Question

In making plastics a colourless gas is used and some will leak into the air. The manufacturer has given instructions to halt production if the mean (average ) amount of this gas in the area exceeds 3.0 parts per million (ppm). Every day the safety manager plans to select a large sample n=50 air specimens and to perform a one tailed test Ha µ > 3 at significance level α = 0.01. Determine the lower limit of the rejection region of this test. Give your answer to 2 decimal places.

Answer 1

Hey need help? ^_^

Answer 2

yes, please. I'm so lost.. I don't even know how to start!

Answer 3

I think this should be in Science dont you think?

Answer 4

wait o_o idk what this is... @pooja195 this is for you

Answer 5

this is a question in my intro to stats textbook

Answer 6

@Bookworm14 @Data_LG2 @Zale101 please help my friend here ^_^ thank you

Answer 7

idk anything about it o-o

Answer 8

@Nnesha @pooja195 @TheSmartOne

Answer 9

We're a little short on information. Is there an assumed population distribution?

Answer 10

thats all the question says, nothing else.that is all the info Im given

Answer 11

Or is it the case where the equipment will return a yes/no answer to whether the sample has more than 3.0 ppm of gas for each specimen?

Answer 12

Tell your teacher to answer it for you ^_^ most teachers give out work and dont even know the answer them selves.

Answer 13

hi

Answer 14

I can't get a hold of my teacher, I need help.

Answer 15

Yeah, I'm not seeing it. I'd love to see the solution, since some of the necessary information will have to leak into the correct answer from an unknown source.

Answer 16

the next question is really similar but not related in anyway….
In making plastics a colourless gas is used and some will leak into the air. The manufacturer has given instructions to halt production if the mean (average ) amount of this gas in the area exceeds 3.0 parts per million (ppm). Every day the safety manager plans to select a large sample n= 74 air specimens and to perform a one tailed test Ha µ > 3 at significance level α = 0.01. Suppose the sample give a sample mean of 3.41 ppm and a s = 0.51 ppm. What would be the value of the z - test when the null hypothesis is Ho : µ = 3.0 and Ha : µ > 3.0 ? Give your answer to 2 decimal places, do not round up or down.

Answer 17

Don't know if this might help:
http://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704_HypothesisTest-Means-Proportions/BS704_HypothesisTest-Means-Proportions3.html

Answer 18

Okay, I changed my mind.

Given \(H_{a}: \mu > 3\)
We have \(H_{0}: \mu \le 3\)

At least we know the expected Mean of the distribution. Without a clue as to the population standard deviation, it's not clear how to proceed.

Also, I'm very concerned about to proximity of 3 to 0. Whatever the distribution, it CANNOT produce a negative result. I suspect it is quite unlikely that we are talking about a Normal Distribution. What section are you studying? Maybe a Poisson Distribution? I suppose we could shoehorn that in there.

Answer 19

I really don't know. He says these questions are supposed to be simple in class but I don't have a clue and obviously I'm not the only one.

Answer 20

The first question is missing something. The second question might be answerable.

Answer 21

Is the first question a word-by-word reproduction?

Answer 22

literally copied and pasted from online textbook version

Answer 23

I don't think the gas ppm distribution is normal since you clearly can not have negative values of ppm of that gas in air.

Answer 24

Another one: A random sample of n = 59 consumers taste-tested a new snack food. Of the 59 observations, x = 22 liked the product.

Use a 80% confidence interval to estimate the proportion of consumers who like the snack food.

Provide as your answer the width of the confidence interval. give your answer to 3 decimal places , do not round up or down.

I have no idea how to even start any of these and apparently they're supposed to be easy??

Answer 25

Right, since we have a sample standard deviation, we might be able to use that.

More and more I despise online mathematics and statistics courses. We're just not getting good enough fast enough and students are paying the price. {sigh}

Answer 26

yeah, tell me about it… Last online stats course i EVER take… not only because the class time is so helpful, you can actually get a response form the profs!!

Answer 27

It doesn't help to solve the problem to know that you are not alone. Sorry.

Answer 28

has anyone had any luck of figuring out how to do at least one of these questions?

Answer 29

??????

Answer 30

A large company wishes to estimate the average number µ of sick leave days taken by its employees. The firm keeps a complete record of sick leave days. From the records a sample of n = 107 employees is selected and the number of sick leave days taken over the history of the employee is recorded. The company finds that for the sample the average number of days is 13.2, That is , the sample average is

x¯
= 13.2 and the sample standard deviation s = 10 days. Construct a estimate of the population mean µ using a 90% confidence interval. For your answer give the width of your confidence interval. Give your answer to 3 decimal places , do not round up or down.

Answer 31

A random sample of n = 4 observations is selected from a normal distribution to test the null hypothesis that µ = 10 versus the alternative Ha : µ < 10 . The test is to be carried using significance level α = 0.05 . What is the upper value on the rejection region. Give your answer to 3 decimal places.

Answer 32

@tkhunny

Answer 33

@thomas5267

Answer 34

@thesmartone

Answer 35

Have you been taught the t distribution?

Answer 36

yes

Answer 37

@pooja195 help!!

Answer 38

this question is still opened?

Answer 39

yes, still no answer

Answer 40

Im not sure about how to do this :( im sorry

Answer 41

@amyvincent12
are you there?

Answer 42

@Michele_Laino

Answer 43

@amyvincent
For the second and third question, we can assume normal distribution for large n, usually when n>30. Since n=59 and n=107 exceeds this threshold, we can safely assume (approximate) normality by the CLT (central limit theorem).
We can also assume \(\sigma=s/\sqrt n\), and you can proceed to calculate the estimate in the third problem.
For the first two problems without standard deviation \(\sigma\) nor sample standard deviation s, you will need to use Chebyshev's inequality to estimate the upper bound of \(\sigma\). But before you do that, please make sure the question has been transcribed correctly.

Answer 44

for question #1, we can use the fact that the \(poisson\) distribution, can be approximated by the \(gaussian\) distribution with the same values for mean and width.
Now from your data, we can assume:
mean: \( \mu=3\), then standard deviation: \(\sigma=\sqrt \mu = \sqrt 3\)
so, we have: standard deviation of mean: \(\sigma^* = \sigma/ \sqrt {50}\)
the requested upper limit \(L\) is given, therefore, by the subsequent expression:
\(L=\mu + 2.33 \cdot \sigma^*\)

Answer 45

since the confidence level is \(c.l.=1\%\)

Answer 46

@amyvincent12

Answer 47

Thanks. Didn't realize this question was still open.