postillan4
2021-06-23
Answered

Let p be the population proportion for the situation. (a) Find point estimates of p and q, (b) construct 90% and 95% confidence intervals for p, and (c) interpret the results of part (b) and compare the widths of the confidence intervals. In a survey of 1035 U.S. adults, 745 say they want the U.S. to play a leading or major role in global affairs.

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BleabyinfibiaG

Answered 2021-06-24
Author has **118** answers

(a)$\stackrel{\u23de}{p}=0.7198,\stackrel{\u23de}{q}=0.2802$

(b)90% confidence interval: (0.6968, 0.7428)

95% confidence interval: (0.6924, 0.7472)

(c)The 95% confidence interval is wider than the 90% confidence interval.

(b)90% confidence interval: (0.6968, 0.7428)

95% confidence interval: (0.6924, 0.7472)

(c)The 95% confidence interval is wider than the 90% confidence interval.

asked 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:

asked 2021-01-10

The average zinc concentration recovered from a sample of measurements taken in 36 different locations in a river is found to be 2.6 grams per liter. Find the 95% confidence intervals for the mean zinc concentration in the river. Assume that the population standard deviation is 0.3 gram per liter.

asked 2021-08-04

A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers.

Assume the distribution of measurements to be approximately normal.

a) Construct a$99\mathrm{\%}$ confidence interval for the average number of kilometers an automobile is driven annually in Virginia.

b) What can we assert with$99\mathrm{\%}$ confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?

Assume the distribution of measurements to be approximately normal.

a) Construct a

b) What can we assert with

asked 2020-12-27

Consider the next 1000 98% Cis for mu that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of $\mu ?$

What isthe probability that between 970 and 990 of these intervals conta the corresponding value of ? (Hint: Let

Round your answer to four decimal places.)

‘the number among the 1000 intervals that contain What king of random variable s 2) (Use the normal approximation to the binomial distribution

What isthe probability that between 970 and 990 of these intervals conta the corresponding value of ? (Hint: Let

Round your answer to four decimal places.)

‘the number among the 1000 intervals that contain What king of random variable s 2) (Use the normal approximation to the binomial distribution

asked 2021-02-11

I have a question:

You intend to estimate a population mean$\mu$ with the following sample.

54.7

48.9

56.3

43.5

41.5

41.2

54.2

You believe the population is normally distributed. Find the$99.9\mathrm{\%}$ confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).

$99.9\mathrm{\%}C.I.=(35.68,61.54)$ Incorrect

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

You intend to estimate a population mean

54.7

48.9

56.3

43.5

41.5

41.2

54.2

You believe the population is normally distributed. Find the

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

asked 2021-08-02

A group of scientists studied the adhesion of various biofilms to solid surfaces for possible use in environmental technologies. Adhesion assay is conducted by measuring absorbance at A590. If the scientists want the confidence interval to be no wider than 0.55 $dy\ne -c{m}^{2}$ at $98\mathrm{\%}$ confidence interval, how many observations should they take? Assume that the standard deviation is known to be 0.66 $dy\ne -c{m}^{2}$

CHOOSE THE RIGHT ANSWER:

a.39

b.21

c.16

d.25

e.18

f.32

CHOOSE THE RIGHT ANSWER:

a.39

b.21

c.16

d.25

e.18

f.32

asked 2022-03-16

Calculating the Confidence Interval

It is found that in a random sample of 100 Science students, there are 48 studying statistics . To test whether the true proportion of students in statistics is 50% or not, suitable null and alternative hypotheses are:

(a)$H0:p=0.48;H1:p\u0338=0.48$

(b)$H0:p=0.5;H1:p>0.5$

(c)$H0:p=0.5;H1:p<0.5$

(d)$H0:p=0.5;H1:p=0.48$

(e)$H0:p=0.5;H1:p\u0338=0.5$

Find a 95% Confidence Interval for p in Q1.

I have determined the answer to Q

1), which I believe is e).

For question 2), I have been told the CI= estimator +/- (table value)(Standard error). The estimator is 0.48, and I have found the SE to be 0.05. However, I am unsure about how to calculate the "table value".

It is found that in a random sample of 100 Science students, there are 48 studying statistics . To test whether the true proportion of students in statistics is 50% or not, suitable null and alternative hypotheses are:

(a)

(b)

(c)

(d)

(e)

Find a 95% Confidence Interval for p in Q1.

I have determined the answer to Q

1), which I believe is e).

For question 2), I have been told the CI= estimator +/- (table value)(Standard error). The estimator is 0.48, and I have found the SE to be 0.05. However, I am unsure about how to calculate the "table value".