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Recent questions in Confidence intervals

Josalynn 2021-05-21

Consider a normal population distribution with the value of $σ$ known?
a. What is the confidence level for the interval $\overset{―}{X} \pm 2.81 \frac{σ}{\sqrt{n}} ?$
b. What is the confidence level for the interval $\overset{―}{X} \pm 1.44 \frac{σ}{\sqrt{n}} ?$

necessaryh 2021-05-19

Construct two 99% confidence intervals for $μ_{X} = μ_{Y}$ using the data of Case Study 9.2.3, first assuming the variances are equal, and then assuming they are not.

Wotzdorfg 2021-05-16

Construct the indicated confidence intervals for (a) the population variance $σ^{2}$ and (b) the population standard deviation $σ$ . Assume the sample is from a normally distributed population. c = 0.90, s = 35, n = 18

Ramsey 2021-05-14

Use the one-standard-deviation x2 $x^{2}$ -test and the one-standard-deviation $x^{2}$ -interval procedure to conduct the required hypothesis test and obtain the specified confidence interval. s=3 and n=10.
a. $H_{0} : σ = 4, H_{3} : σ < 4, α = 0.05$ .
b. 90% confidence interval

nagasenaz 2021-05-14

When σ is unknown and the sample size is $n \geq 30$ , there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When $n \geq 30$ , use the sample standard deviation s as an estimate for σσ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σσ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?

FizeauV 2021-05-13

Assume the sample is from a normally distributed population and construct the indicated confidence intervals for (a) the population variance $σ^{2}$ and (b) the population standard deviation σ. Interpret the results. The record high daily temperatures (in degrees Fahrenheit) of a random sample of 64 days of the year in Grand Junction, Colorado, have a sample standard deviation of 16.8∘F. Use a 98% level of confidence.

Maiclubk 2021-05-09

If you have a 99% confidence interval for $μ; μ$ ; based on a simple random sample, is it correct to say that in the long run, if you computed many, many confidence intervals using the prescribed method, about 99% of such intervals would contain $μ$ ;? Explain.

ruigE 2021-05-09

The two intervals (114.4, 115.6) and (114.1, 115.9) are confidence intervals (The same sample data was used to calculate both values) for $μ =$ true average resonance frequency (in hertz) for all tennis rackets of a certain type.
a. What is the value of the sample mean resonance frequency?
b. The confidence level for one of these intervals is 90% and for the other it is 99%. Which is which, and how can you tell?

Annette Arroyo 2021-05-08

When σ is unknown and the sample is of size $n \geq 30$ , there are two methods for computing confidence intervals for μ. Method1:Method1: Use the Student's t distribution with d.f. = n - 1. Method2:Method2: When $n \geq 30$ , use the sample standard deviation s as an estimate for σ,σ, and then use the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. Compute 90%, 95%, and 99% confidence intervals for μμ using Method 2 with the standard normal distribution. Use s as an estimate for σσ. Round endpoints to two digits after the decimal.

defazajx 2021-05-05

What is meant by the term “90% confident” when constructing a confidence interval for a mean? a. If we took repeated samples, approximately 90% of the samples would produce the same confidence interval. b. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the sample mean. c. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean. d. If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples.

ruigE 2021-03-11

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".

floymdiT 2021-03-11

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.

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

Daniaal Sanchez 2021-03-09

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: $0.127 < p < 0.191$ . What do you conclude? a. Construct a 90% confidence interval. Express the percentages in decimal form. ___

Globokim8 2021-03-08

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 $H_{0}$ then what should it be in $H_{1}$ ?
Also construct a $(1 - α) %$ interval estimate for the true proportion P.
Make sure to state the meaning of P first…and completely interpret the CI for P.

Tobias Ali 2021-03-07

The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l.
a) If $X_{1}, X_{2}, \dots, X_{n}$ 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 $1.8, 1.2, 0.8, 1.4, 1.2, 0.9, 0.6, 1.2, 1.2, 0.8, 1.5, and 0.6$ 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.

Chesley 2021-03-06

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 $μ$ (Round your answer to three decimal places.)

Tyra 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

aflacatn 2021-03-05

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 $95 %$ confidence interval of students mean mark. b) Calculate for the critical regions of the $95 %$ 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.

abondantQ 2021-03-05

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

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