You wish to test the following claim (Ha) at a significance level of = 0.001.H.:µ = 67.4Ha:µ > 67.4You believe the population is normally distributed, but you do not know the standard deviation. Youobtain the following sample of data:data58.593.991.991.9What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic =What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value =The p-value is...O less than (or equal to) aO greater than aThis test statistic leads to a decision to...O reject the nullO accept the nullO fail to reject the nullAs such, the final conclusion is that...O There is sufficient evidence to warrant rejection of the claim that the population mean isgreater than 67.4.O There is not sufficient evidence to warrant rejection of the claim that the population mean isgreater than 67.4.O The sample data support the claim that the population mean is greater than 67.4.O There is not sufficient sample evidence to support the claim that the population mean isgreater than 67.4.

Answered: Image /qna-images/answer/3f467659-605b-4930-bfd3-167c161fe0f7.jpg

Answered: You wish to test the following claim…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Calculate the z-score for the largest value and interpret it in terms of standard deviations. Do the same for the smallest value N: 50 Mean:28.766 SE Mean: 0.476 St.Dev: 3.369 Min: 21.300 Q1: 26.375 Median: 29.400 Q3: 31.150 Max: 35.100
Calculate Z-score and Probabilty We have collected data on the total number of minutes people spent in the local emergency room. The times follow a normal model with a mean of 166.9 min and a standard deviation of 55.4 minutes. Find the following: What percent of ER patients spent between 2.5 and 3.5 hours there?
In a study on running and happiness, runners indicated the number of miles they ran in a day and their subsequent score on a Happiness Survey (higher score = greater happiness). Runner # of miles (x) Happiness Score (y) 1 16 12 2 4 3 12 14 4 2 8 For these data: The mean of miles run is: 8.50 The mean of the Happiness scores is: 10.00 The standard deviation of the Miles scores is: 5.72 The standard deviation of the Happiness scores is: 3.16 In your submission, please include a copy of the SPSS or Excel output and answer the following: A. What is the r and r? value? B. Based on Pearson's r and r2 value, describe the relationship (direction/strength) that exists between # of miles run and Happiness scores. C. Create a scatterplot for the data using Excel or SPSS (include output in your submission). D. Insert the prediction line from Excel/SPSS on your scatterplot.
The distribution of heights in a population of women is approximately normal. Sixteen percent of the women have heights less than 62 inches. About 97.5% of the women have heights less than 71 inches. Use the empirical rule to estimate the mean and standard deviation of the heights in this population. Mean: K inches Standard Deviation: inches
Cereals sodium values have a mean of 167 and a standard deviation of 77.3. Find the z-score for the cereal that has a sodium value of 0. How would you interpret this z-score?
Watching TV: The 2012 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 3.09 with a standard deviation of 2.9. Assume that in a sample of 37 teenagers, the sample standard deviation of daily TV time is 2.2 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers differs from 2.9? Use the a=0.10 level of significance.
Researchers sampled fish from a lake and measured the lengths (mm) of all individuals of two species, bream and perch. They were interested in testing whether there was a difference in the mean lengths of the two species in the lake. Based on the output below, what would be the correct conclusion to draw? Mean n Std deviation Std error of mean Bream 531.00 6 309.60 126.39 Perch 154.82 11 78.76 23.75 options: Bream are significantly larger on average than perch in this lake. There is no significant difference in average sizes of the two species of fish from this lake. The variances are different so it is not possible to do a statistical test comparing means. Perch are significantly larger on average than bream in this lake.
The ages of undergraduate students at two universities (one in the east and one in the west) are being compared. Researchers want to know if there is a difference in the mean age of students at the two universities. The population standard deviations are known. The following data shows the results of samples collected at each institution: School Location n (sample size) sample mean ( ) population standard deviation West 33 25.78 6.29 East 35 23.16 7.52 What is the value of the test statistic for the previous problem? (Enter your answer as a positive value even if you calculated a negative test statistic value i.e. if you calculated -1.311 you should enter 1.311). Round your answer to 3 decimal places.
Consider the following sample data. Sample A: 11, 22, 33 Sample B: 81, 92, 103 Sample C: 1,100; 1,111; 1,122 (a) Find the mean and standard deviation for each sample. Sample A: Sample B: Sample C: Mean Sample Standard Deviation (b) What does this exercise show about the standard deviation? multiple choice The idea is to illustrate that the standard deviation is not a function of the value of the mean. The idea is to illustrate that the standard deviation is a function of the value of the mean.
I need an APA write up description of the descriptives of this data set of the number of male and females who preferred hamburgers or hot dogs. (EXAMPLE: “The total sample was n=…, demographic breakup is ......The mean was...the Standard Deviation was…”) etc.

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