If displaystyle{n}={13},{left(overline{{x}}right)}={31},{quadtext{and}quad}{s}={17}, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.

Question
Confidence intervals
asked 2021-03-08
If \(\displaystyle{n}={13},{\left(\overline{{x}}\right)}={31},{\quad\text{and}\quad}{s}={17},\) construct a confidence interval at a \(99\%\) confidence level. Assume the data came from a normally distributed population.
Give your answers to one decimal place.

Answers (1)

2021-03-09
Step 1
Given:
Sample size, \(\displaystyle{n}={13}\)
Sample mean, \(\displaystyle\overline{{x}}={31}\)
Sample standard deviation, \(\displaystyle{s}={17}\)
The degree of confidence, \(\displaystyle{c}={99}\%{\quad\text{or}\quad}{0.99}\)
Level of significance,
\(\displaystyle\alpha={1}-{c}\)
\(\displaystyle={1}-{0.99}\)
\(\displaystyle={0.01}\)
Let the population mean to be estimated using confidence intervals be \(\displaystyle\mu\)
Step 2
The \(99\%\) confidence interval for the population mean is:
\(\displaystyle\mu=\overline{{x}}\pm{t}_{{\frac{a}{{2}},{d}{f}}}\frac{s}{\sqrt{{n}}}\)
\(\displaystyle={31}\pm{t}_{{\frac{0.01}{{2}},{12}}}\frac{17}{\sqrt{{13}}}\)
\(\displaystyle={31}\pm{3.055}\times\frac{17}{\sqrt{{13}}}\)
\(\displaystyle={31}\pm{14.4}\)
\(\displaystyle={\left({16.6},{45.4}\right)}\)
Thus, the required \(99\%\) confidence interval for the population mean is:
\(\displaystyle{16.6}<\mu<{45.4}\)</span>
0

