# 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 =

Question
Confidence intervals
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 =

2021-01-11
Step 1 a. It is given that, Sample mean = 108. Population standard deviation, $$\displaystyle\sigma={26}$$. The sample size, n is 43. The z-critical value for 95% confidence interval is 1.96. The 95% confidence interval can be calculated as follows: $$\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}$$
$$\displaystyle={108}\pm{1.96}{\left({\frac{{{26}}}{{\sqrt{{{43}}}}}}\right)}$$
$$\displaystyle={108}\pm{7.7713}$$
$$\displaystyle={\left({100.2287},{115.7713}\right)}$$ The 95% confidence interval is (100.2287, 115.7713). Step 2 b. Population standard deviation, \sigma is 58. The 95% confidence interval can be calculated as follows: $$\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}$$
$$\displaystyle={108}\pm{1.96}{\left({\frac{{{58}}}{{\sqrt{{{43}}}}}}\right)}$$
$$\displaystyle={108}\pm{17.3360}$$
$$\displaystyle={\left({90.664},{125.336}\right)}$$ The 95% confidence interval is (90.664,125.336). Step 3 c. Population standard deviation,SPK \sigma=8ZSK. The 95% confidence interval can be calculated as follows: $$\displaystyle{C}{I}=\overline{{{x}}}\pm{z}{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}$$
$$\displaystyle={108}\pm{1.96}{\left({\frac{{{8}}}{{\sqrt{{{43}}}}}}\right)}$$
$$\displaystyle={108}\pm{2.3911}$$
$$\displaystyle={\left({105.6089},{110.3911}\right)}$$ The 95% confidence interval is (105.6089,110.3911).

### Relevant Questions

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?
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 \$________
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.
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 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.
You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 14 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.
$$9 6 10 15 19 6 23 26 19 16 11 25 16 11$$
a. To compute the confidence interval use a t or z distribution.
b. With 95% confidence the population mean number of visits per physical therapy patient is between ___ and ___ visits.
c. If many groups of 14 randomly selected physical therapy patients 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 number of visits per patient and about ___ percent will not contain the true population mean number of visits per patient.
Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let j: denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence intervals (7-8, 9.6)
(a) Would 2 90%% confidence interval calculated from this same sample have been narrower or wider than the glven interval? Explain your reasoning.
(b) Consider the following statement: There is 9 95% chance that Is between 7.8 and 9.6. Is this statement correct? Why or why not?
(c) Consider the following statement: We can be highly confident that 95% of al bottles ofthis type of cough syrup have an alcohol content that is between 7.8 and 9.6. Is this statement correct? Why or why not?
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
The marks of DMT students results in June 2020 sessional examinations were normally disyributed with a mean pass mark of 9 and a standard deviation pass mark of 0.15. After moderation, a sample of 30 papers was selected to see if the mean pass mark had changed. The mean pass mark of the sample was 8.95. a) Find the $$\displaystyle{95}\%$$ confidence interval of students mean mark. b) Calculate for the critical regions of the $$\displaystyle{95}\%$$ confidence intervals. c) Using your results in "a" and "b" above, is there evidence of a change in the mean pass mark of the DMT students.