Which below are types of estimation in statistics? i. Point Estimation ii. Interval Estimation iii. Suspended Estimation i only Both i and ii Both i and iii Both ii and iii
Suspended Estimation is not a valid statistical procedure. Sounds like "suspended animation" to through you off!
so b?
yep
http://en.wikipedia.org/wiki/Point_estimation http://en.wikipedia.org/wiki/Interval_estimation
Soda Tak claims that Diet Tak has 40mg of sodium per can. You work for a consumer organization that tests such claims. You take a random sample of 60 cans and find that the mean amount of sodium in the sample is 42.4mg. The standard deviation in all cans is 7.2mg. You suspect that there is more than 40mg of sodium per can. Find the z-score. 0.2 2.582 2.727 5.143
i already worked this ut with @agent0smith we both said it was b but i wanted to see if you had the same answer
I said it was 1/3, i said it wasn't an option
same 0.33
actually so wouldnt it be 0.2?
-1/3 actually
I dont understand so what would be a logical answer?
none unless the sample mean has something to do with it
@ybarrap ?
what would you say?
$$ Z = \frac{\bar{X}-\operatorname{E}[X]}{\sigma(X)/\sqrt{n}}=\cfrac{(42.4-40)}{7.2/\sqrt{60} }=2.36682... $$ http://www.wolframalpha.com/input/?i=%2842.2-40%29%2F%287.2%2Fsqrt%2860%29%29 http://en.wikipedia.org/wiki/Z_score#Standardizing_in_mathematical_statistics None of the options match
i guess we can go b?
Technically, they are all wrong. Question wasn't asking for the closest answer
this is pretty easy i think
The table above gives data about Magic Sugar cereal. Which is the value used for chi-squared cdf? (1.5179,999,3) (0.4882,999,4) (0.6287,999,4) (0.4882,999,3)
yes, VERY easy
just calculations
i would say it is b?
did you do the calculations?
im doing them now
A. :)
:)
You test calories for a food item. The brand name has a mean of 158.706 and a sample standard deviation =25.236, when seventeen are tested. The generic item has a mean of 122.471 and a sample standard deviation =25.183, when seventeen are tested. Which is a confidence interval of 95%? 17.21 to 55.26 17.79 to 54.67 18.622 to 53.848 18.12 to 54.35
maybe c?
i dont know my calculations are different
These number are too small. The averages are 122.471 and 158.706, which were given. So, the 95% interval should include one of these. No option works.
none of the options are correct?
Obviously! The intervals should include the average!
okie dokie!
Hotdog toppings in a box are mixed in a large tray near the customer service area. The packets include 13 ketchup, 11 salt, 9 honey, 14 mustard, 10 BBQ, and 11 relish. When completing a goodness of fit, what is the degree(s) of freedom? 1 3 5 6
It should be 5 because once you compute the average and you have 5 terms, you can easily compute the 6th term. So their is NO degree of freedom for the 5th term once you have that average.
C?
Yes
Which statement below is true about a 95% confidence interval where the poputation mean is contained within the interval of 3.2 to 3.8? The width of this interval is 3.8. The sample mean is 1.19. The margin of error is 0.3. The CI of 95% uses 1.960.
@ybarrap
3.8-3.2 = .6 This is the full length of the interval. Half of this is the distance from the mean, so margin of error is half of this.
C!
what is half of .6 ?
haha oh shut up you ;)
:)
The table above gives data about Magic Marshmallows cereal. Which value belongs in the bottom of the horseshoes column? 0.0238 0.0909 0.111 1
this is the same table from before but different question
$$ \cfrac{(O-E)^2}{E}=\cfrac{(12-9)^2}{9} $$ Whatever this is...
D!
Find the expected frequency for cell (space) #2 of the contingency table. 38.19 42.75 44.77 46.17
help!
383/9 = 42.6
B!
Should be exact, none match
There is a new treatment for smokers to stop smoking. In an experiment, 85% of 300 smokers quit after 10 days of the treatment. What is the reasonable range for the success rate p of our new treatment? Which is the 95% confidence interval? 0.755 to 0.845 0.8096 to 0.8904 0.8048 to 0.8952 0.749 to 0.851
C?
Look for an interval where .85 is right smack in the middle
okie dokie
b or c?
Take the average of each number and get the one that is closest to .85
how do i do that?
B!
rihgt?
the 3rd one works also, we need to find the 95% for proportion
darnit you're right i checked wulf and i dont know what to choose from b or c :(
help!
80.96% to 89.04%
so b would be a final answer!
yep
In a survey done about voting and gender, it was found that 75 of 200 males voted in a recent election while 124 of 220 females voted. What is the difference between the proportion of females who vote and the proportion of males who vote? That is, which is a 95% confidence interval for the difference? 0.225 to 0.402 0.095 to 0.282 0.044 to 0.233 0.050 to 0.147
brb
alrighty @ybarrap :) tel me when your're back!
back
^ i need help on the question above
@ybarrap
what would you sayit would be ?
difference in proportion is is 0.18863636363636363636363636363636363636363636363636... So 95% conf interval is .0932 to .2792 http://vassarstats.net/prop2_ind.html
B! it really doesnt matter it doesnt have to be exact. retricelong as its arond that number
so yeah b
Find the 99% confidence interval for the population mean when the population standard deviation is 3. The sample mean is 6. We assume that the population has a normal distribution. Ten items are taken in the sample. 3.52 to 8.48 2.74 to 9.26 3.56 to 8.44 4.14 to 7.86
C!
The number of candy pieces in a bag is calculated for a sample of 40 bags of a particular manufacture. Based on this data, the sample mean is 25.30 pieces. The population standard deviation is to be 3.2 (remember this is the unrealistic part, we would not really know this). Which below is a 95% confidence interval for the mean number of pieces of candy per bag? Note: 95% of all possible confidence intervals for a sample size of 30 will contain the population mean. Use 1.960 for a 95% CI and 2.576 for a 99% CI. (23.76,26.84) (24.31,26.29) (20.17,20.57) (23.99,26.60)
Which below is a reasonable rule of check in order to use t-procedures when the sample size is 15 < n < 40? The sample distribution should be pretty normal. There should not be any outliers and very little skewness. There should not be any large outliers and there should not be extreme skewness. The sample size is large enough to use t-procedures regardless. There is no reasonable rule for checking a sample size of this size.
The only one that makes sense to me is The sample size is large enough to use t-procedures regardless.
I can tell you why, but that would be a lecture
Too long! haha
Your younger brother is getting stickers from a vending machine at the local supermarket. What is the proportion that a sticker has the message 'Girls Rule’? Let’s take a sample to try to determine this. let’s say the vending machine stickers represent a random sample of size n from the population of these stickers. In the machine you count 6 of them with the message 'Girls Rule' among the 42 total stickers. Which is 95% confidence interval when dealing with proportions? 0.025 to 0.225 0.031 to 0.235 0.037 to 0.249 0.041 to 0.213
3.71% to 24.89% http://www.wolframalpha.com/input/?i=6%2F42 https://www.mccallum-layton.co.uk/tools/statistic-calculators/confidence-interval-for-proportions-calculator/ wulff
C!
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