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Math  /  Data & Statistics

QuestionA businesswoman is considering whether to open a coffee shop in a local shopping center. Before making this decision, she wants to know how much money people spend per week at coffee shops in that area. She took a random sample of 26 customers from the area who visit coffee shops and asked them to record the amount of money (in dollars) they would spend during the next week at coffee shops. At the end of the week, she obtained the following data (in dollars) from these 26 customers: \begin{tabular}{lllllllll} 16.91 & 38.63 & 15.22 & 14.34 & 5.05 & 63.69 & 10.28 & 13.21 & 32.20 \\ 36.04 & 16.29 & 65.93 & 10.27 & 37.13 & 3.15 & 6.81 & 34.67 & 6.47 \\ 36.25 & 27.66 & 38.71 & 13.17 & 9.64 & 9.39 & 1.30 & 5.16 & \end{tabular}
Assume that the distribution of weekly expenditures at coffee shops by all customers who visit coffee shops in this area is approximately normal.
Round your answers to cents. a. What is the point estimate of the corresponding population mean? xˉ=$\bar{x}=\$ i □\square b. Make a 95\% confidence interval for the average amount of money spent per week at coffee shops by all customers who visit coffee shops in this area. \$ i 1 ! to \$ i

Studdy Solution

STEP 1

1. The sample size is n=26 n = 26 .
2. The sample data are: 16.91,38.63,15.22,14.34,5.05,63.69,10.28,13.21,32.20,36.04,16.29,65.93,10.27,37.13,3.15,6.81,34.67,6.47,36.25,27.66,38.71,13.17,9.64,9.39,1.30,5.16 16.91, 38.63, 15.22, 14.34, 5.05, 63.69, 10.28, 13.21, 32.20, 36.04, 16.29, 65.93, 10.27, 37.13, 3.15, 6.81, 34.67, 6.47, 36.25, 27.66, 38.71, 13.17, 9.64, 9.39, 1.30, 5.16 .
3. The distribution of weekly expenditures is approximately normal.
4. We will use the sample mean as the point estimate for the population mean.
5. We will use the t-distribution to construct the confidence interval since the population standard deviation is unknown.

STEP 2

1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Determine the t-value for a 95% confidence interval.
4. Calculate the margin of error.
5. Construct the confidence interval.

STEP 3

Calculate the sample mean (xˉ\bar{x}):
xˉ=1n∑i=1nxi\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i
Substitute the values:
xˉ=126(16.91+38.63+15.22+14.34+5.05+63.69+10.28+13.21+32.20+36.04+16.29+65.93+10.27+37.13+3.15+6.81+34.67+6.47+36.25+27.66+38.71+13.17+9.64+9.39+1.30+5.16)\bar{x} = \frac{1}{26} (16.91 + 38.63 + 15.22 + 14.34 + 5.05 + 63.69 + 10.28 + 13.21 + 32.20 + 36.04 + 16.29 + 65.93 + 10.27 + 37.13 + 3.15 + 6.81 + 34.67 + 6.47 + 36.25 + 27.66 + 38.71 + 13.17 + 9.64 + 9.39 + 1.30 + 5.16)
Calculate the sum and divide by 26:
xˉ=536.2326≈20.62\bar{x} = \frac{536.23}{26} \approx 20.62

STEP 4

Calculate the sample standard deviation (ss):
s=1n−1∑i=1n(xi−xˉ)2s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}
Calculate each deviation, square it, sum them, divide by 25, and take the square root:
\[ s \approx 17.98 $

STEP 5

Determine the t-value for a 95% confidence interval with n−1=25 n-1 = 25 degrees of freedom. Using a t-table or calculator, find:
\[ t_{0.025, 25} \approx 2.060 $

STEP 6

Calculate the margin of error (ME):
ME=t×snME = t \times \frac{s}{\sqrt{n}}
Substitute the values:
ME=2.060×17.9826≈7.26ME = 2.060 \times \frac{17.98}{\sqrt{26}} \approx 7.26

STEP 7

Construct the confidence interval:
Lower limit=xˉ−ME=20.62−7.26=13.36\text{Lower limit} = \bar{x} - ME = 20.62 - 7.26 = 13.36
Upper limit=xˉ+ME=20.62+7.26=27.88\text{Upper limit} = \bar{x} + ME = 20.62 + 7.26 = 27.88
The 95% confidence interval is:
\[ \$13.36 \text{ to } \$27.88 $
The point estimate of the corresponding population mean is:
\[ \bar{x} = \$20.62 $
The 95% confidence interval is:
\[ \$13.36 \text{ to } \$27.88 $

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