A schoolteacherworried that the concentration of dangerous, cancer-causing radon gas in her classroom is greater than the safe level of 5 pCi/L. The school samples the air for 16 days and finds anaverage concentration of 5.4 pCi/L with a standard deviation of 1 pCi/L. Calculate the value of the test statistic.Multiple Cholcet15 =-160Z=-1.60t15 =1.60z= 1.60

Answered: Image /qna-images/answer/692f1b64-4989-49f6-88ac-d88e6eb48752.jpg

Answered: A schoolteacher average concentration…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Erin is a nutritionist and cardiologist. She is testing whether the average total cholesterol level of adults in the region is less than 175mgdL. Erin randomly selects 25 people in the region who willingly allowed their blood to be drawn for the survey. Based on several previous studies, Erin assumes that the population standard deviation for the total cholesterol level is 30.22mgdL. For this test: H0:μ=175; Ha:μ<175, which is a left-tailed test. The test results are: z0=−1.158, p-value is = 0.123 Using a 5% level of significance, which of the following are appropriate conclusions for this hypothesis test? Select all that apply. Select all that apply: We reject H0. We fail to reject H0. At the 5% significance level, the data provide sufficient evidence to conclude that the average total cholesterol level of adults in the region is less than 175mgdL. At the 5% significance level, the data do not provide sufficient evidence to conclude that…
Find the area, square unit, to the nearest thousandth of the standard deviation between the given z-scores. Z equals 1.12 and z equals 1.73.
Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than 1.95 and draw a sketch of the region. Sketch the region. Choose the correct graph below OA 1.95 The probability is 9744 (Round to four decimal places as needed.) OB. +1.35 L OC. -195 C D. C
A distribution of values is normal with a mean of 38.3 and a standard deviation of 84.2.Find P98, which is the score separating the bottom 98% from the top 2%.P98 =
Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of the table, Click to view page 2 of the table. Step 1 теan(u) s tan dard o 90 115 The area of the shaded region is (Round to four decimal places as needed.) Step 2 Required pr X-µ = P = P (Z < 1 = P (Z < 2 = 0.9773 Enter your answer in the answer box and then click Check Answer. %3D All parts showing Clear All Final Check P Type here to search hp
Solve this for me. Thank you.
You collected data from a random sample of students on their impressions of the university. On a particular opinion item item, the mean for male students was 15.2 -- with a standard deviation of 3.4. The mean on that item for women was 14.4, with a standard deviation of 3.8. Calculate Cohen's d effect size. IN THE SPACE BELOW, write the effect size AND your interpretation of it.
A manufacturer of light bulbs finds that one light bulb model has a mean life span of 1025 h with a standard deviation of 83 h.Find the z-scores for lightbulbs which last 850 and 900 h. (Round your answers to two decimal places.) 850 h: z = 900 h: z =
The mean height of females in a certain country is 66 inches with a standard deviation of 4 inches. Find the height that correspond to a z score of 0.8. Round your answer to one decimal place, if needed.
A distribution of values is normal with a mean of 100.9 and a standard deviation of 60.6.Find P72, which is the score separating the bottom 72% from the top 28%.
Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P70, the 70-percentile. This is the temperature reading separating the bottom 70% from the top 30%.P70 = °C(Round answer to three decimal places)
Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P50, the 50-percentile. This is the temperature reading separating the bottom 50% from the top 50%. Round to 3 decimal places.

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