# 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 95% confidence interval for the true mean service life. 2) 2. Find a 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? Question
Confidence intervals 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? 2021-02-10
Step 1 Given information Sample size $$\displaystyle{\left({n}\right)}={25}$$ Sample mean $$\displaystyle\overline{{{x}}}={30},{000}\ {k}{m}$$ Population standard deviation $$\displaystyle={5},{000}\ {k}{m}$$ 3) Significance level $$\displaystyle{\left(\alpha\right)}={1}\ -\ {0.95}={0.05}$$ $$\displaystyle{95}\%$$ Confidence interval is given by $$\displaystyle{C}.{I}=\ \overline{{{x}}}\ \pm\ {Z}_{{{\frac{{\alpha}}{{{2}}}}}}\ \times\ {\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle={30.000}\ \pm\ {1.9600}\ \times\ {\frac{{{5000}}}{{\sqrt{{{25}}}}}}={\left({28040},\ {31390}\right)}$$
$$\displaystyle{Z}_{{{\frac{{{0.05}}}{{{2}}}}}}={Z}_{{{0.025}}}=\ \pm\ {1.9600}\ \text{(From Excel = NORM.S.INV(0.025))}$$ Step 2 2) Significance level $$\displaystyle{\left(\alpha\right)}={1}\ -\ {0.75}={0.25}$$
$$\displaystyle{75}\%$$ Confidence interval is given by $$\displaystyle{C}.{I}=\ \overline{{{x}}}\ \pm\ {Z}_{{{\frac{{\alpha}}{{{2}}}}}}\ \times\ {\frac{{{5000}}}{{\sqrt{{25}}}}}={\left({28849.7},\ {31150.3}\right)}$$
$$\displaystyle{Z}_{{{\frac{{{0.25}}}{{{2}}}}}}={Z}_{{{0.125}}}=\ \pm\ {1.1503}\ \text{(From Excel = NORM. S. INV(0.125))}$$ Step 3 3) Width of interval for $$\displaystyle{95}\%$$ confidence interval $$\displaystyle={31960}\ -\ {28040}={3920}$$ Width of interval for $$\displaystyle{75}\%$$ confidence interval $$\displaystyle={31150.3}\ -\ {28849.7}={2300.6}$$ On decreasing Confidence level. Width of confidence level decreases.

### 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 random sample of size $$\displaystyle{n}={25}$$ from a normal distribution with mean $$\displaystyle\mu$$ and variance 36 has sample mean $$\displaystyle\overline{{X}}={16.3}$$
a)
Calculate confidence intervals for $$\displaystyle\mu$$ at three levels of confidence: $$80%, 90% \text{and} 99%$$. How to the widths of the confidence intervals change?
b) How would the CI width change if n is increased to 100? 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. (a) The company's production equipment produces metal discs weighing 200 g. It should be noted that the weight of the discs corresponds to the normal distribution. To check machine consistency, 20 discs are randomly selected with an average weight of 205 g and a standard deviation of 7 g. What is the 99% confidence interval for the average weight of the selected discs?
(b) A company launched a new model of golf ball. It claimed that the driving distance is at least 300m. A sample of 20 balls yields a sample mean of 295m and sample standard deviation of 8m. It is assumed that the driving distance is normally distributed.
Conduct an appropriate hypothesis testing at the 0.05 level of significance. Is there any evidence that the average travel distance stated by the company is true? 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. 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. 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) 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.
With $$90\%$$ confidence the population mean minutes of concentration is between ____ and ____ minutes. 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 \$________