# A random sample of size displaystyle{n}={25} from a normal distribution with mean displaystylemu and variance 36 has sample mean displaystyleoverline{{X}}={16.3} a) Calculate confidence intervals for displaystylemu 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?

Question
Confidence intervals
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?

2020-11-30
Step 1
a)
Computation of $$80\%$$ confidence interval is obtained below:
It is assumed that the population is approximately normal.
Here, $$\displaystyle\alpha\ {i}{s}\ {0.20}{\left(={1}-{0.80}\right)}$$.
Degrees of freedom is $$\displaystyle{24}={\left({n}-{1}\right)}.$$
Using EXCEL formula $$\displaystyle“={T}.{I}{N}{V}{.2}{T}{\left({0.2},{24}\right)}\text{"}$$, the critical value is obtained as 1.3178. That is, $$\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.3178}.$$
The $$80\%$$ confidence interval is:
$$\displaystyle{C}{I}={16.3}\pm{\left({1.3178}\right)}\frac{6}{\sqrt{{25}}}$$
$$\displaystyle={\left(\begin{matrix}{16.3}-{1.5814}\\{16.3}+{1.5814}\end{matrix}\right)}$$
$$\displaystyle={\left({14.7186},{17.8814}\right)}$$
Step 2
Computation of $$90\%$$ confidence interval is obtained below:
It is assumed that the population is approximately normal.
Here, $$\displaystyle\alpha\ {i}{s}\ {0.10}{\left(={1}-{0.90}\right)}.$$
Degrees of freedom is $$\displaystyle{24}{\left(={n}-{1}\right)}.$$
Using EXCEL formula $$\displaystyle“={T}.{I}{N}{V}{.2}{T}{\left({0.10},{24}\right)}\text{"}$$, the critical value is obtained as 1.7108. That is, $$\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.7108}.$$
The $$90\%$$ confidence interval is:
$$\displaystyle{C}{I}={16.3}\pm{\left({1.7108}\right)}\frac{6}{\sqrt{{25}}}$$
$$\displaystyle={\left(\begin{matrix}{16.3}-{1.7108}\\{16.3}+{1.7108}\end{matrix}\right)}$$
$$\displaystyle={\left({14.589},{18.011}\right)}$$
Step 3
Computation of $$99\%$$ confidence interval is obtained below:
It is assumed that the population is approximately normal.
Here, $$\displaystyle\alpha\ {i}{s}\ {0.01}{\left(={1}-{0.99}\right)}.$$
Degrees of freedom is $$\displaystyle{24}{\left(={n}-{1}\right)}$$
Using EXCEL formula $$\displaystyle“={T}.{I}{N}{V}{.2}{T}{\left({0.01},{24}\right)}\text{"}$$, the critical value is obtained as 2.7969. That is, $$\displaystyle{t}_{{\frac{\alpha}{{2}}}}={2.7969}.$$
The $$99\%$$ confidence interval is:
$$\displaystyle{C}{I}={16.3}\pm{\left({1.7108}\right)}\frac{6}{\sqrt{{25}}}$$
$$\displaystyle={\left(\begin{matrix}{16.3}-{2.053}\\{16.3}+{2.053}\end{matrix}\right)}$$
$$\displaystyle={\left({14.247},{18.353}\right)}$$
Step 4
The confidence level is increased, and then the length of the confidence interval gets wider.
Therefore, the confidence interval gets wider.

### Relevant Questions

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
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?
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The​ $$90\%$$ confidence interval is
The​ $$95\%$$ confidence interval is
Which interval is​ wider?
Interpret the results.

A population of values has a normal distribution with $$\displaystyle\mu={204.3}$$ and $$\displaystyle\sigma={43}$$. You intend to draw a random sample of size $$\displaystyle{n}={111}$$.
Find the probability that a single randomly selected value is less than 191.2.
$$\displaystyle{P}{\left({X}{<}{191.2}\right)}=$$?
Find the probability that a sample of size $$\displaystyle{n}={111}$$ is randomly selected with a mean less than 191.2.
$$\displaystyle{P}{\left({M}{<}{191.2}\right)}=$$?

A CI is desired for the true average stray-load loss A (watts) for a certain type of induction motor when the line current is heldat 10 amps for a speed of 1500 rpm. Assume that stray-load loss isnormally distributed with A = 3.0.
In this problem part (a) wants you to compute a 95% CI for A when n =25 and the sample mean = 58.3.

A population of values has a normal distribution with $$\displaystyle\mu={29.3}$$ and $$\displaystyle\sigma={65.1}$$. You intend to draw a random sample of size $$\displaystyle{n}={142}$$.
Find the probability that a sample of size n=142 is randomly selected with a mean between 27.7 and 35.3.
$$\displaystyle{P}{\left({27.7}{<}\overline{{{X}}}{<}{35.3}\right)}=$$?

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If $$q(X,0)\sim N(0,1)$$ is a pivotal function for 0, explain how you would use this result to obtain a symmetrical 95% confidence interval for 0.