A new fertilizer was applied to the soil of 292 bean plants. 44% showed increased growth. Find the margin of error and 95% confidence interval for the percentage of all bean plants which show increased growth after application of the fertilizer. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval:            %  to           %

Q: Use the sample data and confidence level to construct the confidence interval estimate of the…

A: We have given that the sample size 550, favourable number of cases X=385 and confidence level is…

Q: Use the confidence interval to find the margin of error and the sample mean. (0.498,0.640) The…

A: Confidence interval is 0.498, 0.640 To find :  (i) margin of error (ii) sample mean Margin of error…

Q: Use the sample data and confidence level given below to complete parts  (c).   A research institute…

A: It is given that, the number of respondents said yes (x) is 587 and the sample size (n) is 912.  The…

Q: Use the confidence interval to find the margin of error and the sample mean. (0.228,0.390) The…

A: Given, C.I is (0.228,0.390) Margin of error(E)=?

Q: se the given degree of confidence and sample data to construct a confidence interval for the…

A:

Q: Use the confidence interval to find the margin of error and the sample mean. (0.346,0.490) The…

A: Given,Lower limit=0.346Upper limit=0.490ME=?x=?

Q: (e) Explain why you might want to potentially use the formula n > 4h? instead of n 2 2P (1 – p) when…

A: when there is no prior estimate of the sample proportion () is known, then we use, putting the…

Q: find the margin of error for a sample mean, given n=16, s=4, Xbar=17. for a 95% confidence interval.

A: The margin of error for sample mean is given by,

Q: Use the confidence interval to find the margin of error and the sample mean. (.228, 0.390)

A: We know that the Margin of error = (upper_limit - lower_imit )/2 and Sample mean = (upper limit +…

Q: se the sample data and confidence level given below to complete parts​ (a) through​ (d). A research…

A: Consider that p is the true population proportion of the respondents who felt vulnerable to identify…

Q: T.J. wants to estimate the percentage of college students who have taken out student loans. He…

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Q: Use the confidence interval to find the margin of error and the sample mean. (0.062,0.200) ..... The…

A: The random variable X follows normal distribution. The confidence interval for population mean is…

Q: The. Nielsen television ratings used a random sample of 2,000 homes and found that 45% of the homes…

A:

Q: Use the confidence interval to find the margin of error and the sample mean. (0.328,0.430) The…

A: Given data,Lower limit=0.328Upper limit=0.430

Q: Use the confidence interval to find the margin of error and the sample mean. (0.186,0.350) The…

A: Solution: From the given information, the confidence interval is (0.186, 0.350).

Q: Emma wants to estimate the percentage of people who use public transportation in a city. She surveys…

A: Obtain the margin of error. The margin of error is obtained below: The formula for the margin of…

Q: 5. Nancy surveyed 35 randomly selected students in her school about the number of hours they spend…

A: Here out of all students in the school Nancy selected 35 students and obtained a measure 1.2, we…

Q: Find the confidence level and α for a. an 85% confidence interval. b. a 95% confidence interval.

A:

Q: In order to make the decision for one business, your team constructed a 90% confidence interval, the…

A: For 90% confidence interval, the width of that confidence interval is 500.We have to reduce the…

Q: Use the confidence interval to find the margin of error and the sample mean.   (0.342, 0.460) The…

A: Here we have given confidence interval to find the margin of error and the sample mean.  As (0.342,…

Q: The 90% confidence interval for the population proportion p is (,) The 95% confidence interval for…

A: here n = 634 p^ = proportion of males gone to dentist pin past year = 390634 = 0.6151 z-critical for…

Q: the sample proportion of females is  x=118  n=164 P = .720  #1 Make a 95% confidence interval for…

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Q: In a previous election, the confidence interval for the proportion of likely voters who would vote…

A: given;confidence interval for the proportion;(0.4401, 0.5199)i.e. (p^-E, p^+E)where  p^ : point…

Q: Use the confidence interval to find the margin of error and the sample mean. (0.264,0.420) ......…

A: Given confidence interval is(0.264  ,  0.420)

Q: Education Television In a random sample of 200 people, 154 said that they watched educational…

A: Given that, X = 154 N = 200 90% confidence level

Q: In a survey, the planning value for the population proportion is .25. How large a sample should be…

A: We have to find sample size..

