Samuel orders DVD’s from an online music store. Each DVD costs $9.99. He has a 20% discount code, and sales tax is 6.75%. What i
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Answer:
a. ; b. The 95% interval is USD [45.82, 56.86].
Step-by-step explanation:
The mean amount a customer spends at the grocery store
The mean is the sum of all of the data divided by the number of observations for this case, that is, n = 50.
The sum of all of the data is: 5.24+11.97+13.59+15.52+15.69+19.96+20.54+22.59+27.25+27.29+28.25+30.00+31.63+33.01+37.21+40.36+40.90+42.83+45.66+46.67+47.49+49.32+50.12+51.71+52.2 1+53.57+56.16+56.36+56.88+58.65+59.42+60.20+60.48+62.23+62.50+67.95+70.45+73.08+74.15+74.48+74.78+75.75+76.27+76.47+78.92+84.09+84.91+85.06+88.45+88.91 = 2567.17.
The mean is
Then, <em>the mean amount a customer spends at the grocery store is</em> USD 51.34.
The 95% confidence interval for the mean amount spent
For the 95% confidence interval for the mean, we have the value for the <em>mean = </em>51.34, as we have just found, the value for the<em> standard deviation</em> is already given as USD 20.00. There are 50 observations, so n = 50.
Moreover, for a 95% confidence, the value for the corresponding z-score is z = 1.96 (this is the confidence coefficient). Then, the formula for finding the "limits" values for the confidence interval is as follows:
For the upper limit
For the lower limit
Thus
The 95% confidence is USD [45.82, 56.86].
Mathematically
There is a probability of 0.95 that the values are between [45.82, 56.86], both inclusive.
We have to remember here that we are dealing with a sample. This sample has a standard deviation, called standard error of the sample, and is .
So, the population mean is, with a 95% confidence, between the values [45.82, 56.86], both inclusive.
We can see the graph below that represents the former.
