A researcher believes that children who attend elementary school in a rural setting differ in their amount of  physical activity than children who attend elementary school in an urban setting. The researcher collects a random sample from each population and records 63 of the 93 children from rural schools and 74 of the 88 children from urban schools participate in at least one hour of rigorous activity a day. Compute a 95% confidence interval for the difference in proportions of physical activity of rural and urban children.   a. Name of Interval: b. State the random variable(s) and the parameter(s) in words. c. State and check the assumptions for a hypothesis test/interval. d. Find the sample statistics and the confidence interval.

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Q: 2. Out of a sample of 760 people, 367 own their homes.. Construct a 95% confidence interval for the…

A:

Q: In a survey, 10 people were asked how much they spent on their child's last birthday gift. The…

A: Here we need to construct an 80% CI for the mean amount of money spent on child's last birthday…

Q: In a survey of 3005 adults aged 57 through 85 years, it was found that 81.7% of them used at least…

A: Given :  n = 3005 p̂=0.817

Q: The National Health and Nutrition Examination Survey is conducted each year and contains a variety…

A: There are two independent samples which are year 2003 and year 2018. Both samples come from…

Q: a survey of 3088 adults aged 57 through 85​ years, it was found that 87.1​% of them used at least…

A: The formula for computing confidence interval is,  n is the sample size and P is the proportion.

Q: Use the sample data and confidence level given below to complete parts (a) through (d). In a study…

A: a.

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Q: Estimate the variability in such sample proportions by finding the standard error. Among 1815…

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Q: Generate a 95% confidence interval estimate of the true BMI

A: The descriptive analysis is conducted using EXCEL. The software procedure is given below: Enter the…

Q: 2. To estimate how often someone runs a red light in Cyber City, a student watched traffic for a…

A:

Q: According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the…

A: Reviewing information, Group A p1=7.8%=0.078 n1=11506 Group B p2=9.1%=0.091 n2=4792

Q: A study of 420,072 cell phone users found that 130 of them developed cancer of the brain or nervous…

A: The 90% confidence interval can be calculated as:

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A: The objective of this question is to calculate the 95% and 99% confidence intervals for the average…

Q: An apartment hunter is shopping for either a one-bedroom or two-bedroom apartment in different…

A: Here the new variable is  X= difference  = (Two bed- One bed)  Then mean = ΣXi/n X̄ = (200 + 290 +…

Q: In a survey of 3115 adults aged 57 through 85​ years, it was found that 86.1​% of them used at…

A:

Q: A developmental psychologist would like to determine how much fine motor skill changes for children…

A: In statistical inference, there are two types of estimation,  Point estimation and Interval…

Q: A study of 420,098 cell phone users found that 0.0324% of them developed cancer of the brain or…

A: sample size(n)=420098sample proportion()=0.0324%=0.000324

Q: A study of 420,074 cell phone users found that 138 of them developed cancer of the brain or nervous…

A:

Q: ) Find the 99% confidence interval (CI) and margin of error (ME) used to estimate the population…

A: Given data,n=175p=72%=0.72z-value at 99% confidence is Zc=2.576

Q: Bad Dawgz is hosting its annual hot dog eating competition which involves eating 14 hot dogs, six…

A: ****Run the code in r(a)

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A:

Q: state your conclusion in context b. creat and interpret a 95% confidence interval for the difference…

A: Standard words per minute Ergonomic words per minute 12 43 38 35 21 35 28 69 18 52 58 38…

Q: A study of 420,010 cell phone users found that 0.0319% of them developed cancer of the brain or…

A: Solution: Given information: n= 420010 Sample size  p^=0.0319100= 0.000319 Sample Proportion

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A: Sample mean and sample standard deviation: Sample mean and sample standard deviation are calculated…

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Q: company manufactures a certain over-the-counter drug. The company samples 80 pills and finds that…

A: We have given that  Sample size n = 80 Sample mean = 325.5 Standard deviation s=10.3 90%confidence…

Q: A study of 420,007 cell phone users found that 136 of them developed cancer of the brain or…

A: Given, n = 420007 X = 136 p^ = Xn = 136420007 = 0.000324 Confidence level = 0.90     To find, a. 90%…

Q: We take a sample of 218 men and find they have an average commute time of 32 minutes with a standard…

A: Given: a sample of 218 men and find they have an average commute time of 32 minutes with a standard…

Q: A simple random sample of 1000 people age 18 or over is taken in a large town. It turns out that 223…

A: From the provided information, Sample size (n) = 1000 223 people in the sample are newspaper…

Q: A study of 420,098 cell phone users found that 0.0324% of them developed cancer of the brain or…

A: The question is about confidence interval.Given :Sample no. of cell phone users ( n ) = 420098Sample…

Q: A statistics instructor randomly selected four bags of oranges, each bag labeled 10 pounds, and…

A: Confidence Rate = 95 Value Level = Î ± = (100-95)% = 0.05 Degrees of freedom = n-1 = 4 -1 = 3 Value…

Q: study of 420,048 cell phone users found that 137 of them developed cancer of the brain or nervous…

A: Given that, n=420048 x=137 p¯=xn=137420048=0.000326 90% C.I. for p P(-z<p<z)=0.90 z=1.645 […

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.