tests one possible value for the true mean difference.