Sick Days in Bed A researcher wishes to see if the average number of sick days a worker takes per year is less than 5. AEspañolrandom sample of 29 workers at a large department store had a mean of 4.7. The standard deviation of the population is1.3. Is there enough evidence to support the researcher's claim at a= 0.10? Assume that the variable is normallydistributed. Use the P-value method with tables.Part 1 of 5aloState the hypotheses and identify the claim.Ho:(Choose one) ▼O Part 4 of 5Make the decision.(Choose one)the null hypothesis.Part: 4 / 5Part 5 of 5Summarize the results.There is (Choose one)that the average number of sickdays is less than 5.

From the provided information, Sample size (n) = 29 Sample mean (x̄) = 4.7 Population standard…

Answered: Sick Days in Bed A researcher wishes to…

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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See pictures for question D and E and (A, B and Cand was submitted on a different question. According to a report done by S & J Power, the mean lifetime of the light bulbs it manufactures is 44 months. A researcher for a consumer advocate group tests this by selecting 16 bulbs at random. For the bulbs in the sample, the mean lifetime is 47 months. It is known that the population standard deviation of the lifetimes is 5 months. Assume that the population is normally distributed. Can we conclude, at the 0.10 level of significance, that the population mean lifetime, μ, of light bulbs made by this manufacturer differs from 44 months? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below.
When using 1-Var Stats on a scientific or graphing calculator, the resulting statistical output includes two values for the standard deviation of the data set. These values are labeled as Sxand σx in the statistical output. Why is it necessary to give two values for the standard deviation? How would you choose which one to use as the standard deviation of a data set you are working with? Be sure to explain the difference between these two values as you answer these questions
Assume that adults have IQ scores that are normally distributed with a mean of 102.1 and a standard deviation 16.3. Find the first quartile Q1, which is the IQ score separating the bottom 25% from the top 75%. (Hint: Draw a graph.)
The grade point average of the students at university A is 2.80 with a standard deviation of 1.092. The grade point average of students at university B is 2.4 with a standard deviation of 0.96. Which university shows a more dispersed grade distribution? The coefficient of variation for university A is ______%, while the coefficient of variation for university B is _________ %. Therefore, the ages in (A OR B?) show a more dispersed distribution.
Colonial Funds claims to have a bond fund that has maintained a mean share price of $⁢9.00. They claim that the standard deviation of the share price is 0.2. To test this claim, the investor randomly selects 22 days during the last year. He finds an average share price of $⁢8.80 with a standard deviation of 0.1734. Can the investor conclude that the share price of the bond fund varies from Colonial Funds claims at α=0.05? Step 3 of 5 : Determine the value of the test statistic. Round your answer to three decimal places.
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The ability to determine the age of some individuals can be difficult if there are not quality government records of birth. Bone growth takes place at the growth plates at the end of long bones. Once all growth plates fuse, growth stops, and an individual is considered a biological adult. The age at which growth plates fuse for males is approximately normally distributed with a mean of 19 years and a standard deviation of 15.1 months. Complete parts (a) through (d). e this (a) What is the probability a male's growth plates fuse after age 21? The probability a male's growth plates fuse after age 21 is (Round to four decimal places as needed.) View an example Get more help. 4 CE 4 PrtScn Home End F10 PgUp Clear all ^O S ↓ PgDn Check answer 12:55 PM 11/21/2022
I need help with question b
Test the claim that on average, heights (in feet) of oak trees in a forest are less than heights of pine trees. Use a = 0.01. It is known that tree heights follow a normal distribution, and the population standard deviation for both species is known to be 10.3 feet. The table below shows the results of two random samples. Туре 21 n Oaks 56.7 Pines 17 60.7 1. The fact that heights are known to be normally distributed is important in this test because: O this is a Z-test O both sample sizes are less than thirty O both standard deviations are less than thirty O this is a T-test The hypotheses are: O Ho:p1 2 P2; Ha:p1 µ2 O Ho:µ1 = µ2; Ha: µ1 + p2 O Ho:p = P2; Ha:p1 # P2 O Ho:pi p2 O Ho:µ1 2 H2; Ha:p < p2 2. This is a O leftO rightO two tailed test and the distribution used is Oz since both o values are known 13 MacBook Air esc F1 80 F3 F2 F4 FS F6 %23 & 2 3 5 6 tab R.
A company that develops an automated customer service model is interested in knowing whether two versions, Version A and Version B, will get different ratings from customers. Participants in a focus group are taken through samples from both versions, then take a survey to rate each version. A summary of the data obtained from the study is given below. Assume ratings from the different surveys generally have the same standard deviation. Mean Observations Pearson Correlation Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Confidence Level Version A Version B 33.133 28.67 15 15 -0.325 0.00 14 n = Ex: 9 -1.426 0.088 -1.345 0.176 -1.761 95% Degrees of freedom: df Point estimate for Version A: XA = Ex: 1.234 Point estimate for Version B: x B = 1 = من t = −1 Ex: 1.234 0 1 t = P = = Ex: 1.234 2 3 2
not grade. one question. answer all part
There are frogs and koi in a pond, and the number of frogs and the number of koi in the pond are independent. Let X represent the number of frogs in any given week, and let Y represent the number of koi in any given week. X has a mean of 28 with a standard deviation of 2.7, and Y has a mean of 15 with a standard deviation of 1.6. Which answer choice correctly calculates and interprets the standard deviation of the difference, D = X - Y? = 1.05; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean. = 1.1; this pond can expect the difference of frogs and koi to vary by approximately 1.1 from the mean. = 3.1; this pond can expect the difference of frogs and koi to vary by approximately 3.1 from the mean. = 13; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean.

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