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How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from [tex]$20^{\circ} F$[/tex] to [tex]$45^{\circ} F$[/tex]. A random sample of prices for sleeping bags in this temperature range is given below. Assume that the population of [tex]$x$[/tex] values has an approximately normal distribution.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
60 & 105 & 100 & 110 & 75 & 95 & 30 & 23 & 100 & 110 \\
\hline
105 & 105 & 105 & 60 & 110 & 120 & 105 & 90 & 100 & 70 \\
\hline
\end{tabular}
\][/tex]

(a) Use a calculator with mean and sample standard deviation keys to find the sample mean price [tex]$\bar{x}$[/tex] (in dollars) and sample standard deviation [tex]$s$[/tex] (in dollars). (Round your standard deviation to four decimal places.)
[tex]\[
\bar{x} = \square
\][/tex]
[tex]\[
s = \square
\][/tex]

(b) Using the given data as representative of the population of prices of all sleeping bags, find a [tex]$90\%$[/tex] confidence interval for the mean price [tex]$\mu$[/tex] (in dollars) of all sleeping bags. (Enter your answer in the form: lower limit to upper limit. Include the word "to." Round your numerical values to two decimal places.)
[tex]\[
\mu = \square \text{ to } \square
\][/tex]


Sagot :

### (a) Calculate the Sample Mean and Sample Standard Deviation

To find the sample mean ( [tex]\(\bar{x}\)[/tex] ) and sample standard deviation ( [tex]\(s\)[/tex] ), we first need to list all the given prices:

[tex]\[ 60, 5, 100, 110, 75, 95, 30, 23, 100, 110, 105, 5, 105, 60, 110, 120, 5, 90, 0, 70 \][/tex]

Sample Size ( [tex]\(n\)[/tex] ):

[tex]\[ n = 20 \][/tex]

Sample Mean ( [tex]\(\bar{x}\)[/tex] ):

The sample mean [tex]\(\bar{x}\)[/tex] is calculated by summing all the prices and then dividing by the number of prices:

[tex]\[ \bar{x} = \frac{60 + 5 + 100 + 110 + 75 + 95 + 30 + 23 + 100 + 110 + 105 + 5 + 105 + 60 + 110 + 120 + 5 + 90 + 0 + 70}{20} \][/tex]

Upon calculation, we get:

[tex]\[ \bar{x} = 68.9 \][/tex]

Sample Standard Deviation ( [tex]\(s\)[/tex] ):

The sample standard deviation [tex]\(s\)[/tex] is calculated using the formula for the sample standard deviation, which corrects for bias by using [tex]\( n-1 \)[/tex] in the denominator:

[tex]\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \][/tex]

Upon calculation, we find:

[tex]\[ s \approx 42.4015 \][/tex]

### (b) Find the 90% Confidence Interval for the Mean Price

To determine a 90% confidence interval for the mean price [tex]\( \mu \)[/tex], we use the t-distribution because the sample size is small ( [tex]\( n < 30 \)[/tex] ) and the population standard deviation is unknown.

Confidence Level ( [tex]\(\gamma\)[/tex] ):

[tex]\[ \gamma = 0.90 \][/tex]

This means the level of significance [tex]\( \alpha \)[/tex] is:

[tex]\[ \alpha = 1 - \gamma = 0.10 \][/tex]

Degrees of Freedom ( df ):

[tex]\[ df = n - 1 = 20 - 1 = 19 \][/tex]

Using the t-distribution table, for a 90% confidence level and 19 degrees of freedom, the t-critical value [tex]\(t_{ \text{critical} } \)[/tex] is approximately:

[tex]\[ t_{ \text{critical} } \approx 1.729 \][/tex]

Next, we calculate the margin of error ( [tex]\( E \)[/tex] ):

[tex]\[ E = t_{ \text{critical} } \times \left( \frac{s}{\sqrt{n}} \right) \][/tex]

[tex]\[ E = 1.729 \times \left( \frac{42.4015}{\sqrt{20}} \right) \][/tex]

Upon calculation, we find:

[tex]\[ E \approx 16.39 \][/tex]

Finally, the confidence interval for the mean price [tex]\(\mu\)[/tex]:

[tex]\[ (\bar{x} - E) \text{ to } (\bar{x} + E) \][/tex]

[tex]\[ 68.9 - 16.39 \text{ to } 68.9 + 16.39 \][/tex]

[tex]\[ 52.51 \text{ to } 85.29 \][/tex]

### Conclusion
(a) The sample mean price [tex]\(\bar{x}\)[/tex] is [tex]$68.90 and the sample standard deviation \(s\) is approximately $[/tex]42.4015.

(b) The 90% confidence interval for the mean price of all sleeping bags in the given temperature range is:

52.51 to 85.29 dollars.