Join StudySoup for FREE

Get Full Access to
GWU - Study Guide - Final

Description

Reviews

Topics and Examples

Measures of Central Tendency and Variability:

Mean: Denoted by x (“x-bar”). Is the sum of the measurements divided by the number of measurements.

∑n

xi

x= n

i=1

Median: the middle number of a data set when arranged in an ascending or descending order. Range: In a selection of data is equal to the largest measurement subtracted by the smallest measurement.

Sample Variance: is the sum of the squared divisions of from the mean divided by n-1.

∑ (x −x) i2

2n−1

s (sample variance) =

S (sample standard deviation)= √s2

____________________________________________________________ 1. Calculate the mean, median, range, and standard deviation of the data set. 1. 30, 25, 27, 34, 37, 41, 26, 43, 35

2. Explain why each of these measures would/would not change if each measure in the first set increased by 5.

Probability (Unions and Intersections):

A⋃ B: the union of two events A and B is the event that occurs of either A or B (or both) occur on a single performance of the experiment.

A⋂ B: the intersection of A and B is the event that occurs if both A and B occur on a single performance of the experiment.

Complement: the complement of an event A is the event that A does not occur. Denoted by A .c c Ac

P( A )= 1 - P(A) P(A) + P( ) = 1

A l B: the conditional probability that event A occurs given that event B occurs. Divide the probability of A and B by the probability of B.

____________________________________________________________________________ 3. Suppose that each college student is classified as underclassman or upperclassmen. One student is selected at random. The probability that the student is an upperclassman is 0.63. The probability that the student is a male underclassman is 0.17. The probability that the student is male is 0.47. Find the probability that the student is a male or an underclassman (or both).

If you want to learn more check out yuild

Multiplicative Rule:

The probability that both events A and B are equivalent to the probability that event A occurs multiplied by the probability that event B occurs, given that A has occurred.

P(A⋂ B)= P(A) P(B l A) or P(A⋂ B)= P(B) P(A l B)

Events A and B are independent events if the occurrence of B does not alter the probability that A has occurred (or occurs). We also discuss several other topics like which of the following statements about hadley cells is true?

P(A l B)= P(A) or P(B l A)= P(B)

____________________________________________________________________________ 4. The probability that a randomly selected female is a brunette is 0.79. Suppose the conditional probability that the female has blue eyes, given that she is brunette is 0.35. Find the probability that a randomly selected female is brunette and has blue eyes.

Probability Distribution:

The mean or expected value of a discrete random variable is:

μ = E(X)= Σ r p(r)

The variance of a random variable is:

2 [(X − p)2] Σ(x − μ)2

σ = E = p(x)

The standard deviation is the square root of the variance:

σ = √σ2

____________________________________________________________________________ 5. You observe 4 students at a university. X is the number of students that attended a private high school. Find the probability that less than 3 of the students went to a private high school. And find the expected value of X and the variance of X. We also discuss several other topics like which expression adds 1 to the element of array arrayname at index i?

If you want to learn more check out linguistics midterm quizlet

P(X=3)= 0.15 P(X=2)=0.31 P(X=4) = 0.05 P(X=1) = 0.23

Binomial Random Variable:

Consider a sequence that consists of H’s and (n-r) T’s, in whatever order. The probability of r qn−r pr(1 − p)n−r pr qn−r

such a sequence must be p = and the probability P(X=r) = multiplied by the number of such sequences or arrangements of (r)H’s and (n-r)T’s.

This multiplier is called “n choose r” denoted by ( )rn

n n!

It is the ratio (r) = where n! = n(n-1)(n-2) … 3 2 1

r!(n−r)!× ×

____________________________________________________________________________ 6. In a population of college students, a proportion of 0.47 has a mental illness. A random sample of 10 college students is selected. Find the probability that exactly 8 of them have a mental illness. Also, find the probability that fewer than 9 have a mental illness.

Normal Distribution:

(X−μ)

If X is a normal random variable with mean μ and sd σ , then is a linear transformation. σ When we convert a normal to a standard normal by subtracting the mean μ and dividing by the sd σ , we call it standardizing.

