Stats Test #3

We are 86% to 94% confident 56% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation reasonable?

The interpretation is flawed. It indicates that the level of confidence is varying.

IN a survey of 2055 adults in a certain country conducted during a period of economic uncertainty, 56% thought that wages paid to workers in industry were too low. The margin of error was 4% points with 90% confidence. Is the interpretation reasonable?

The interpretation is flawed. It suggests that this interval sets the standard for all the other intervals, which is not true.

We are 90% confident the proportion of adults in the country during the period of economic uncertainty who believe wages paid to workers in industry were too low was between .52 and .6. Is the interpretation reasonable?

The interpretation is reasonable.

P(90<x<110)- n=10 P(90<x<110)- n=20 Which has higher probability?

The probability with the higher sample size will have a higher probability. As n increases, the standard deviation decreases.

If the sample size is 30 or larger what does that mean?

The sampling distribution is approximately normal.

(T or F): The distribution of the sample mean will be normally distributed if the sample is obtained from a population that is normally distributed, regardless of the sample size.

True

(T or F): The mean of the sampling distribution of p^ is p.

True

When is the population mean normally distributed?

When the population says its normal or when sample size is at least 30

How do you find the sampling standard deviation on the graph?

You subtract the mean by a number either on the right or the left.

What is the sampling distribution of sample mean if n=33, mean=68 and standard deviation =9?

mean=68 (no math required for this answer) Standard deviation is the formula attached: - insert deviation and n

A _____________________ is the value of a statistic that estimates the value of a parameter.

point estimate

The sample portion, denoted p is given by the formula p=

x/n where x is the number of individuals with a specified characteristic in a sample of n individuals.

Z beta/2 -corresponds to an 84% level of confidence

(1-beta)* 100% .84= (1-beta)* 1

What is the probability a random sample of size 17 will have a mean gestation period within 8 days of the mean? mean (u)= 149 days deviation (o)= 11 days

(149-8)- 149 ------------- 11 square root 17 (149+8)- 149 ------------- 11 square root 17 * answers: -3.0 and 3.0 - find their z-scores - .0013 and .9987 -Subtract .9987- .0013= .9974 [.9974 is the answer]

There's a 5% chance the mean speed of 25 students will exceed what value?

* 5%= .05= z-score 1.645 * 1.645 = x - 92 ---------- (10 square root 25) = 95.29

What is P[ (sample mean) < or equal to 75.55]? - n= 36 - Mean(u) = 82 - Deviation(o)= 18

- 18/ square root 36= 3 z-score: (75.55- 82)/ 3 =-2.15 find the value of z-score -2.15 = .0158. .0158 is the answer

What is P[ 78.1 < x < 88.75]? - n= 36 - Mean(u) = 82 - Deviation(o)= 18

- 18/ square root 36= 3 z-score: (78.1- 82)/ 3 = -1.3 * -1.3= .0968 z-score: (88.75- 82)/ 3 = 2.25 * 2.25= .9878 .9878- .0968= .891 .891 is the answer

P(x > or equal to 68.4)=? - n= 38 - Mean (u)= 67 - Deviation(o)= 18

- 18/ square root 38= 2.91998558 - (68.4- 67)/ 2.91998558 = .48 - .48= .6844= .3156

What is P [(sample mean > 86.95)? - n= 36 - Mean(u) = 82 - Deviation(o)= 18

- You need to know the (Ox) before you calculate the standard deviation in the z-score formula Ox= O/ square root n 1. Find the z-score: (sample mean- mean) divided by standard deviation Ox * (86.95-82) divided by calculated Ox (3)= 1.65 2. Find the z-score of 1.65= .9505 3. If it's ">" then its the area to the right and you need to subtract by 1 from the z-score * (1- .9505)= .0495

P(x>70)= ? - n= 1000 - Mean (u)= -.05 - Deviation(o) = 5.76

- to find deviation (x) you divide the deviation 5.76/ square root of (n) 1000= .182147193 0-(-.05)= .05 - divide .05/ .182147193= .274503269 * .274503269 rounds to .27 * .27 as the z-score = .6064 Since ">" you have to (1- .6064)= .3936 .3936 is the answer

P(x>70)= ? - n= 200 - Mean (u)= -.05 - Deviation(o) = 5.76

- to find deviation (x) you divide the deviation 5.76/ square root of (n) 200= .40729 0-(-.05)= .05 - divide .05/ .40729= .12276 * .12276 rounds to .12 * .12 as the z-score = .5478 Since ">" you have to (1- .5478)= .4522 .4522 is the answer

Katrina- 100 sample size, 95% confidence Matthew- 400 sample size, 99% confidence Who will have a smaller margin of error?

-Matthew after calculating the larger sample size compensates for the higher confidence

What is true regarding the distribution of the population?

-Since the sample size is large enough, the population distribution does not need to be normal.

o(p)=? n=200 and p=.5

.035

(u)p=? p=.5

.5, same as p

What is the probability a simple random sample of 60 gm results in at least 3.7 insect fragments? -acceptable level= 3 insect fragments - deviation calculated in question prior (.224)

3.7- 3= .7 .7/ .224= 3.125 3.125 z-score is .9991 1-.9991= .0009 its unusual because the prob. is small and can assume the pop. mean is higher than 3

Population Standard Deviation, find the standard deviation (o) given Population standard deviation- ox= [40 from subtracting the mean by the number to left of right on the graph] and n=9 given to you in the question?

40 * (square root of 9) 3= 120

what is the probability student can read 96 words per minute?

96- 92= 4 4/ 10=.4 -find .4 on the chart= .6554 * (1-.6554)= .3446 [.3446 is the answer}

Describe a sampling distribution of population 10,000, n=200 and p=.5

Approximately normal because n < .05 and np(1-p) >10

What is the relationship between probability of being ahead and the number of games played?

As the probability of being ahead decreases, the number of games played increases.

(T or F): The population proportion and sample proportion always have the same value.

False

(T or F): To cut the standard error of the mean in half, the sample size must be doubled.

False. The sample size must be increased by a factor of 4 to cut the standard error in half.

In the game of roulette... 38 slots , if it falls into one of the slots you win $35 otherwise you lose $1.

P(35)= 1/38= .0263 P(-1)= .(1-.0263) =9737 - plug in 35 and -1 into L1 and .0263 and .9737 into L2 to find the u and o. * u= -.05 * o= 5.76 - always approx. normal because the distribution doesn't matter - the mean stays the same (-.05) - plug o into the standard deviation formula

The standard deviation of the sampling distribution of (X with a line over it) , denoted (standard deviation with a subscript of x is called _________ _________ of the _________

Standard Error of the Mean

What effect does increasing the sample size have on the probability?

Increasing the sample size decreases the probability because: deviation decreases as sample size increases.

What is the value of sampling distribution mean (ux) on the graph?

It's the middle number or the regular mean

The _____ __ _____________ represents the expected proportion of intervals that will contain the parameter if a large number of different samples of size n is obtained. It is denoted _______________.

Level of Confidence and (1- beta)*100%

Does the population need to be normally distributed for the sampling distribution of (sample mean) to be approximately normally distributed?

No because the Central Limit Theorem states that regardless of the shape of the underlying population, the sampling distribution of sample mean becomes approximately normal as the sample size "n" increases.