You intend to estimate a population mean

Question

with the following sample.
77
44.4
33.6
63.1
68.1
70.6
52
43.5
26.4
72.6
30.9
29.4

You believe the population is normally distributed. Find the 90% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

90% C.I. =

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

To find the 90% confidence interval for the population mean based on the given sample, we will follow these steps:

1. Calculate the sample mean ($\overline{x}$).
2. Calculate the sample standard deviation ($s$).
3. Determine the critical value ($t^{\ast }$) for a 90% confidence level.
4. Calculate the margin of error.
5. Construct the confidence interval.

Step 1: Calculate the sample mean ($\overline{x}$)

The sample data is: $$77,44.4,33.6,63.1,68.1,70.6,52,43.5,26.4,72.6,30.9,29.4$$

First, calculate the sum of the sample values:

$$\text{Sum}=77+44.4+33.6+63.1+68.1+70.6+52+43.5+26.4+72.6+30.9+29.4=683.8$$

Then, the sample mean ($\overline{x}$) is:

$$\overline{x}=\frac{\text{Sum}}{n}=\frac{683.8}{12}\approx 56.9833$$

Step 2: Calculate the sample standard deviation ($s$)

We will first calculate the sum of squared deviations from the mean:

$$\text{Sum of Squared Deviations}=\sum (x_i-\overline{x}{)}^2$$

Each term $(x_i-\overline{x}{)}^2$ is calculated as follows:

• For 77: $(77-56.9833{)}^2\approx 400.6338$
• For 44.4: $(44.4-56.9833{)}^2\approx 158.4526$
• For 33.6: $(33.6-56.9833{)}^2\approx 568.6851$
• For 63.1: $(63.1-56.9833{)}^2\approx 37.8409$
• For 68.1: $(68.1-56.9833{)}^2\approx 125.8235$
• For 70.6: $(70.6-56.9833{)}^2\approx 178.1312$
• For 52: $(52-56.9833{)}^2\approx 4.8311$
• For 43.5: $(43.5-56.9833{)}^2\approx 179.8831$
• For 26.4: $(26.4-56.9833{)}^2\approx 926.0673$
• For 72.6: $(72.6-56.9833{)}^2\approx 236.0750$
• For 30.9: $(30.9-56.9833{)}^2\approx 679.5150$
• For 29.4: $(29.4-56.9833{)}^2\approx 743.0652$

Now summing these squared deviations:

$$\text{Total}=400.6338+158.4526+568.6851+37.8409+125.8235+178.1312+4.8311+179.8831+926.0673+236.0750+679.5150+743.0652\approx 3140.0968$$

Now, calculate the sample variance ($s^2$):

$$s^2=\frac{\text{Sum of Squared Deviations}}{n-1}=\frac{3140.0968}{12-1}\approx \frac{3140.0968}{11}\approx 285.463$$

Then take the square root to find the sample standard deviation ($s$):

$$s\approx \sqrt{285.463}\approx 16.91$$

Step 3: Determine the critical value ($t^{\ast }$)

For a 90% confidence level with $n-1=11$ degrees of freedom, we look up $t^{\ast }$ in the t-distribution table.

$t_{0.05,11}\approx 1.796$ (rounded to 3 decimal places)

Step 4: Calculate the margin of error (ME)

The margin of error is calculated as:

$$\text{ME}=t^{\ast }\times \frac{s}{\sqrt{n}}=1.796\times \frac{16.91}{\sqrt{12}}\approx 1.796\times 4.886=8.796$$

Step 5: Construct the confidence interval

Finally, the confidence interval is given by:

$$(\overline{x}-\text{ME},\overline{x}+\text{ME})=(56.9833-8.796,56.9833+8.796)$$

Calculating the endpoints:

$$\text{Lower Limit}=56.9833-8.796\approx 48.1873$$ $$\text{Upper Limit}=56.9833+8.796\approx 65.7793$$

Thus, the 90% confidence interval is:

$$\text{90\% C.I.}=(48.19,65.78)$$

Final result with two decimal places as requested:

$$\text{90\% C.I.}=(48.19,65.78)$$