Question 7 A new sneaker claims that it can make male athletes jump higher. A sample of 34 male athletes were asked to jump once with their own sneakers on and once wearing the new sneakers. The average jump increased by 1.95 inches with the new sneakers with a standard deviation of 1.13 inches. What is the upper bound of the 95% confidence interval for the average jump increase? Note: Round your answers to 4 decimal places. Add your answer

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A: Solution

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A:

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A: Given data,For 95% CILower limit=0.1238Upper limit =0.2036Proportion (p)=?

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A: The point estimate of population proportion is sample proportion.Margin of error=zα2p^×q^n…

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A:

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A: Given Information : In a survey , 29 people were asked how much they spent on their child’s…

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A: Given Information: Sample size (n) = 28 Mean x¯= 49 Standard deviation s= 8 Confidence level = 80% =…

Q: mpling process is repeated many time sers who use this social media platfor

A: Option b) is correct.   The estimate is given by p^ =0.458 + 0.5282 = 0.52

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A: The mean is 49 and the standard deviation is 11.

Q: Find a 95% confidence interval for a mean, µ, where n = 324, x̄ = 2.7, s = .91, and se = .06.

A: Given: n = 324, x̄ = 2.7, s = .91, se = .06.

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A: Answer: From the given data, 99% confidence interval,

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A:

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A: The confidence level is 0.90, the standard deviation is 310 and margin of error E is 50.

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A: Answer: From the given data, 92%confidence interval,

Q: Part A: Make a 90% confidence interval for given that you have a sample of size 49 which produced a…

A: Given,margin of error(E)=0.2standard error(σ)=3.2α=1-0.90=0.10α2=0.05Z0.05=1.645      (from Z-table)

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A: Here we have given the regression output of the regression analysis of happiness score on life…

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A:

Q: Answer part b

A: Solution Weight can be written in fraction

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A: Here, the parent distribution from which the sample is taken is unknown and also, the population…

Q: Kim wants to determine a 80 % confidence interval for the true proportion ? of high school students…

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A: Givensample size(n)=20mean(x¯)=46standard deviation(s)=10α=1-0.99=0.01α2=0.005Degrees of…

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A: Given,Confidence interval(CI)=(3038  ,  4613)

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A: Given information: Number of college men surveyed, n1=688 Number of college women surveyed, n2=579…

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A: Given that, n = 166 x = 82 Confidence level 99%

Q: one with a 95% confidence level and the other with a 99.7% confidence level. Which interval is…

A: Confidence interval is probability, which refers that population parameter will fall within set of…

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Q: Solve this for me

A: Given:sample size n=35sample mean x̄=1.41sample standard deviation s=1.19population standard…

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A: As per given by the question, There are 25 people were asked how much they spent on their child's…

Q: A random sample of 328 medical doctors showed that 160 had a solo practice. (a) Let p represent the…

A: It is given that, x = 160 and n = 328.

Q: Question 3 If n = 28, a = 31, and s = 15, construct a confidence interval at a 90% confidence…

A:

Q: Find the following. Enter your answer rounded to at least two decimals. Z/2 for the 88% confidence…

A: GIVEN DATA,WE HAVE TO FIND OUT THE VALUE OF OF Z CRITICAL VALUE FOR GIVEN DATA HERE CONFIDENCE…

Q: Part D only

A: From the provided information, The mean and standard deviation of the data can be obtained as:

Q: QUESTION 10 A confidence interval was found to be (2.4, 7.9). What was the point estimate? OA. 5.5…

A: CI is (2.4,7.9) Point estimate= ?

Q: In a survey, 25 people were asked how much they spent on their child's last birthday gift. The…

A: Givensample size(n)=25Mean(x)=36standard deviation(s)=3confidence level = 99%

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A: Givensample size = n = 25Mean price = $20,500Standard deviation = $2500 The distribution is normal.…

Q: 4. Which one of the following is used for categorical data? a. Histogram b. Pie Chart C. Stem and…

A:

Q: Give a 95% confidence interval for the true mean time students spent practicing for the play.…

A: Given,   Sample size = 120 Sample mean = 45.6 Sample standard deviation = 13.3   Since the…

Q: In a survey, 16 people were asked how much they spent on their child's last birthday gift. The…

A: Given Mean=36 Standard deviations=16 n=16

Q: You measure 30 watermelons' weights, and find they have a mean weight of 79 ounces. Assume the…

A: Mean =79Standard deviation = 13.7Sample size = 3099% confidence level

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A: Given that,  23 people were asked how much they spent on their child's last birthday gift.  Sample…

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A: From the provided information, The data values are as follow: 9, 12, 13, 10 and 10 Sample size (n) =…