Q: Use the given degree of confidence and sample data to construct a confidence interval for the…

A: It is given that n=195 and x=162. The sample proportion is 162/195 = 0.8308 The sample size is 195…

Q: Cora wants to determine a 90 percent confidence interval for the true proportion of high school…

A:

Q: Use the confidence interval to find the margin of error and the sample mean. (1.59,1.89)

A: Given data,Lower limit=1.59Upper limit=1.89Margin of error=?

Q: Find the level of a two-sided confidence interval for t = 2.776 with sample size 5. Express the…

A: According to the provided information, sample size (n) = 5 The degrees of freedom is: df =n-1 = 5-1…

Q: In a survey, the planning value for the population proportion is p* 0.16. How large a sample should…

A: Obtain the sample size of a preliminary estimate for p is 0.16 with the 90% confidence.   The…

Q: Describe when you would use a t-score instead of a z-score when finding a confidence interval.

A: Describe in what conditions t-score would be used instead of a z-score when finding a confidence…

Q: The 85.7% confidence interval estimate for p was (0.14, 0.16). Find ?′ and ME

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Q: The Nielson television ratings used a sample of 2,000 homes to estimate the proportion of all homes…

A:

Q: At one large college, 1867 out of the 2500 students surveyed said they plan to take another science…

A: The known values are, p^=1,867/2,500=0.7468, n=2,500. Using standard normal table, the critical…

Q: Suppose a 90 percent confidence interval to estimate a population proportion was calculated from a…

A: Solution: From the given information, the sample proportion is 18% and a margin of error is 4…

Q: Use the given degree of confidence and sample data to construct a confidence interval for the…

A: Given data,n = 85x = 49 Sample proportion(p) = xn =4985=0.576Ƶ-value at 98% confidence is Ƶc = 2.326

Q: Use the confidence interval to find the margin of error and the sample mean.   (.364,0.500)

A:

Q: = 360 and p = 0.09, find the margin of error at a 99% confidence level

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Q: For each confidence interval procedure, provide the confidence level. (Round the answers to the…

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Q: a). Find a 93% confidence interval for p, the true crash risk (probability) for novice drivers, also…

A: For novice drivers, sample size n = 371, Number of crashes, X = 24 For expert drivers, sample size n…

Q: Find the sample size need to find the Population Proportion if the margin of error is 4%, the…

A:

Q: Find the confidence level and α for an 80​% confidence interval.

A: Given confidence level =80%

Q: Use the confidence interval to find the margin of error and the sample mean. (0.196,0.300) The…

A: Given confidence interval is (0.196  , 0.300)Upper limit(UL)=0.300Lower limit(LL)=0.196

Q: Use the given degree of confidence and sample data to construct a confidence interval for the…

A: Given data, n=195 x=126 p=x/n=126/195=0.646   z-value at 95% confidence is Zc=1.96

Q: (d) Use the combined data of 80 students, construct a 95% confidence interval estimate of the…

A: Given that  Total students=n=40+40=80 Number of pass students=X=28+32=60

Q: %3D Elsie made a one-sample z interval for a proportion and used the critical value z* = 1.405. What…

A: Identify the correct option of the confidence level that has been used.   The correct option of the…

Q: c) i: Calculate the average and median to the results. ii: Find a 95% confidence interval for µ.

A: S.no. x (x-xbar)^2 1 11.79 0.03092 2 11.82 0.02127 3 11.87 0.00918 4 11.92 0.00210 5 11.93…

Q: Consider a 90 percent confidence interval to estimate a population proportion that is constructed…

A:

Q: Keisha has over 500 songs on her cell phone. She wants to estimate the proportion of songs that are…

A: We have given that the simple random sample of size 50 and the favourable number of cases X is 20.…

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.