A random sample of the price of gasoline from 30 gas stations in a region gives the statistics below. Complete parts​ a) through​ c).
y overbarequals​$3.39​, sequals​$0.29
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Part 1
​a) Find a 95​% confidence interval for the mean price of regular gasoline in that region.
left parenthesis 3.282 comma 3.498 right parenthesis
  
  
 
​(Round to three decimal places as​ needed.)
Part 2
​b) Find the​ 90% confidence interval for the mean.
left parenthesis 3.300 comma 3.480 right parenthesis
select: 3.300
  
​(Round to three decimal places as​ needed.)
Part 3
​c) If we had the same statistics from a sample of 60 ​stations, what would the​ 95% confidence interval be​ now?
left parenthesis nothing comma nothing right parenthesis
  
  
​(Round to three decimal places as​ needed.)

Let's solve each part step by step, using the provided sample data.

### Given Data:
- Sample mean, $\bar{y} = 3.39$
- Sample standard deviation, $s = 0.29$
- Sample size, $n = 30$ for parts (a) and (b), and $n = 60$ for part (c).

### Part 1: 95% Confidence Interval for the Mean
To calculate a confidence interval (CI), we use the following formula:
\[
CI = \bar{y} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}
\]
Where:
- $\bar{y}$ is the sample mean,
- $t_{\alpha/2}$ is the critical value from the $t$-distribution,
- $s$ is the sample standard deviation,
- $n$ is the sample size.

For a 95% confidence level and $n = 30$, the degrees of freedom $df = n - 1 = 29$.

From the $t$-distribution table, $t_{0.025, 29} \approx 2.045$.

Now calculate the margin of error:
\[
ME = 2.045 \times \frac{0.29}{\sqrt{30}} \approx 0.108
\]

So, the 95% confidence interval is:
\[
CI = 3.39 \pm 0.108 = (3.282, 3.498)
\]

### Part 2: 90% Confidence Interval for the Mean
For a 90% confidence level, we use a different critical value. From the $t$-distribution table, $t_{0.05, 29} \approx 1.699$.

Now calculate the margin of error:
\[
ME = 1.699 \times \frac{0.29}{\sqrt{30}} \approx 0.090
\]

So, the 90% confidence interval is:
\[
CI = 3.39 \pm 0.090 = (3.300, 3.480)
\]

### Part 3: 95% Confidence Interval for $n = 60$
If we increase the sample size to 60, the formula remains the same, but now $n = 60$. The degrees of freedom $df = 59$.

From the $t$-distribution table, $t_{0.025, 59} \approx 2.000$.

Now calculate the margin of error:
\[
ME = 2.000 \times \frac{0.29}{\sqrt{60}} \approx 0.075
\]

So, the new 95% confidence interval is:
\[
CI = 3.39 \pm 0.075 = (3.315, 3.465)
\]

### Summary:
- **a)** 95% CI for $n = 30$: $ (3.282, 3.498) $
- **b)** 90% CI for $n = 30$: $ (3.300, 3.480) $
- **c)** 95% CI for $n = 60$: $ (3.315, 3.465) $

Do you want more details or have any questions about these calculations?

Here are 5 related questions to deepen your understanding:
1. How does increasing the sample size affect the confidence interval?
2. What is the relationship between confidence level and the width of the confidence interval?
3. Why do we use the $t$-distribution instead of the normal distribution here?
4. What happens to the margin of error if the sample standard deviation is larger?
5. How would the confidence interval change if the sample mean were lower?

**Tip:** Larger sample sizes tend to decrease the margin of error, leading to narrower confidence intervals, meaning more precise estimates.

A random sample of the price of gasoline from 30 gas stations in a region gives the statistics below. Complete parts​ a) through​ c). y overbarequals​$3.39​, sequals​$0.29. Part 1: ​a) Find a 95​% confidence interval for the mean price of regular gasoline in that region. (3.282, 3.498). ​b) Find the​ 90% confidence interval for the mean. (3.300, 3.480). ​c) If we had the same statistics from a sample of 60 ​stations, what would the​ 95% confidence interval be​ now?

Learn how to calculate 95% and 90% confidence intervals for the mean gasoline price from a sample of 30 gas stations, including how the sample size affects the interval width. Understand key concepts such as confidence intervals, margin of error, and t-distribution.

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