Therefore a probability problem about a normal random variable X with mean μ and sd σ , can be easily written as a probability problem about a standard normal variable Z. Look at Z tables to find probabilities for the Z variable.

Using the normal (Z) table in reverse(the “inverse problem”)

Suppose you are given a probability and have to find the corresponding value; for example, find the value of w such that P(Z>w)=0.10 so P(Z ≤ w)= 0.9

The area beneath the curve from 0 to w must be equal to 0.5-0.1=0.4 so P(0<Z<w)=0.4. Then you would need to look for .4000, or a probability closest to it, in the body of the Z table. The closest probability was 0.3997=1.28 so the value of this is Z=1.28. Don't forget about the age old question of jmu assessment day

____________________________________________________________________________ 7. Suppose that scores on a test are normal with a mean of 500 and a sd of 100. Find the probability that a person scores a 625 on the test.

Then find the probability that a person scores between 450 and 700 on the test.

8. Suppose that scores on a test are normal with a mean of 500 and a sd of 100. Find the score “t” such that the probability is 0.15 that the person’s score is greater than t.

Sampling Distribution:

The probability of the distribution of a sample statistic (a random quantity that can vary from sample to sample)

For the sample mean for a sample of size n from a population with mean μ and standard deviation σ,

μx = μ

σx =σ√n

____________________________________________________________________________ 9. Suppose a population of GPA scores has a population mean of μ = 2.8 and sd σ = 0.4 In a sample of n=100 students, the sample mean varies from sample to sample, but we can think of its average value and its variability. Find μ and . x σx We also discuss several other topics like usc mat

Central Limit Theorem:

For large n, the sample mean has (approximately) a normal distribution.

We can calculate probabilities after standardizing to a Z, using

μx = μ

σx =σ√n

____________________________________________________________________________ 10. Find the probability that, for a random sample of 100 students, find P(2.72<X<2.88), P(2.76X<2.84), and P(2.70<X<2.90)

Note: the population mean μ= 2.8 and population sd σ= 0.4.

Confidence Intervals:

The probability that the random interval ( x - 2 σ , + 2 ) contains the population mean . x x σx C Mathematical fact:

x-c<K<x+c is equivalent to K-c<x<K+c. Thus, the event x - 2 σx < μx < x + 2 σxis the same as the event μ - 2 < < +2 , so they have the same probability. x σx x μx σx

Hence, P( x - 2 σx < μx < x + 2 σx) = P( μx- 2 σx < x < μx +2 σx)= P(-2<Z<2)= 2 × 0.4772= .9544 You can get this by standardizing; subtract μ and divide by . x σx

P( x - 2 σx < μx < x + 2 σx) = 0.9544

The interval ( x - 2 σx, x + 2 σx) is said to be a confidence interval for μ with confidence coefficient 0.9544, or

The interval ( x - 2 σ , + 2 ) is a 95.44% confidence interval for

x x σx μ

____________________________________________________________________________ 11. For the GPA example, suppose μ is unknown, but we have σ =0.4 and n=100 then σ =0.04 and if the mean of a sample is 2.95, Find the confidence interval.

Large Sample CI for μ :

Thus, 100(1- α )% CI for μ is ( x - z , + ) where is the appropriate value from the z

2σ σx x z2σ σxz2σ

table, x is the sample mean and σx = 2σ

If the population sd σ is not provided in the problem, use the sample standard deviation s

z s 2σ

z s 2σ

(formula from chapter 2) instead; the CI formula becomes (- , + )

√n

√n

____________________________________________________________________________ 12. 1- α =0.93, x =2.95, σ=0.4 n=100 Find the confidence interval

Confidence Interval for p:

Population proportion p

︿

Sample proportion p (from a sample size n)

︿

For large n, p has approximately normal distributions

μ = p = (calculate first, then square root the whole thing) p︿ σp︿ √ npqnpq ︿z2σ σp︿ p︿z2σ σp︿ σp︿ √ npq

