# The average zinc concentration recovered from a sample of measurements taken in 36 different locations in a river is found to be 2.6 grams per liter. Find the 95% confidence intervals for the mean zinc concentration in the river. Assume that the population standard deviation is 0.3 gram per liter. (Z_{0.025} = 1.96, Z_{0.005} = 2.575) Question
Confidence intervals The average zinc concentration recovered from a sample of measurements taken in 36 different locations in a river is found to be 2.6 grams per liter. Find the 95% confidence intervals for the mean zinc concentration in the river. Assume that the population standard deviation is 0.3 gram per liter. (Z_{0.025} = 1.96, Z_{0.005} = 2.575) 2021-01-11
Step 1 The $$\displaystyle{100}{\left({1}-\alpha\right)}\%$$ confidence interval for the population mean, $$\displaystyle\mu$$ when the population standard deviations, $$\displaystyle\sigma$$ is known the z-confidence interval as follows: $$\displaystyle{\left(\overline{{{x}}}-{Z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}},\overline{{{x}}}+{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}$$ In this case,$$\displaystyle\overline{{{x}}},\sigma,\mu,$$ are the sample mean, population standard deviation, population mean and sample size of the population, respectively and $$\displaystyle{z}_{{\frac{\alpha}{{2}}}}$$ is the critical value of the standard normal distribution, above which,$$\displaystyle{100}{\left(\frac{\alpha}{{2}}\right)}\%{P}{S}{K}{\quad\text{or}\quad}{P}{S}{K}\frac{\alpha}{{2}}$$ proportion of the observations lie. Step 2 The sample mean is, $$\displaystyle\overline{{{x}}}={2.6}$$, the population standard deviation is $$\displaystyle\sigma={0.3}$$ and the sample size is $$\displaystyle{n}={36}$$. If 95% of the observations must lie within an interval, the remaining 5% must lie outside the interval. Further, due to symmetry, 2.5% will lie above the upper limit and the remaining 2.5% will lie below the lower limit of the interval. Thus, the upper limit of the interval is such that, 97.5% lie below it. As a result,$$\displaystyle{z}_{{\frac{\alpha}{{2}}}}={z}_{{{0.025}}}\approx{1.96}$$ [Using Excel formula: =NORM.S.INV(0.975)]. Thus, the confidence interval is, $$\displaystyle{\left(\overline{{{x}}}-{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}},\overline{{{x}}}+{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}={\left({2.6}-{\left({1.96}\right)}{\left({\frac{{{0.3}}}{{\sqrt{{{36}}}}}}\right)},{2.6}+{\left({1.96}\right)}{\left({\frac{{{0.3}}}{{\sqrt{{{36}}}}}}\right)}\right)}$$
$$\displaystyle={\left({2.6}-{\left({1.96}\right)}{\left({0.05}\right)},{2.6}+{\left({1.96}\right)}{\left({0.05}\right)}\right)}$$
$$\displaystyle{\left({2.6}-{0.098},{2.6}+{0.098}\right)}$$
PSK(2.502, 2.698) Thus, the 95% confidence interval for the mean zinc concentration in the river is (2.502 grams per liter, 2.698 grams per liter).

### 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? 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) 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 \$________ 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. Construct 98% confidence intervals for the a) population variance and the b)population standard deviation $$\displaystyle\sigma$$ The number of hours of reserve capacity of 10 randomly selected automotive batteries. 1.71, 1.87, 1.58, 1.61, 1.78, 1.98, 1.36, 1.58, 1.47, 2.05, Assume the sample is taken from a normally distributed population. 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? 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? 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. $$9 6 10 15 19 6 23 26 19 16 11 25 16 11$$ $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{H}{o}{u}{s}{e}{w}{\quad\text{or}\quad}{k}{H}{o}{u}{r}{s}\backslash{h}{l}\in{e}{G}{e}{n}{d}{e}{r}&{S}{a}\mp\le\ {S}{i}{z}{e}&{M}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {D}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{W}{o}{m}{e}{n}&{473473}&{33.133}{.1}&{14.214}{.2}\backslash{h}{l}\in{e}{M}{e}{n}&{488488}&{18.618}{.6}&{15.715}{.7}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ a. Based on this​ study, calculate how many more hours per​ week, on the​ average, women spend on housework than men. b. Find the standard error for comparing the means. What factor causes the standard error to be small compared to the sample standard deviations for the two​ groups? The cause the standard error to be small compared to the sample standard deviations for the two groups. c. Calculate the​ 95% confidence interval comparing the population means for women Interpret the result including the relevance of 0 being within the interval or not. The​ 95% confidence interval for ​$$\displaystyle{\left(\mu_{{W}}-\mu_{{M}}​\right)}$$ is: (Round to two decimal places as​ needed.) The values in the​ 95% confidence interval are less than 0, are greater than 0, include 0, which implies that the population mean for women could be the same as is less than is greater than the population mean for men. d. State the assumptions upon which the interval in part c is based. Upon which assumptions below is the interval​ based? Select all that apply. A.The standard deviations of the two populations are approximately equal. B.The population distribution for each group is approximately normal. C.The samples from the two groups are independent. D.The samples from the two groups are random.