# Suppose you have selected a random sample of ? displaystyle={14} measurements from a normal distribution. Compare the standard normal ? values with the corresponding ? values if you were forming the following confidence intervals. a) 95% confidence interval displaystyle?= displaystyle?= b) 80% confidence interval displaystyle?= displaystyle?= c) 98% confidence interval displaystyle?= displaystyle?= Question
Confidence intervals Suppose you have selected a random sample of ? $$\displaystyle={14}$$ measurements from a normal distribution. Compare the standard normal ? values with the corresponding ? values if you were forming the following confidence intervals.
a) $$95\%$$ confidence interval
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b) $$80\%$$ confidence interval
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c) $$98\%$$ confidence interval
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$$\displaystyle?=$$ 2021-02-11
Step 1
Given:
A random sample of 14 measurements from a normal distribution is selected. That is,
Sample size: $$\displaystyle{n}={14}$$
Let, X be the measurement values.
If some data {x} are normally distributed, the corresponding {z} will be normal with mean 0 and standard deviation 1 where the correspondence between the {x} and {z} is given by $$\displaystyle{z}=\frac{{{x}-\mu}}{\sigma}$$. Such z's are called z-scores.
A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution. and Z is called as standard normal variate.
That is, $$\displaystyle{Z}~\text{Normal}{\left({0},{1}\right)}$$
It's formula is given as:
$$\displaystyle{\left|{{z}}\right|}=\frac{{{x}-\mu}}{\sigma}$$
Therefore, $$\displaystyle{x}=\mu\pm{z}\sigma$$
Confidence interval for x is: $$\displaystyle{\left(\mu-{z}\sigma,\mu+{z}\sigma\right)}$$
Step 2
(a) $$95\%$$ confidence interval:
Using standard normal table ,z value at probability 0.95 corresponding to 1.96.
Therefore , it's confidence interval will be: $$\displaystyle{\left(-{1.96},+{1.96}\right)}$$
A $$95\%$$ confidence interval means that if we were to take 100 different samples and compute a $$95\%$$ confidence interval for each sample, then approximately 95 of the 100 confidence intervals will contain the true mean value $$\displaystyle{\left(\mu\right)}$$
. (b) $$80\%$$ confidence interval:
Using standard normal table ,z value at probability 0.80 corresponding to 1.28.
Therefore, it's confidence interval will be: $$\displaystyle{\left(-{1.28},+{1.28}\right)}$$
(c) $$98\%$$ confidence interval:
Using standard normal table , z value at probability 0.98 corresponding to 2.33.
Therefore, it's confidence interval will be: $$\displaystyle{\left(-{2.33},+{2.33}\right)}$$
Result:
From the above (a), (b) and (c), it can be observe that:
A higher confidence level is tend to produce a broader confidence interval.

### 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? 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 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. 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. 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. 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? 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? Consider the next 1000 98% Cis for mu that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of $$\mu?$$