( p - , + ), = contains the unknown parameter p, so in this instance, ︿q︿p︿

the formula is not useful. In practice instead, use p in place of p and = (1- ) in place of q. ︿z2σ √ npq

︿︿p︿z2σ √ npq

︿︿

( p - , + )

____________________________________________________________________________ 13. n=1600 x=800 Find the confidence interval for 94.1% and for 96.6%

T Confidence Interval (Small Sample CI for μ )

μx = μ and σx =σ√n

Suppose that n is small and the population sd σ is unknown. The Central Limit Theorem does not apply if n is small. Instead, we assume that the population is normal

x−μ

x−μ

Instead of ( σ ) which is a z statistic, we use ( ) which has the so-called t distribution with s

√n

√n

df = n-1 (degrees of freedom - new parameter)

If n is small: assume the population is normal ↳ if σ is known the previous CI method works ↳ if σ is unknown then use the small sample CI

αs√nx t2αs√nt2α

100(1- α )% CI: ( x - t , + ) where comes from the t table and df= n-1 2

*note the t table has a different layout from the standard normal

Ex: 1- α = 0.9 so α = 10, = 0.05 = 2αt2α t0.05

____________________________________________________________________________ 14. Suppose that, in a random sample of 14 college students, the average score on a standardized test is 1040.6, and the standard deviation is 196.3. Find the 99% Confidence Interval for the population mean score on the standardized test.

CI for μ (the difference in population sample means): 1 − μ2

If you’re interested in a CI for μ , use sample statistic 1 − μ2 x1 − x2

μ = x −x 1 2μ1 − μ2

x −x 1 2 √n1

σ21 + n2

σ22 σxσ√n √ nσ2

σ = because = =

σ21 + n2

σ22

Note: in the σ formula, calculate first, then take the square root x −x 1 2 n1

*Use σ , if it is unknown then you must use s (the sample sd). If s is unknown, note that you can calculate it with information from the data.

CI formula:

( x ) 1 − x2 ± z2σ σx −x 1 2

This formula uses the population sd σ & , which are typically unknown. If they are not 1 σ2 known, use the sample, s1&s2. Thus, calculate σ = as = x −x 1 2 √n1

σ22 sx −x 1 2 √s21n1+s22

σ21 + n2

n2

(In a real problem s1&s2 can be calculated from the sample data, just as the sample means x &x can be calculated) 1 2

____________________________________________________________________________ 15. There was a study done to measure the test scores in a school. In 2011 there was a random sample of 400 students that found an average score of 1120.9 and a sd of 71.2. The similar but

independent study was done in 2016 with a random sample of 350 students which found an average score of 1070.5 and a sd of 66.4. Find the 93% CI for the population mean test score in 2011 minus the population mean score in 2016.

Large Sample CI for p : 1 − p2

Population proportions: p1 and p2

︿p2︿

Sample proportions: p1 and

Sample sizes: n1 and n2

*We can’t use the unknown population quantities p1 and p2in the calculation ︿p2︿

Solution: Use sample proportions p1 and for an approximation

p q1︿1︿+ n2

︿

︿

( p ) → ( ) 1︿− p2︿± z2σ σp −p 1︿2

︿ p1︿− p2︿± z2σ √ n1 p q1︿1︿+ n2

p q2

2

︿

︿

p q2

2

*First calculate then take the square root n1

____________________________________________________________________________ 16. In a random sample of 800 students admitted into a college, 212 of them scored above average on a 2011 standardized test. In an independent sample of 600 students admitted into a

college, 120 of them scored above average on a 2016 standardized test. Find the 94% CI for p . 1︿− p2︿

*Note 2011 sample is population 1 and 2016 sample is population 2

Hypothesis Testing:

Steps in testing hypotheses

1. State hypotheses, describing parameters

H : = 200 (null hypothesis) 0 μ

Ha : μ > 200 (alternative hypothesis)

2. Level of significance

α is called the level of significance or probability of Type I error of the test. It is a pre-set small probability, often chosen as α=0.05 or α =0.01. It is the probability that even if H0is true, one gets data that rejects H0. Finding α:

