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Confidence intervals

### Do piano lessons improve the spatial-temporal reasoning of preschool children? A study designed to investigate this question measured the spatial-temporal reasoning of a random sample of 34 preschool children before and after 6 months of piano lessons. The differences (After - Before) in the reasoning scores have mean 3.618 and standard deviation 3.055.

Confidence intervals

### The local energy company claims the average annual electricity bill for its subscribers is just $600. A consumer watchdog group wants to dispute this claim. All agree that the standard deviation sigma of annual electricity bills is$150. Some time later, a wealthy activist provides funding for a simple random sample of 250 households. The average annual electricity bill for this sample is \$622. Find a $$95\%$$ confidence interval for the true mean annual electric bill, based on this sample.

Confidence intervals

### Let $$f(x)=4−\frac{2}{x}+\frac{6}{x^{2}}$$. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima). f is increasing on the intervals f is decreasing on the intervals The relative maxima of f occur at x = The relative minima of f occur at x = Notes: In the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf, 2), (3,4), or the word "none". In the last two, your answer should be a comma separated list of x values or the word "none".

Confidence intervals

### 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) $$\displaystyle\alpha={0.01},\ {n}_{{{1}}}={25},\ {s}_{{{1}}}={5.2},\ {n}_{{{2}}}={20},\ {s}_{{{2}}}={6.8}$$

Confidence intervals

### In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: $$\displaystyle{0.127}{<}{p}{<}{0.191}$$. What do you​ conclude? a. Construct a 90​% confidence interval. Express the percentages in decimal form. ___

Confidence intervals

### The homebuilder’s association reports that $$75\%$$ of home buyers would like a fireplace in their new home. A local builder finds that 20 out of 30 of his customers wanted a fireplace. Using the p value decision rule determine at the 5% level of significance is there is enough evidence from the sample to conclude that the homebuilders report is too high? If P is too high in $$\displaystyle{H}_{{0}}$$ then what should it be in $$\displaystyle{H}_{{1}}$$? Also construct a $$\displaystyle{\left({1}-\alpha\right)}\%$$ interval estimate for the true proportion P. Make sure to state the meaning of P first…and completely interpret the CI for P.

Confidence intervals

### 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. Give your answers to one decimal place.

Confidence intervals

### The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l. a) If $$\displaystyle{X}_{{1}},{X}_{{2}},\ldots,{X}_{{n}}$$ are the times, in minutes, between successive customers selected randomly, estimate the parameter of the distribution. b) The randomly selected 12 times between successive customers are found as $$\displaystyle{1.8},{1.2},{0.8},{1.4},{1.2},{0.9},{0.6},{1.2},{1.2},{0.8},{1.5},{\quad\text{and}\quad}{0.6}$$ mins. Estimate the mean time between successive customers, and write down the distribution function. c) In order to estimate the distribution parameter with 0.3 error and $$4\%$$ risk, find the minimum sample size.

Confidence intervals

### A sample of 51 research cotton samples resulted in a sample average percentage elongation of 8.19 and a sample standard deviation of 1.45. Calculate a $$95\%$$ large-sample CI for the true average percentage elongation $$\displaystyle\muZSL. (Round your answer to three decimal places.) Confidence intervals asked 2021-03-06 ### Based on a sample of 80 recent Masters graduates (40 male and 40 female), the following information was made available regarding their annual salaries. The standard deviation of salaries for the male graduates was 40,000 and that for the female graduates was 25,000. a) For the male graduates, what is the probability of obtaining a sample mean salary within 10,000 of the population mean? b) Consider the same question in (a) but for the female graduates. In which case, males or females, do we have a higher probability of obtaining a sample estimate within 10,000 of the population mean? Why? c) Suppose that the sample mean salary of females is 125,000. Develop a \(95\%$$ confidence interval estimate for the mean salary of all female graduates

Confidence intervals

### 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.

Confidence intervals

### A random sample of $$\displaystyle{n}={400}$$ students is selected from a population of $$\displaystyle{N}={4000}$$ students to estimate the average weight of the students. The sample mean and sample variance are found to be $$\displaystyle{x}={140}{l}{b}{\quad\text{and}\quad}{s}^{2}={225}.\ \text{Find the}\ {95}\%{\left({z}={2}\right)}$$ confident interval.

Confidence intervals

### Consider the function $$f(x) = \sin x$$ on th interval [0, 3]. Let P be a uniform partition of [0, 3] with 4 sub-intervals. Compute the left and right Riemann sum of f on the partition. Enter approximate values, rounded to three decimal places.

Confidence intervals

### Is America’s romance with movies on the wane? In a Gallup Poll of n = 800 randomly chosen adults, 45% indicated that movies were getting better whereas 43% indicated that movies were getting worse. Find a 98% confidence interval for p, the overall proportion of adults who say that movies are getting better.

Confidence intervals

### Given sample informatio: $$\displaystyle\overline{{{x}}}={34},\sigma={4},{n}={10}$$ to calculate the following confidence intervals for $$\displaystyle\mu$$ assuming the sample is from a normal population. 99 percent confidence. (Round your answers to 4 decimal places.)

Confidence intervals

### A psychologist is interested in constructing a $$99\%$$ confidence interval for the proportion of people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain. 68 of the 702 randomly selected people who were surveyed agreed with this theory. Round answers to 4 decimal places where possible. a) With $$99\%$$ confidence the proportion of all people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain is between ____ and ____. b) If many groups of 702 randomly selected people are surveyed, then a different confidence interval would be produced from each group. About ____ percent of these confidence intervals will contain the true population proportion of all people who accept the theory that a person’s spirit is no more than the complicated network of neurons in the brain and about ____ percent will not contain the true population proportion.

Confidence intervals

### Construct a 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty, if it was found that the average monthly salary earned by a sample of 19 employees of the company was R18 500, with a standard deviation of R1 750. Interpret your answer

Confidence intervals

### 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?

Confidence intervals

### 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?

Confidence intervals

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