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 83.43^{circ}F. Assume the population standard deviation is 14.02^{circ}F. 90%= 95%= Which interval is wider?

Question
Confidence intervals
asked 2020-12-15
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?

Answers (1)

2020-12-16
Step 1 Given Data: The sample size is: \(\displaystyle{n}={58}\) The mean temperature of the city is: \(\displaystyle\mu={83.43}^{{\circ}}{F}\) The standard deviation of the temperature is: \sigma=14.02^{\circ}F 95% Population: The expression calculate the 95% of the Population is, \(\displaystyle{1}-{2}\alpha={0.95}\)
\(\displaystyle\alpha={\frac{{{1}-{0.95}}}{{{2}}}}\)
\(\displaystyle={0.025}\) Corresponds to probability 0.025, the z value from the right tail end of the z-table 1.96. Step 2 The expression to calculate the interval of the population is, \(\displaystyle{C}{I}=\overline{{{X}}}\pm{M}.{E}\)
\(\displaystyle=\overline{{{X}}}\pm{Z}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\) Substitute values in the above expression. \(\displaystyle{C}{I}={83.43}\pm{1.96}\times{\frac{{{14.02}}}{{\sqrt{{{58}}}}}}\)
\(\displaystyle={83.43}\pm{3.60}\)
\(\displaystyle={\left({83.43}-{3.60},{83.43}+{3.60}\right)}\)
\(\displaystyle{\left({79.83},{87.03}\right)}\) Thus, the interval width of the 95% level is 3.6. and the interval is (79.83, 87.03). Step 3 90% Population: The expression calculate the 95% of the Population is, \(\displaystyle{1}-{2}\alpha={0.90}\)
\(\displaystyle\alpha={\frac{{{1}-{0.90}}}{{{2}}}}\)
\(\displaystyle={0.050}\) Corresponds to probability 0.050, the z value from the right tail end of the z-table is 1.64. The expression to calculate the interval of the Population is, \(\displaystyle{C}{I}=\overline{{{X}}}\pm{M}.{E}\)
\(\displaystyle=\overline{{{X}}}\pm{Z}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\) Substitute values in the above expression. \(\displaystyle{C}{I}={83.43}\pm{1.96}\times{\frac{{{14.02}}}{{\sqrt{{{58}}}}}}\)
\(\displaystyle={83.43}\pm{3.02}\)
\(\displaystyle={\left({83.43}-{3.60},{83.43}+{3.02}\right)}\)
\(\displaystyle={\left({80.41},{86.45}\right)}\) Thus, the interval width of 90% population level is 3.02, and the interval is (80.41, 59.45). Thus, 95% population has higher interval width.
0

Relevant Questions

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 2020-12-30
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)
asked 2021-02-23
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?
asked 2020-11-22
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 __,__
asked 2021-01-10
A statistics practitioner took a random sample of 43 observations from a population whose standard deviation is 26 and computed the sample mean to be 108. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 95% confidence, changing the population standard deviation to 58, Confidence Interval = C. Estimate the population mean with 95% confidence, changing the population standard deviation to 8, Confidence Interval =
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?
asked 2021-02-06
You may need to use the appropriate appendix table or technology to answer this question.
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 $________
asked 2021-02-09
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?
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 2020-11-09
A researcher is interested in finding a \(90\%\) confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture. The study included 117 students who averaged 40.9 minutes concentrating on their professor during the hour lecture. The standard deviation was 11.8 minutes. Round answers to 3 decimal places where possible.
a.
To compute the confidence interval use a ? distribution.
b.
With \(90\%\) confidence the population mean minutes of concentration is between ____ and ____ minutes.
c.
If many groups of 117 randomly selected students are studied, then a different confidence interval would be produced from each group. About ____ percent of these confidence intervals will contain the true population mean minutes of concentration and about ____ percent will not contain the true population mean number of minutes of concentration.
...