Which of the following is the confidence length of the interval estimation at the 90% confidence level of the proportion of women who respond to a sudden event in less than 1 second?
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A: a) It is given that n=1518 and x=82.
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Q: A smart phone manufacturer is interested in constructing a 95% confidence interval for the…
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Q: Which of the below Z scores is the appropriate z* when calculating a 98% confidence interval? (a) Z…
A: We hvae to find corrcet answer..
Q: A smart phone manufacturer is interested in constructing a 99% confidence interval for the…
A: Givenx=86sample size(n)=1594p^=xn=861594=0.0540confidence level=99%
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Q: None
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Q: A smart phone manufacturer is interested in constructing a 90% confidence interval for the…
A: It is given that sample size (n) is 1,558 and x= 92.
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A: p = 92/1508 = 0.061
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Q: 42. Show that in Example 4.4.7, page 153, the difference between the two candidates is statistically…
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Q: 507 children
A: Nearest value = 2.010
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A: It is given that n = 1,585 and x = 88.
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A: Given that Sample size n =600 Favorable cases x =540 Sample proportion p^=x/n =540/600 =0.90
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A:
Q: A survey of 503 women shoppers found that 46% of them shop on impulse. What is the 99% confidence…
A: Given : Sample Size, n = 503 confidence level,c=0.99
Q: uppose that a recent poll found that 53%of adults believe that the overall state of moral values is…
A: Given data, p=53%=0.53 n=400 Compute Mean and Standard deviation.
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In a study investigating the reaction times of women in the face of a sudden
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- A smart phone manufacturer is interested in constructing a 90% confidence interval for the proportion of smart phones that break before the warranty expires. 86 of the 1686 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible. a. With 90% confidence the proportion of all smart phones that break before the warranty expires is between and . b. If many groups of 1686 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about percent will not contain the true population proportion.A smart phone manufacturer is interested in constructing a 90% confidence interval for the proportion of smart phones that break before the warranty expires. 87 of the 1578 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible. a. With 90% confidence the proportion of all smart phones that break before the warranty expires is between and b. If many groups of 1578 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about percent will not contain the true population proportion. Hint: Hints Video 2 [+]A smart phone manufacturer is interested in constructing a 90% confidence interval for the proportion of smart phones that break before the warranty expires. 94 of the 1521 randomly selected smart phones broke before the warranty expired. a. With 90% confidence the proportion of all smart phones that break before the warranty expires is between and b. If many groups of 1521 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about percent will not contain the true population proportion.
- С. Consider another brand C of car tyres. From a random sample of 36 car tyres, it has been found that 26 of these car tyres last longer than 40,000 km. Find a 95% confidence interval corresponding to the sample proportion of brand C car tyres which last longer than 40,000 km. Give values correct to three decimal places.a) Let p be the proportion of persons who like pineapple on their pizza. A random sample of n = 1530 people yielded 612 respondents who like pineapple on their pizza. Find an approximate 99% confidence interval for p. (Give your answers to 3 decimal places.)smart phone manufacturer is interested in constructing a 90% confidence interval for the proportion of smart phones that break before the warranty expires. 91 of the 1608 randomly selected smart phones broke before the warranty expired. Round answers 5 decimal places where possible. a. With 90% confidence the proportion of all smartphones that break before the warranty expires is between ? and . ? b. If many groups of 1608 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About ? percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about ? percent will not contain the true population proportion.
- In a sample of 510 adults, 464 had children. Construct a 95% confidence interval for the true population proportion of adults with children.Give your answers as decimals, to three places __< p < __1A linguist at a large university was studying the word length of papers submitted by students enrolled in humanities programs. From a random sample of 25 papers, the linguist counted the number of words used in each paper. The 95 percent confidence interval was calculated to be (20,995, 22,905).
- ments In a simple random sample of 1200 Americans age 20 and over, 138 of them were found to have Type I diabetes. Round all answers to three decimal places. a) Report a 95% confidence interval for the proportion of all Americans age 20 and over with diabetes. b) According to the Centers for Disease Control and Prevention, nationally, 10.7% of all Americans age 20 and over have diabetes. Does the confidence interval you found support or refute this claim? Explain. 3. Michael wants to make a confidence interval estimate for the proportion of community members in favor of a tax increase for more local school funding. If he wants the margin of error to be no more than 0.025 at a 90% confidence level, a) What is the required sample size required to obtain the desired margin of error? 331A random sample of high school seniors were asked whether they were applying for college. The resulting confidence interval for the proportion of students applying for college is (0.65,0.69). What is the margin of error?A smart phone manufacturer is interested in constructing a 95% confidence interval for the proportion of smart phones that break before the warranty expires. 88 of the 1586 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible. a. With 95% confidence the proportion of all smart phones that break before the warranty expires is between and .