- For .03 zα = z.03 →0.5-α = 0.5 - .03 = 0.47 → 1.88 zα = 1.88

- For .01 zα = z.01 →0.5-α = 0.5 - .01 = 0.49 → 2.33 zα = 2.33

- For .05 zα = z.05 →0.5-α = 0.5 - .05 = 0.45 → 1.65 zα = 1.65

- For .025 zα = z.025 →0.5-α = 0.5 - .025 = 0.465 → 1.96 zα = 1.96 3. Test statistic (formula)

(x−μ )0 s

zobs = where s=

s

√n

4. Critical (or rejection) region

The test that rejects H0in favor of Haif zobs > zαis called a level α test. 5. Calculation and conclusion

(x−μ )0s

(x−μ )0 s

z= zobs = where s= σ

√n

____________________________________________________________________________ 17. Suppose that in a random sample of 250 college students, the average hours spent in class per week is 13.6 and the standard deviation is 4.3. Do the data provide significant evidence at the 6.5% level, that the population mean hours spent in class per week is less than 14?

Another one-sided test

H0: μ= μ0 vs

Ha: μ< μ0

For testing H0: μ= μ0 vs. Ha: μ< μ0 we reject H0(and favor the alternative Ha) if the sample mean is much below μ0.

Specifically, reject H0 at level α in favor of alternative Haif zobs < -zα.

Suppose α=0.03; we would reject H0if zobs < -1.88.

____________________________________________________________________________ 18. In a random sample of 64 boxes of cereal, the average weight of contents is 15.9 ounces and the standard deviation is 0.2 ounces. Do the data provide significant evidence at the 3% level that the net population mean weight of contents of cereal box is less than 16 ounces?

Two-sided test

H0: μ= μ0 vs

Ha: μ≠ μ0.

The alternative hypothesis is called a two-sided alternative.

Idea: for testing H0: μ= μ0 vs. Ha: μ≠ μ0 we reject H0(and favor the alternative Ha) if the sample mean is either much larger than μ0 or much smaller than μ0.

Specifically, (change in step 4: rejection region):

reject H0 at level α in favor of the alternative Haif zobs > or if z z obs < - .

α z

α

2

2

Finding a P-Value:

Instead of just deciding between hypotheses, suppose we want to quantify the strength of the evidence against the null hypothesis.

p-value: a measure of agreement/disagreement between the null hypothesis and the data. We illustrate it for testing hypotheses about the population mean μ. Referring to the example in the last class, where σ =80 and n=100.

Suppose we observe sample mean 220.

The probability P(≥ 220) assuming H0: μ= μ0 holds

is called the p-value for testing H0 against Ha: μ> μ0.

H0: μ= μ0

Ha: μ> μ0

220−200

20

Zobs = 80 = = 2.5 → 0.4938 0.5 - 0.4938 = 0.0062

8

√100

Intuition supported by probability calculation:

The p-value=P( ≥220) = .0062 if μ= 200.

The small probability means that there is strong evidence against H0 and in favor of Ha.

*The smaller the p-value, the greater the disagreement between the null H0 and the data, and the more we would favor the alternative.

Testing hypotheses about population proportion p

H0: p= p0 vs

Ha: p< p0

Idea: would reject H0in favor of Haif the sample proportion is much smaller than p0. ︿

The test statistic is zobs =(p−p )

0

p q0 0

In general, σ = √ npq

√ n

pq√ n

assuming H0is true, we have √ n =

p q0 0

____________________________________________________________________________ 19. In a random sample of 1000 households, 210 tune in to a show. Do the data provide significant evidence, at the 3% level, that the population proportion of households that tune in to the show is less than 0.25?

Two-sided version of the above problem:

____________________________________________________________________________ 20. In a random sample of 1000 households, 210 tune in to a show. Do the data provide significant evidence, at the 3% level, that the population proportion of households that tune in to the show is different from 0.25?

Small-sample test of hypotheses about population mean

We came across t distributions in small-sample confidence intervals (CI) for μ. Suppose that n is small (Central Limit theorem doesn’t apply) and population sd σ is unknown. *Need to assume that the population is normal.

