On Assignment 11.15, you found the mean and standard deviation for X, the number of dots facing up when a single die is rolled. You may want to pause and look over the solutions to MAC15. These are copied, for your convenience, on the back of this page. We found that the mean is u = 3.5 and the standard deviation is o = 1.7078. Now, suppose that instead of tossing one die and counting the number of dots facing up we toss two dice and let X =the mean number of dots facing up on the two dice. Note that when tossing 2 dice, x could end up being 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, or 6.0. X is a random variable. We call the expected value (also known as the mean) of this random variable E( X ) or H. Your text uses the symbol A. The standard deviation of , also called the standard error of the mean is denoted by the symbol O. It is easy to calculate once you know o, which you found on MAC15. Review Chapter 7 if you've orgotten the formula. Then complete the following sentence: When tossing two dice and finding the mean, Xwe find that has a mean or expected value, ミク and a standard deviation Finally, suppose that instead of tossing two dice and finding the mean we were to toss five dice and record the mean for those five dice. Consider how this might (or might not) change the expected value and the standard deviation of the sample mean. Repeat this for larger samples of 10, 20, and 50 dice. Complete the table below using decimals rounded to four decimal places where necessary. Mean or expected Standard deviation of Number of dice tossed value of x. Note that x. Note that for n=1, from which we'll find for n=1, this is the this is the standard the sample mean, I mean for the roll of a deviation for the roll of single die. a single die. n or 1 3.5 1.7078 5 10 20 50

A recent study evaluated how addicted teenagers become to nicotine once they start smoking. The response variable was the number of yes answers on a questionnaire called the Hooked on Nicotine Checklist (HONC). Of teenagers who had tried tobacco, the mean HONC score was 3.8 (s = 4.4) for the 141 females and 2.4 (s = 3.5) for the 178 males. Complete parts a through c below. a. Find the standard error comparing the sample means. Interpret. The standard error is se= 0.4540. (Round to four decimal places as needed.) What does the standard error indicate? ... OA. The standard error is the standard deviation of the sample for this study. OCT 28 OB. The standard error is the difference in standard deviations for the two populations. OC. The standard error is the standard deviation of the difference between x₁-x₂. D. The standard error describes the spread of the sampling distribution of x₁-x₂. b. Find the test statistic and P-value for Ho: H₁ H₂ and Ha: H₁ H₂. Interpret, and explain what (if…

complete the following details please

Please answer part (a) and part (b)

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I need the shade of the part of the curve corresponding to the calculation on both. On #1 and #2 since it was given thru this app when I submitted it. Thank you

given the following histogram what of the following could be tru ? (Coulf be multiple) Given those outliers, this one may not ever get to normalNothing seems odd with this number of points.• Looks like it could be bimodalLooks like it's normalData has outliers* Data offset significantly from customer's desired range of 20-21 #/ream

I need help and directions on how to solve the two subparts of this question on TI84 calculator. Thank you. The average fruit fly will lay 424 eggs into rotting fruit. A biologist wants to see if the average will change for flies that have a certain gene modified. The data below shows the number of eggs that were laid into rotting fruit by several fruit flies that had this gene modified. Assume that the distribution of the population is normal. 426, 445, 417, 410, 415, 418, 412, 439, 452, 447, 407, 423, 447, 441 What can be concluded at the the alpha = 0.10 level of significance level of significance?  The p-value =  (Please show your answer to 4 decimal places.) Based on this, we should__________the null hypothesis.

Assume that the readings on the thermometers are normally distributed with a mean of 0° and standard deviation of 1.00°C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to P34, the 84th percentile. This is the temperature reading separating the bottom 84% from the top 16%. Click to view page 1 of the table. Click to view page 2 of the table. Which graph represents P ? Choose the correct graph below. 84 A. В. OC. D. Ay X The temperature for P84 is approximately (Round to two decimal places as needed.)

I'm not sure where is my mistake solving this problem.  I have revised my calculation three times already and I'm not sure if the step to solve it is wrong or if it is my rounding the problem.   Thank you for your help