Relevant Questions

asked 2021-02-12
What is the optimal time for a scuba diver to be on the bottom of the ocean? That depends on the depth of the dive. The U.S. Navy has done a lot of research on this topic. The Navy defines the "optimal time" to be the time at each depth for the best balance between length of work period and decompression time after surfacing. Let \(\displaystyle{x}=\) depth of dive in meters, and let \(\displaystyle{y}=\) optimal time in hours. A random sample of divers gave the following data.
\(\begin{array}{|c|c|} \hline x & 13.1 & 23.3 & 31.2 & 38.3 & 51.3 &20.5 & 22.7 \\ \hline y & 2.78 & 2.18 & 1.48 & 1.03 & 0.75 & 2.38 & 2.20 \\ \hline \end{array}\)
(a)
Find \(\displaystyleΣ{x},Σ{y},Σ{x}^{2},Σ{y}^{2},Σ{x}{y},{\quad\text{and}\quad}{r}\). (Round r to three decimal places.)
\(\displaystyleΣ{x}=\)
\(\displaystyleΣ{y}=\)
\(\displaystyleΣ{x}^{2}=\)
\(\displaystyleΣ{y}^{2}=\)
\(\displaystyleΣ{x}{y}=\)
\(\displaystyle{r}=\)
(b)
Use a \(1\%\) level of significance to test the claim that \(\displaystyle\rho<{0}\). (Round your answers to two decimal places.)
\(\displaystyle{t}=\)
critical \(\displaystyle{t}=\)
Conclusion
Reject the null hypothesis. There is sufficient evidence that \(\displaystyle\rho<{0}\).Reject the null hypothesis. There is insufficient evidence that \(\displaystyle\rho<{0}\).
Fail to reject the null hypothesis. There is sufficient evidence that \(\displaystyle\rho<{0}\).Fail to reject the null hypothesis. There is insufficient evidence that \(\displaystyle\rho<{0}.\)
(c)
Find \(\displaystyle{S}_{{e}},{a},{\quad\text{and}\quad}{b}\). (Round your answers to four decimal places.)
\(\displaystyle{S}_{{e}}=\)
\(\displaystyle{a}=\)
\(\displaystyle{b}=\)
asked 2021-02-11
I have a question:
You intend to estimate a population mean \(\displaystyle\mu\) with the following sample.
54.7
48.9
56.3
43.5
41.5
41.2
54.2
You believe the population is normally distributed. Find the \(99.9\%\) confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).
\(\displaystyle{99.9}\%{C}.{I}.={\left({35.68},{61.54}\right)}\)Incorrect
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.
asked 2020-11-01
The owner of a large equipment rental company wants to make a rather quick estimate of the average number of days a piece of ditch-digging equipment is rented out per person per time. The company has records of all rentals, but the amount of time required to conduct an audit of all accounts would be prohibitive. The owner decides to take a random sample of rental invoices. Fourteen different rentals of ditch-diggers are selected randomly from the files, yielding the following data. She wants to use these data to construct a \(99\%\) confidence interval to estimate the average number of days that a ditch-digger is rented and assumes that the number of days per rental is normally distributed in the population. Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 2 decimal places, using conventional rounding rules.
DATA: 3 1 3 2 5 1 2 1 4 2 1 3 1 1
asked 2021-01-05
You are given the sample mean and standard deviation of the population. Use this information to construct the​ \(\displaystyle{90}\%{\quad\text{and}\quad}​{95}\%\) confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.
From a random sample of 66 ​dates, the mean record high daily temperature in a certain city has a mean of \(\displaystyle{85.69}^{\circ}{F}\). Assume the population standard deviation is \(\displaystyle{13.60}^{\circ}{F}.\)
The​ \(90\%\) confidence interval is
The​ \(95\%\) confidence interval is
Which interval is​ wider?
Interpret the results.
asked 2020-10-23
A random sample of \(\displaystyle{n}_{{1}}={16}\) communities in western Kansas gave the following information for people under 25 years of age.
\(\displaystyle{X}_{{1}}:\) Rate of hay fever per 1000 population for people under 25
\(\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}\)
A random sample of \(\displaystyle{n}_{{2}}={14}\) regions in western Kansas gave the following information for people over 50 years old.
\(\displaystyle{X}_{{2}}:\) Rate of hay fever per 1000 population for people over 50
\(\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}\)
(i) Use a calculator to calculate \(\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.\) (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use \(\displaystyle\alpha={0.05}.\)
(a) What is the level of significance?
State the null and alternate hypotheses.
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}\)
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value \(\displaystyle>{0.250}\)
\(\displaystyle{0.125}<{P}-\text{value}<{0},{250}\)
\(\displaystyle{0},{050}<{P}-\text{value}<{0},{125}\)
\(\displaystyle{0},{025}<{P}-\text{value}<{0},{050}\)
\(\displaystyle{0},{005}<{P}-\text{value}<{0},{025}\)
P-value \(\displaystyle<{0.005}\)
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
asked 2021-01-15
Use technology to construct the confidence intervals for the population variance \(\displaystyle\sigma^{{{2}}}\) and the population standard deviation \(\displaystyle\sigma\). Assume the sample is taken from a normally distributed population. \(\displaystyle{c}={0.99},{s}={37},{n}={20}\) The confidence interval for the population variance is (?, ?). The confidence interval for the population standard deviation is (?, ?)
asked 2021-03-05
A random sample of \(\displaystyle{n}={400}\) students is selected from a population of \(\displaystyle{N}={4000}\) students to estimate the average weight of the students. The sample mean and sample variance are found to be \(\displaystyle{x}={140}{l}{b}{\quad\text{and}\quad}{s}^{2}={225}.\ \text{Find the}\ {95}\%{\left({z}={2}\right)}\) confident interval.
asked 2020-11-07
Here are summary stastistics for randomly selected weights of newborn girls: \(\displaystyle{n}={224},\overline{{{x}}}={28.3}\text{hg},{s}={7.1}\text{hg}\). Construct a confidence interval estimate of mean. Use a 98% confidence level. Are these results very different from the confidence interval \(\displaystyle{26.5}\text{hg}{<}\mu{<}{30.7}\text{hg}\) with only 14 sample values, \(\displaystyle\overline{{{x}}}={28.6}\) hg, and \(\displaystyle{s}={2.9}\) hg? What is the confidence interval for the population mean \(\displaystyle\mu\)? \(\displaystyle?{<}\mu{<}?\) Are the results between the two confidence intervals very different?
asked 2021-02-21
We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 192 had kids. Based on this, plot a \(99\%\) confidence interval for the proportion of adult residents who are parents in a given county.
Express your answer in the form of three inequalities. Give your answers in decimal fractions up to three places \(\displaystyle<{p}<\) Express the same answer using a point estimate and a margin of error. Give your answers as decimals, to three places.
\(\displaystyle{p}=\pm\pm\)
asked 2021-02-16
A poll in 2017 reported that 699 out of 1027 adults in a certain country believe that marijuana should be legalized. When this poll about the subject was first conducted in 1969, only 12% of rhe country supported legalizztion. Assume the conditions for using the CLT are met.
a) Find and interpet a 99% confidence interval for the proportion of adults in the country 2017 that believe marijuana should be legalized is (0.643, 0.718)
b) Find and interpret a 90%confidence interval for this population parameter. The 90% confidence interval for the proportion of adults in the country 2017 that believe marijuana should be legalized is (0.657, 0.705)
c)Find the margin of error for each of the confidence intervals found The margin of error of the 99% confidence interval is 0.039 and the margin of error of the 90% confidence interval is 0.025
d) Without computing it, how would the margin of error of an 80% confidence interval compare with the margin of error for 90% and 99% intervals? Construct the 80% confidence interval to see if your production was correct
How would a 80% interval compare with the others in the margin of error?
...