The procedure is quite similar to the large sample test of μ, except that in step 3, the test statistic is

(x−μ )0

tobs = s and

√n

in step 4, critical region, we have

t α or with df=n-1 in place of z t α or .

α z

α

2

2

____________________________________________________________________________ 21. In a random sample of 10 students from a population, the average GPA is 3.20 and the sd is 0.41. Do the data provide significant evidence at the 5% level, that the population mean GPA is greater than 3.0?

Hypothesis test for difference of population means

H : 0 μ1 − μ2 = D0

Ha : μ1 − μ2 > D0

μ1& μ2 are the population means and D0is the numerical difference between the population means

x1−x2−D0 σx1−x2= √n1

σ21 + n2

zobs = where σx −x 1 2

σ22

If the population sd’s σ1&σ2 are unknown, use the sample sd’s s1&s2 and calculate σx1−x2= √s21n1+s22

n2

____________________________________________________________________________ 22. In a random sample of 48- female shoppers at a grocery store, the average amount spent is $38.75, and the standard deviation is $9.90. In an independent random sample of 390 male

shoppers, the average amount is $26.30 and the standard deviation is $7.50. Do the data provide significant evidence at the 1.7% level that the population mean amount spent by female shoppers exceeds the mean amount spent by male shoppers by more that $10?

Hypothesis testing for comparing population proportions

A key difference is that the null hypothesis is always H0: p1 − p2 = 0 where the alternative hypothesis could be H , , or 0: p1 − p2 < 0 H0: p1 − p2 > 0 H0: p1 − p2 =/ 0 H says that are equal, but does not specify the common value. 0 p &p 1 2

︿

︿

p −p −0

︿︿p︿=x +x 1 2

︿︿+ n2

2 √n1

pq

pq

z where = and obs = σp −p

2 σp −p 1

︿1︿2

︿ 1

︿

n +n 1 2

____________________________________________________________________________ 23. In a random sample of 800 students admitted to a college in year 2011, 212 of them scored above twelve hundred in a scholastic test. In an independent random sample of 600 students admitted to a college in year 2016, 120 of them scored above twelve hundred in the scholastic test. Do the data provide significant evidence at the 3% level that the population proportion of year 2011 students who had a scholastic test score above twelve hundred is greater than the population proportion of year 2016 students who had a scholastic test score above twelve hundred?

Regression and Correlation:

Least squares regression equation is: ︿

0 β︿1

Y = β + x

︿

SSxyβ︿0 y β︿1 x

β = = -

1 SSxx

Model assumptions:

1. E(ε )= 0

2. Variance of ε is constant (=σ2) for all values of x.

3. Probability distribution of ε is normal

4. Values of ε associated with any two observations are independent SSE = Σ (yi – - xi)2 β︿0 β︿1

*This is also the smallest possible sum of squared differences for any straight line equation.

To calculate it, don’t have to actually work out all the

fitted values and subtract them. We have a formula

SSE = SSyy- SSxy β︿1

2

(∑ yi)2∑ xi2 − n (∑ x )i

∑ x )( )

i ∑ yiβ︿1 SSxx

where SSyy = Σ yi2- , SSxx = ,SSxy = , and = n

∑ x yi i − n

Example:

Xi Yi Xi Yi Xi2 Yi2

3.3 3.2 10.56 10.89

3.5 3.3 11.55 12.25

3.6 3.6 12.96 12.96

3.7 3.6 13.32 13.69

3.4 3.3 11.22 11.56

__________________________

Total:

17.5 17 59.61 61.33

(Won’t need Yi2till later)

∑ x )( )

i ∑ yi

SSxy = ∑ xiyi − = 0.11 n

2

(∑ x )i

SSxx = ∑ x = 0.10 i2 − n

︿

1 β︿0

β = 1.1 & =3.4 - (3.5)(1.1) = -0.45

Testing hypotheses about slope β of the model 1

︿

SSxyβ︿0

(We can calculate based on the sample data but we don’t know the slope 1 β for the whole

1

population)

Important: Testing hypotheses questions are phrased differently.

