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Chapter 10: Problem 14

The main stem growth measured for a sample of seventeen 4 -year-old red pine trees produced a mean and standard deviation equal to 11.3 and 3.4 inches, respectively. Find a \(90 \%\) confidence interval for the mean growth of a population of 4 -year-old red pine trees subjected to similar environmental conditions.

Short Answer

Expert verified
Answer: The 90% confidence interval for the mean growth of a population of 4-year-old red pine trees subjected to similar environmental conditions is (9.86, 12.74) inches.

Step by step solution

01

Set up the problem

Given the information, we have the sample mean (\(\bar{x}\)) and the sample standard deviation (s) for the 17 red pine trees. Sample mean (\(\bar{x}\)): 11.3 inches Sample standard deviation (s): 3.4 inches Sample size (n): 17 Confidence level: 90% Our goal is to calculate the 90% confidence interval for the mean growth of the population of 4-year-old red pine trees subjected to similar environmental conditions.
02

Find the degrees of freedom and t-score

Degrees of freedom (df) are calculated using the formula: $$ df = n - 1 $$ Where n is the sample size. In this case, the degrees of freedom df is: $$ df = 17 - 1 = 16 $$ Now we need to find the t-score that corresponds to a 90% confidence level for 16 degrees of freedom. To do this, use a t-table or a calculator with t-distribution functions. In this case, the t-score is approximately 1.746.
03

Calculate the margin of error

The margin of error (E) is calculated using the formula: $$ E = t \times \frac{s}{\sqrt{n}} $$ Where t is the t-score, s is the sample standard deviation, and n is the sample size. In this case, the margin of error is: $$ E = 1.746 \times \frac{3.4}{\sqrt{17}} \approx 1.44 $$
04

Calculate the confidence interval

Finally, we can calculate the 90% confidence interval using the sample mean and the margin of error: $$ (\bar{x} - E, \bar{x} + E) $$ In this case, the confidence interval is: $$ (11.3 - 1.44, 11.3 + 1.44) = (9.86, 12.74) $$
05

Interpret the confidence interval

The 90% confidence interval for the mean growth of a population of 4-year-old red pine trees subjected to similar environmental conditions is (9.86, 12.74) inches. This means that we are 90% confident that the true mean growth of these trees lies within this interval.

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Most popular questions from this chapter

A random sample of size \(n=7\) from a normal population produced these measurements: \(1.4,3.6,1.7,\) \(2.0,3.3,2.8,2.9 .\) a. Calculate the sample variance, \(s^{2}\) b. Construct a \(95 \%\) confidence interval for the population variance, \(\sigma^{2}\). c. Test \(H_{0}: \sigma^{2}=.8\) versus \(H_{\mathrm{a}}: \sigma^{2} \neq .8\) using \(\alpha=.05 .\) State your conclusions. d. What is the approximate \(p\) -value for the test in part \(\mathrm{c} ?\)

The EPA limit on the allowable discharge of suspended solids into rivers and streams is 60 milligrams per liter (mg/) per day. A study of water samples selected from the discharge at a phosphate mine shows that over a long period, the mean daily discharge of suspended solids is \(48 \mathrm{mg} / \mathrm{l}\). but day-to-day discharge readings are variable. State inspectors measured the discharge rates of suspended solids for \(n=20\) days and found \(s^{2}=39(\mathrm{mg} / 1)^{2}\) Find a \(90 \%\) confidence interval for \(\sigma^{2}\). Interpret your results.

To compare the demand for two different entrees, the manager of a cafeteria recorded the number of purchases of each entree on seven consecutive days. The data are shown in the table. Do the data provide sufficient evidence to indicate a greater mean demand for one of the entrees? Use the MINITAB printout. \begin{tabular}{lcc} Day & \(\mathrm{A}\) & \(\mathrm{B}\) \\ \hline Monday & 420 & 391 \\ Tuesday & 374 & 343 \\ Wodnesday & 434 & 469 \\\ Thursday & 395 & 412 \\ Friday & 637 & 538 \\ Saturday & 594 & 521 \\ Sunday & 679 & 625 \end{tabular}.

Measurements of water intake, obtained from a sample of 17 rats that had been injected with a sodium chloride solution, produced a mean and standard deviation of 31.0 and 6.2 cubic centimeters \(\left(\mathrm{cm}^{3}\right),\) respectively. Given that the average water intake for non injected rats observed over a comparable period of time is \(22.0 \mathrm{~cm}^{3},\) do the data indicate that injected rats drink more water than non injected rats? Test at the \(5 \%\) level of significance. Find a \(90 \%\) confidence interval for the mean water intake for injected rats.

In Exercise 10.6 we presented data on the estimated average price for a 6 -ounce can or a 7.06 -ounce pouch of tuna, based on prices paid nationally in supermarkets. A portion of the data is reproduced in the table below. Use the MINI TAB printout to answer the questions. \begin{tabular}{rr|rr} Light Tuna in Water & \multicolumn{2}{|l} { Light Tuna in 0il } \\ \hline .99 & .53 & 2.56 & .62 \\ 1.92 & 1.41 & 1.92 & .66 \\ 1.23 & 1.12 & 1.30 & .62 \\ .85 & .63 & 1.79 & .65 \\ .65 & .67 & 1.23 & .60 \\ .69 & .60 & & .67 \\ 60 & .66 & & \end{tabular} MINI TAB output for Exercise 10.25 Two-Sample T-Test and Cl: Water, Oil Two-sample T for Water vs Oil Mean StDev \(\mathrm{SE}\) Mean 1 er \(0.896 \quad 0.400 \quad 0 .\) 11 1.147 0.679 0.2 Difference \(=\) mu (Water) - mu (Oil) Estimate for difference: -0.251 958 CI for difference: (-0.700,0.198) T-Test of difference \(=0\) (vs not \(=\) ): T-value \(=-1.16\) P-Value \(=0.260 \quad D F=23\) Both use Pooled StDev \(=0.5389\) a. Do the data in the table present sufficient evidence to indicate a difference in the average prices of light tuna in water versus oil? Test using \(\alpha=.05 .\) b. What is the \(p\) -value for the test? c. The MINITAB analysis uses the pooled estimate of \(\sigma^{2}\). Is the assumption of equal variances reasonable? Why or why not?
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