# Use technology to construct the confidence intervals for the population variance sigma^{2} and the population standard deviation sigma. Assume the sample is taken from a normally distributed population. c=0.99, s=37, n=20 The confidence interval for the population variance is (?, ?). The confidence interval for the population standard deviation is (?, ?) Question
Confidence intervals Use technology to construct the confidence intervals for the population variance $$\displaystyle\sigma^{{{2}}}$$ and the population standard deviation $$\displaystyle\sigma$$. Assume the sample is taken from a normally distributed population. $$\displaystyle{c}={0.99},{s}={37},{n}={20}$$ The confidence interval for the population variance is (?, ?). The confidence interval for the population standard deviation is (?, ?) 2021-01-16
Step 1 Solution: We need to construct the 99% confidence interval for the population variance. We have been provided with the following information about the sample variance and sample size: $$\displaystyle{s}^{{{2}}}={1369}$$
$$\displaystyle{n}={20}$$ Step 2 The critical values for $$\displaystyle\alpha={0.01}$$ and $$\displaystyle{d}{f}={19}$$ degrees of freedom are $$\displaystyle{{X}_{{{L}}}^{{{2}}}}={{X}_{{{1}-\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}={6.844}$$ and $$\displaystyle{{X}_{{{U}}}^{{{2}}}}={{X}_{{{1}-\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}={38.5823}$$. The corresponding confidence interval is computed as shown below: $$\displaystyle{C}{I}{\left({V}{a}{r}{i}{a}{n}{c}{e}\right)}={\left({\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{{X}_{{\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}}}},{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{{X}_{{{1}-\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}}}}\right)}$$
$$\displaystyle{\left({\frac{{{\left({20}-{1}\right)}\times{1369}}}{{{38.5823}}}},{\frac{{{\left({20}-{1}\right)}\times{1369}}}{{{6.844}}}}\right.}$$
$$\displaystyle={\left({674.17},{3800.5711}\right)}$$ Now that we have the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then: CI(Standard Deviation) $$\displaystyle={\left(\sqrt{{{674.17}}},\sqrt{{{3800.5711}}}\right)}={\left({25.9648},{61.6488}\right)}$$ Therefore, based on the data provided, the 99% confidence interval for the population variance is $$\displaystyle{674.17}{<}\sigma^{{{2}}}{<}{3800.5711}$$</span>, and the 99% confidence interval for the population standard deviation is $$\displaystyle{25.9648}{<}\sigma{<}{61.6488}$$</span>. Step 3 Ci for population variance (674.17 , 3800.57) Ci for population standard deviation ( 25.96 , 61.65)

### Relevant Questions 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. 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 randomly select and weigh 27 samples of an allergy medication. The sample standard deviation is 1.22 milligrams. Assuming the weights are normally distributed, construct $$90\%$$ confidence intervals for the population variance and standard deviation. Please interpret your result. 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. Calculate the confidence intervals for the ratio of the two population variances and the ratio of standard deviations. Suppose the samples are simple random samples taken from normal populations.
a. $$\displaystyle\alpha={0.05},{n}_{{1}}={30},{s}_{{1}}={16.37},{n}_{{2}}={39},{s}_{{2}}={9.88},$$
b. $$\displaystyle\alpha={0.01},{n}_{{1}}={25},{s}_{{1}}={5.2},{n}_{{2}}={20},{s}_{{2}}={6.8}$$  Calculate confidence intervals for ratio of two population variances and ratio of standard deviations. Assume that samples are simple random samples and taken from normal populations. a) $$\displaystyle\alpha={0.05},\ {n}_{{{1}}}={30},\ {s}_{{{1}}}={16.37},\ {n}_{{{2}}}={39},\ {s}_{{{2}}}={9.88}$$ b) $$\displaystyle\alpha={0.01},\ {n}_{{{1}}}={25},\ {s}_{{{1}}}={5.2},\ {n}_{{{2}}}={20},\ {s}_{{{2}}}={6.8}$$ If $$\displaystyle{n}={13},{\left(\overline{{x}}\right)}={31},{\quad\text{and}\quad}{s}={17},$$ construct a confidence interval at a $$99\%$$ confidence level. Assume the data came from a normally distributed population.