# Here are summary stastistics for randomly selected weights of newborn girls: 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 26.5 text{hg} < mu < 30.7 text{hg} with only 14 sample values, overline{x}=28.6 hg, and s=2.9 hg? What is the confidence interval for the population mean mu? ?

Question
Confidence intervals
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 $$\displaystyle{26.5}\text{hg}{<}\mu{<}{30.7}\text{hg}$$</span> 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 $$\displaystyle\mu$$? $$\displaystyle?{<}\mu{<}?$$</span> Are the results between the two confidence intervals very different?

2020-11-08
Step 1 From the provided information, Sample size $$\displaystyle{\left({n}\right)}={244}$$ Sample mean $$\displaystyle{\left(\overline{{{x}}}\right)}={28.3}$$ hg Sample standard deviation $$\displaystyle{\left({s}\right)}={7.1}$$ hg Step 2 Since, the population standard deviation is unknown, therefore, the t distribution will be used. Confidence level = 98% Level of significance $$\displaystyle{\left(\alpha\right)}={1}-{0.98}={0.02}$$ The degree of freedom $$\displaystyle={n}–{1}={244}–{1}={243}$$ The critical value of t at 243 degree of freedom with 0.01 level of significance from the t value table is 2.34. Step 3 The required 98% confidence interval can be obtained as: $$\displaystyle{C}{I}=\overline{{{x}}}\pm{t}_{{\frac{\alpha}{{2}},{n}-{1}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle={28.3}\pm{\left({2.34}\right)}{\frac{{{7.1}}}{{\sqrt{{{224}}}}}}$$
$$\displaystyle={28.3}\pm{1.1}$$
$$\displaystyle={\left({27.2},{29.4}\right)}$$ Thus, the confidence interval is $$\displaystyle{27.2}{<}\mu{<}{29.4}$$</span>. No, the results between the two confidence intervals are not very different. The confidence interval limits are almost similar.

### 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 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.
Testing for a Linear Correlation. In Exercises 13–28, construct a scatterplot, and find the value of the linear correlation coefficient r. Also find the P-value or the critical values of r from Table A-6. Use a significance level of $$\alpha = 0.05$$. Determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables. (Save your work because the same data sets will be used in Section 10-2 exercises.) Lemons and Car Crashes Listed below are annual data for various years. The data are weights (metric tons) of lemons imported from Mexico and U.S. car crash fatality rates per 100,000 population [based on data from “The Trouble with QSAR (or How I Learned to Stop Worrying and Embrace Fallacy),” by Stephen Johnson, Journal of Chemical Information and Modeling, Vol. 48, No. 1]. Is there sufficient evidence to conclude that there is a linear correlation between weights of lemon imports from Mexico and U.S. car fatality rates? Do the results suggest that imported lemons cause car fatalities? $$\begin{matrix} \text{Lemon Imports} & 230 & 265 & 358 & 480 & 530\\ \text{Crashe Fatality Rate} & 15.9 & 15.7 & 15.4 & 15.3 & 14.9\\ \end{matrix}$$
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
Solve and answer this question Are the results between the two confidence intervals very different
$$27.2 hg < \mu < 29.4 hg$$
What is the confidence interval for the population mean $$\mu?$$
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.
Here is a sample of amounts of weight change (kg) of college students in their freshman year: 10, 7, 6, -7, where -7 represents a loss of 7 kg and positive values represent weight gained. Here are ten bootstrap samples: $$\displaystyle{\left\lbrace{10},{10},{10},{6}\right\rbrace},{\left\lbrace{10},-{7},{6},{10}\right\rbrace},{\left\lbrace{10},-{7},{7},{6}\right\rbrace},{\left\lbrace{7},-{7},{6},{10}\right\rbrace},{\left\lbrace{6},{6},{6},{7}\right\rbrace},{\left\lbrace{7},-{7},{7},-{7}\right\rbrace},{\left\lbrace{10},{7},-{7},{6}\right\rbrace},{\left\lbrace-{7},{7},-{7},{7}\right\rbrace},{\left\lbrace-{7},{6},-{7},{7}\right\rbrace},{\left\lbrace{7},{10},{10},{10}\right\rbrace}$$. a) Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the mean weight change for the population. $$\displaystyle?{k}{g}{<}{m}{y}{<}?{k}{g}$$
Using the Minitab statistical analysis program to enter the data and perform the analysis, complete the following: Construct a one-sided $$\displaystyle{95}\%$$ confidence interval for the true difference in population means. Test the null hypothesis that the population means are identical at the 0.05 level of significance.
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?