H0: β = 0 vs. 1

Ha: >0 or Ha: <0 or Ha β1 β1: β1 ≠ 0

One-sided or two-sided alternatives.

Understand what each of the possible alternatives means – sketch a line with slope >0 or <0 or =0.

The test statistic is of the form

︿

(β – 0 )

tobs= σβ︿1

1

and we use the following expression (it’s an estimate) for the denominator, β︿1s

σ =

√SSxx

The degrees of freedom is n-2 for the critical value tα.

____________________________________________________________________________ 24. Suppose that the following high school (HS) GPA and college GPA are observed for five students, the first number in each pair being the HS GPA. (3.3, 3.2), (3.5, 3.3), (3.6, 3.6), (3.7, 3.6), (3.4, 3.3). Do the data provide significant evidence, at the 1% level, to indicate that as HS GPA increases, on the average college GPA increases?

Correlation coefficient:

r=SSxy

√SSxx SSyy

.11

For the problem above r= = 0.93

√(.10)(.14)

r is the sample correlation coefficient.

-1≤ r ≤1 for any data set.

r=1 (or -1) if and only if the data fall exactly on a straight line with positive slope (or negative slope respectively).

r is a measure of how close the data are to a straight line.

The corresponding measure for the whole population is denoted by ρ.

Answer Key:

1. x = 33.1, median= 37, range = 43-25 = 18, s =

2

2 2 2 222 2 2 2 (30−33.1) + (22−33.1) + (27−33.1) + (34−33.1) + (37−33.1) + (41−33.1) +(26−33.1) +(43−33.1) +(35−33.1)

8=

9.67+123.43+37.33+0.79+15.13+62.25+50.55+97.81+3.37 8

s= √s2 = √50.04 = 7.07

= 50.04

2. Mean: The mean will increase by 5 because the new total will be equivalent to the old total multiplied by n(5).

Median: The median will change because each of the numbers will increase by 5 and they will still be in the same order of increasing value and the median is the value in the middle of the set.

Range: The range does not change because all of the numbers increase by 5 so the maximum and the minimum numbers both increase by 5 leaving the range to be the same. Standard deviation: This does not change because the mean and the value increase by 5 so the

2

∑ (x −x)

i

values obtained in the formula will be the same and will have the same value. n−1 3. P(A) = the event that the student is an upperclassman

P(B) = the event that the student is male

Male

Female

upper

0.30

0.33

under

0.17

0.20

0.47 0.53

c ⋃ B Ac B Ac ⋂ B

P( A ) = P( )+P( ) - P( ) = 0.37 + 0.47 - 0.17 = 0.67

4. B = the event that the female is brunette

E = the event the female had blue eyes

P(B)= 0.79

P(E I G) = 0.35

P( B ⋂ E ) = P(B) P(E I B) = (0.79)(0.35) = 0.2765

5. P(X<3)= P(X=0)+P(X=1)+P(X=2)= 0.26+0.23+0.31 = 0.80

E(X) = 3(0.15)+ 4(0.05)+2(0.31)+1(.23) = 1.5

2 [(3 − 1.5) (.15)

2] [(4 − 1.5) (.05)

2] [(2 − 1.5) (.31)

2] [(1 − 1.5) (.23)

2] [(0 − 1.5) (.26)

σ = + + + + = 0.338+0.313+0.078+0.058+0.585 = 1.372

6. n=10 p=0.47

2]

( ) = = = = =(45) rn pr qn−r( ) 810(.47)8(.53)2 10! 8!2!(.47)8(.53)28!2!

