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# 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) alpha=0.05, n_{1}=30, s_{1}=16.37, n_{2}=39, s_{2}=9.88 b) 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) alpha=0.05, n_{1}=30, s_{1}=16.37, n_{2}=39, s_{2}=9.88 b) alpha=0.01, n_{1}=25, s_{1}=5.2, n_{2}=20, s_{2}=6.8

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Confidence intervals asked 2021-03-09
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) $$\displaystyle\alpha={0.01},\ {n}_{{{1}}}={25},\ {s}_{{{1}}}={5.2},\ {n}_{{{2}}}={20},\ {s}_{{{2}}}={6.8}$$

## Answers (1) 2021-03-10

a) Given: Sample size, $$\displaystyle{n}_{{{1}}}={30}$$ Sample size, $$\displaystyle{n}_{{{2}}}={39}$$ Sample standard deviation 1, $$\displaystyle{s}_{{{1}}}={16.37}$$ Sample standard deviation 2, $$\displaystyle{s}_{{{2}}}={9.88}$$ Let's calculate $$\displaystyle{95}\%$$ confidence interval for the ratio of two population variances. $$\displaystyle{C}{I}=\ {\left({\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{{s}_{{{2}}}^{{{2}}}}}}}\ {F}_{{{1}\ -\ \frac{\alpha}{{2}},\ {n}_{{{2}}}\ -\ {1},\ {n}_{{{1}}}\ -\ {1},}}\ {\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{{s}_{{{2}}}^{{{2}}}}}}}{F}_{{\frac{\alpha}{{2}},\ {n}_{{{2}}}\ -\ {1},\ {n}_{{{1}}}\ -\ {1}}}\right)}$$
$$\displaystyle{C}{I}=\ {\left({\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}\ {F}_{{{1}\ -\ \frac{{0.05}}{{2.39}}\ -\ {1.30}\ -\ {1},}}\ {\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}{F}_{{\frac{{0.5}}{{2.39}}\ -\ {1.30}\ -\ {1}}}\right)}$$
$$\displaystyle{C}{I}=\ {\left({\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}\ {F}_{{\frac{{0.95}}{{2}},\ {38},\ {29},}}\ {\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}{F}_{{\frac{{0.5}}{{2}},\ {38},\ {29}}}\right)}$$ Using the critical value table, $$\displaystyle{C}{I}=\ {\left({\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}\ \times{0.507},\ \ {\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}{2.038}\right)}$$
$$\displaystyle{C}{I}=\ {\left({1.3919},\ {5.5949}\right)}$$ Hence, it is $$\displaystyle{95}\%$$ confidence that the true ratio of population variances lies in interval $$\displaystyle{\left({1.3919},\ {5.5949}\right)}$$

b) Given: Sample size, $$\displaystyle{n}_{{{1}}}={25}$$ Sample size, $$\displaystyle{n}_{{{2}}}={20}$$ Sample standard deviation 1, $$\displaystyle{s}_{{{1}}}={5.2}$$ Sample standard deviation 2, $$\displaystyle{s}_{{{2}}}={6.8}$$ Let's calculate $$\displaystyle{99}\%$$ confidence interval for the ratio of two population variances. $$\displaystyle{C}{I}=\ {\left({\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{{s}_{{{2}}}^{{{2}}}}}}}\ {F}_{{{1}\ -\ \frac{\alpha}{{2}},\ {n}_{{{2}}}\ -\ {1},\ {n}_{{{1}}}\ -\ {1},}}\ {\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{{s}_{{{2}}}^{{{2}}}}}}}{F}_{{\frac{\alpha}{{2}},\ {n}_{{{2}}}\ -\ {1},\ {n}_{{{1}}}\ -\ {1}}}\right)}$$
$$\displaystyle{C}{I}=\ {\left({\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}\ {F}_{{{1}\ -\ \frac{{0.01}}{{2}},\ {39}\ -\ {1},\ {30}\ -\ {1},}}\ {\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}{F}_{{\frac{{0.01}}{{2}},\ {39}\ -\ {1},\ {30}\ -\ {1}}}\right)}$$
$$\displaystyle{C}{I}=\ {\left({\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}\ {F}_{{\frac{{0.99}}{{2}},\ {38},\ {29},}}\ {\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}{F}_{{\frac{{0.01}}{{2}},\ {38},\ {29}}}\right)}$$ Using the critical value table, $$\displaystyle{C}{I}=\ {\left({\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}\ \times{0.3025},\ \ {\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}{3.092}\right)}$$
$$\displaystyle{C}{I}={\left({0.2128},\ {2.175}\right)}$$ Hence, it is $$\displaystyle{99}\%$$ confidience that the ratio of population variances lies in the interval $$\displaystyle{\left({0.2128},\ {2.175}\right)}$$

### Relevant Questions asked 2021-02-09
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. $$\displaystyle\alpha={0.01},{n}_{{1}}={25},{s}_{{1}}={5.2},{n}_{{2}}={20},{s}_{{2}}={6.8}$$ asked 2021-05-05

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
$$\mu_1 - \mu_2$$.
(Round your answers to two decimal places.)
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. asked 2020-11-05

Assume that the random variable Z follows standard normal distribution, calculate the following probabilities (Round to two decimal places)
a)$$P(z>1.9)$$
b)$$\displaystyle{P}{\left(−{2}\le{z}\le{1.2}\right)}$$
c)$$P(z\geq0.2)$$ asked 2021-01-15
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 (?, ?) asked 2020-11-08
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. asked 2021-01-17
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}$$
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings. asked 2021-02-09
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? asked 2021-02-23
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? asked 2020-10-23
1. Find each of the requested values for a population with a mean of $$? = 40$$, and a standard deviation of $$? = 8$$ A. What is the z-score corresponding to $$X = 52?$$ B. What is the X value corresponding to $$z = - 0.50?$$ C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of $$M=42$$ for a sample of $$n = 4$$ scores? E. What is the z-scores corresponding to a sample mean of $$M= 42$$ for a sample of $$n = 6$$ scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: $$a. -2.00 b. 1.25 c. 3.50 d. -0.34$$ 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with $$\mu = 78$$ and $$\sigma = 12$$. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: $$82, 74, 62, 68, 79, 94, 90, 81, 80$$. 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $$12 (\mu = 42, \sigma = 12)$$. You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is$44.50 from tips. Test for a difference between this value and the population mean at the $$\alpha = 0.05$$ level of significance. asked 2020-11-07
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?
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