# You are given the sample mean and standard deviation of the population. Use this information to construct the​ displaystyle{90}%{quadtext{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. Question
Confidence intervals 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. 2021-01-06
Step 1
Given:
This mean recorded temperature is: $$\displaystyle\overline{{x}}={85.69}{F}^{\circ}$$
The population's standard deviation is: $$\displaystyle\sigma={13.60}{F}^{\circ}$$
The number of random samples is: $$\displaystyle{n}={66}$$ dates.
Confidence level $$\displaystyle-{90}\%$$:
The expression for the confidence level is,
$$\displaystyle{1}-{2}\alpha={0.9}$$
$$\displaystyle\alpha=\frac{{{1}-{0.9}}}{{2}}$$
$$\displaystyle={0.05}$$
Step 2
From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,
$$\displaystyle{P}={1}-\alpha$$
$$\displaystyle={1}-{0.05}$$
$$\displaystyle={0.95}$$
From the z-table corresponding to $$\displaystyle{P}={0.95}$$, the value of z is 1.65
The expression to calculate the interval is,
$$\displaystyle{c}.{I}=\overline{{x}}\pm{z}\frac{\sigma}{\sqrt{{n}}}$$
Substitute the values in the above expression.
$$\displaystyle{c}.{l}={85.69}\pm{1.65}\times\frac{13.60}{\sqrt{{66}}}$$
$$\displaystyle={85.69}\pm{2.762}$$
$$\displaystyle={\left({85.69}-{2.762},{85.69}-{2.762}\right)}$$
$$\displaystyle={\left({82.928},{88.452}\right)}$$
Therefore, the interval for confidence leven $$90\%$$ is from 82.928 to 88.452
Step 3
Confidence level $$- 95\%:$$
The expression for the confidence level is,
$$\displaystyle{1}-{2}\alpha={0.95}$$
$$\displaystyle\alpha=\frac{{{1}-{0.95}}}{{2}}$$
$$\displaystyle={0.025}$$
From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,
$$\displaystyle{P}={1}-\alpha$$
$$\displaystyle={1}-{0.025}$$
$$\displaystyle={0.975}$$
From the z-table corresponding to $$\displaystyle{P}={0.975}$$, the value of z is 1.96.
The expression to calculate the interval is,
$$\displaystyle{c}.{l}=\overline{{x}}\pm{z}\frac{\sigma}{\sqrt{{n}}}$$
Substitute the values in the above expression.
$$\displaystyle{c}.{l}-{85.69}\pm{1.96}\times\frac{13.60}{\sqrt{{66}}}$$
$$\displaystyle={85.69}\pm{3.281}$$
$$\displaystyle={\left({85.69}-{3.281},{85.69}-{3.281}\right)}$$
$$\displaystyle={\left({82.409},{88.971}\right)}$$
Hence, the interval for confidence level $$95\%$$ is from 82.409 to 88.971, and the interval for confidence level $$95\%$$ is the widest one.

### Relevant Questions You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 58 ​dates, the mean record high daily temperature in a certain city has a mean of $$\displaystyle{83.43}^{{\circ}}{F}$$. Assume the population standard deviation is $$\displaystyle{14.02}^{{\circ}}{F}$$. $$\displaystyle{90}\%=$$ $$\displaystyle{95}\%=$$ Which interval is wider? A researcher has gathered information from a random sample of 178 households. For each of the following variables, construct confidence intervals to estimate the population mean. Use the 90% level. a. An average of 2.3 people resides in each household. Standard deviation is 0.35. b. There was an average of 2.1 television sets (s  0.10) and 0.78 telephones (s  0.55) per household. c. The households averaged 6.0 hours of television viewing per day (s  3.0) 1. A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 44.7 texts per day. The standard deviation was 16.5 texts. a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean number of texts per day is between and texts. c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day. 2. You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately $$\displaystyle\sigma={40.4}$$ dollars. You would like to be 90% confident that your estimate is within 1.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? n = 3. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $$\displaystyle\sigma={57.5}$$. You would like to be 95% confident that your estimate is within 0.1 of the true population mean. How large of a sample size is required? In a survey of 2566 adults in a recent​ year, 1420 say they have made a New​ Year's resolution. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. The​ 90% confidence interval for the population proportion p is __,__ The​ 95% confidence interval for the population proportion p is __,__ 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? 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 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. Money reports that the average annual cost of the first year of owning and caring for a large dog in 2017 is $1,448. The Irish Red and White Setter Association of America has requested a study to estimate the annual first-year cost for owners of this breed. A sample of 50 will be used. Based on past studies, the population standard deviation is assumed known with $$\displaystyle\sigma=\{230}.$$ $$\begin{matrix} 1,902 & 2,042 & 1,936 & 1,817 & 1,504 & 1,572 & 1,532 & 1,907 & 1,882 & 2,153 \\ 1,945 & 1,335 & 2,006 & 1,516 & 1,839 & 1,739 & 1,456 & 1,958 & 1,934 & 2,094 \\ 1,739 & 1,434 & 1,667 & 1,679 & 1,736 & 1,670 & 1,770 & 2,052 & 1,379 & 1,939\\ 1,854 & 1,913 & 2,163 & 1,737 & 1,888 & 1,737 & 2,230 & 2,131 & 1,813 & 2,118\\ 1,978 & 2,166 & 1,482 & 1,700 & 1,679 & 2,060 & 1,683 & 1,850 & 2,232 & 2,294 \end{matrix}$$ (a) What is the margin of error for a $$95\%$$ confidence interval of the mean cost in dollars of the first year of owning and caring for this breed? (Round your answer to nearest cent.) (b) The DATAfile Setters contains data collected from fifty owners of Irish Setters on the cost of the first year of owning and caring for their dogs. Use this data set to compute the sample mean. Using this sample, what is the $$95\%$$ confidence interval for the mean cost in dollars of the first year of owning and caring for an Irish Red and White Setter? (Round your answers to nearest cent.)$_______ to \$________ In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: $$\displaystyle{0.127}{<}{p}{<}{0.191}$$. What do you​ conclude? a. Construct a 90​% confidence interval. Express the percentages in decimal form. ___ The service life in kilometer of Goody tires is assumed to follow a normal distribution with a standard deviation of 5,000 km. A random sample of 25 tires yielded a mean service life of 30,000 km. 1) Find the $$\displaystyle{95}\%$$ confidence interval for the true mean service life. 2) 2. Find a $$\displaystyle{75}\%$$ confidence interval for the true mean service life. 3) Calculate the widths of the intervals found in 1 and 2. How do these widths change as the confidence level decreases?