Number of Text Messages a Day A random sample of n=755 US cell phone users age 18 and older in May 2011 found that the average number of text messages sent or received per day is 41.5 messages,1 with standard error about 6.1.   (a) Use the information from the sample to give the best estimate of the population parameter. x¯=  (b) Find a 95% confidence interval for the mean number of text messages. The 95% confidence interval is  to .

Q: Tourism is extremely important to the economy of Florida. Hotel occupancy is an often-reported…

A: given data  n1 =1750    x1 = 1470p^1 = x1n1 = 14701750 = 0.8400 n2 = 1800    x2 = 1458 p^2 = x2n2 =…

Q: Ten wind speed readings gave an average speed of 1105 cm/sec. From the previous records the standard…

A: It is given that sample mean is 1,105 and the population standard deviation is 240.

Q: Denise is a professional swimmer who trains, in part, by running. She would like to estimate the…

A: Given,sample size(n)=20sample mean(x¯)=17.5  milesstandard deviation(s)=3.8  milesdegrees of…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…

A: The sample size is 48, the sample mean is 3.2 and the standard deviation is 17.8.

Q: One group of accounting students used simulation programs, while another group received a tutorial.…

A:

Q: Energy drinks: A survey of college students reported that in a sample of 417 male college students,…

A: Formula used Margin of Error (MOE) = Zc*Se Confidence interval of the sample CI = bar X1 - bar X2 ±…

Q: a. What is the appropriate distribution to use to construct a confidence interval based on this…

A: Given that  Sample size = 13 Mean = 40.8 Standard deviation = 2.2 99% confidence interval

Q: A study was conducted to measure the effectiveness of hypnotism in reducing pain. The measurements…

A: From the provided information,n = 8Confidence level = 95%The difference (before-after) table is as…

Q: Adrenomedullin and pregnancy loss. The plasma levels of adrenomedullin for 38 women with recurrent…

A:

Q: Tourism is extremely important to the economy of Florida. Hotel occupancy is an often-reported…

A: given data  n1 =1750    x1 = 1470p^1 = x1n1 = 14701750 = 0.84 n2 = 1800    x2 = 1458 p^2 = x2n2 =…

Q: Working Breath Rate Two random samples of 32 individuals were selected. One sample participated in…

A: Given that : xˉ=14,σ=4.2,n=32α=1−C=1−0.9=0.10zc​=z2a​​=z20.10​​=1.645(critical value is obtain from…

Q: n a test of the effectiveness of garlic for lowering​ cholesterol, 49 subjects were treated with…

A: Obtain the 95%confidence interval estimate of the mean net change in LDL cholesterol after the…

Q: A researcher estimates the 90% CI for a sample with a mean of M = 8.9 and a standard error (sM) of…

A:

Q: ou destroy 121 parts by dropping them from a height of ten feet and you estimate that the average…

A: Mean = 893Standard deviation = 69

Q: Work Time Lost due to Accidents At a large company, the Director of Research found that the average…

A:

Q: Determine the t-score for a 99% confidence interval for a population mean from a sample of size 8.

A: Confidence level 99%  with sample size n=8 then to find t-score for the 99% confidence interval

Q: The tolerable level a certain pollutant is 16 mg/l. A researcher takes a sample of size n = 25, and…

A: Given that, Sample size, n= 25 Mean, X bar= 18.5 Standard deviation, sigma= 7 Confidence interval…

Q: A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in…

A: Answer The mean =98.4The standard deviation =43.6The sample size [n] =24

Q: the study, n = 52 men were put on a diet. The men who dieted lost an average of 6.0 kg, with a…

A: Given that ; Sample size (n) = 52 Sample mean = 6.0 kg  ? = 3.7  kg By using one sample mean test we…

Q: Students at a local elementary school were randomly selected to participate in a reading fluency…

A: It is given that the sample mean difference for the pretest and posttest is 11.The 90 percent…

Q: 1 POINT In a recent survey, a random sample of 450 families visiting a zoo were asked how…

A:

Q: n a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…

A: In the given scenario, the effectiveness of garlic in decreasing the cholesterol level was measured.…

Q: Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts…

A: Given n=55 x¯=18.5s=5.2

Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…

A:

Q: A study was conducted to measure the effectiveness of hypnotism in reducing pain. The measurements…

A: The data shows the effectiveness of hypnotism in reducing pain.

Q: The distribution of study times for first year students follows a Normal distribution with standard…

A:

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.