10×9×8!(.47)8(.53)2290(.47)8(.53)2(.47)8

2

(.53) = 0.03

P(X<9)= 1-P(X≥ 9) = 1- [P(X = 9) + P(X = 10)] = 1-[(9)(.47) (.53) ( )(.47) (.53) =

10 9 1 +101010 0]

1-[(10)(.47) (.53) 1)(.47) (.53) = 1-(.006+ .0005) = 0.0035

9 1 + (10 0]

7. Getting a 625 -

(X−μ)

(625−500)

125

= = = 1.25 Z=1.25 σ

100

100

Probability found=0.3944 P(Z>1.25)= 0.5 - 0.3944 = 0.1056 so P(X>625)=0.1056 Getting a score between 450 and 700 -

(450−500)

(700−500)

(−50)

(200)

P(450<X<700)= < X< = < X< = -0.5< Z< 2 = 0.0199< X< 0.4772 100

100

100

100

So P(450<X<700)= 0.0199 + 0.4772 = 0.4871

8. P(X>t)=0.15 so we would then standardize this by converting to a standard normal (t−μ)

P(Z> ) = 0.15 σ

(t−μ)

Write w= to simplify notation. σ

P(Z>w)= 0.15

P(0<X<w) = 0.5-0.15 = 0.35 Look for .3500 or probability close to it in the body of the Z table Closest probability is 0.3508 = 1.04 so w=1.04

(t−μ)

(t−μ)μ + 1.04σ

w= so 1.04 = so t= = 500 + 1.04(100) = 604

σ

σ

The answer is the score t= 604 9. μx = μ = 2.8

σ

0.4

0.4

σx = = = = 0.04

√n

√100 10

σ

0.4

0.4

10. μx = μ = 3.1 σx = = = = 0.04

√n

√100 10

2.72−μxσx

2.88−μx0.04

2.72−2.80.04

2.88−2.8

P(2.72 < X < 2.88) = P( < Z <) = P( < Z <) σx = P(-2.0 < Z < 2.0) = .4772 + .4772 = .9544

2.76−μxσx

2.84−μx0.04

2.72−2.80.04

2.88−2.8

P(2.76 < X < 2.84) = P( < Z <) = P( < Z <) σx

= P(-1.0 < Z < 1.0) = .3413 + .3413 = .6826

2.70−μxσx

2.90−μx0.04

2.72−2.80.04

2.88−2.8

P(2.70 < X < 2.90) = P( < Z <) = P( < Z <) σx = P(-2.5 < Z < 2.5) = .4938 + .4938 = .9876

11. (-2 σ , +2 σ )=(-0.08, +0.08)= (2.87, 3.03) is a 95% CI for μ . (1−α)z2σ

12. 1- α =0.93 = 0.465 = 1.81 2

z σ

z σ

2σx√n 2σ

(1.81)(0.4)

(1.81)(0.4)

( x - , + ) = (2.95- , 2.95+ ) = (2.95 - 0.0724, 2.95 + 0.0724) =

√n

(2.8776, 3.0224)

√100

√100

1600 σp︿ √ npq √ npq

︿ 800

︿︿√ 1600 (0.5)(0.5)

13. p = = 0.5 = = = = 0.0125

(1−α)z2σ p︿z2σ √ npq

︿︿p︿z2σ √ npq

︿︿

I. 1- α =0.941 = 0.4705 = 1.89 ( - , + ) = 2 (0.5 - (1.89)(0.0125) , 0.5+(1.89)(0.0125) = (0.4764, 0.5236)

(1−α)z2σ p︿z2σ √ npq

︿︿p︿z2σ √ npq

︿︿

II. 1- α =0.966 = 0.483 = 2.12 ( - , + ) = 2

(0.5 - (2.12)(0.0125) , 0.5+(2.12)(0.0125) = (0.4735, 0.5265)

14. n=14 df=13 x = 1040.6 s=196.3

1- α = 0.99 α = 0.01 = 0.005 2α

t .012

α = t.005 = 3

2

αs√nx t2αs√n±√14

196.3 ±

CI: ( x - t , + ) = (1040.6 (3.012)( )) = (1040.6 158.02) = (882.58, 1198.62) 2

The 99% CI for the population mean score is (882.58, 1198.62)

15. x1 = 1120.9 x2 = 1070.5 s1 = 71.2 s2 = 66.4

(1−α)z2σ

1- α =0.93 = 0.465 = 1.81 2

s = = = = 5.03 x −x 1 2 √s21n1+s22

n2 √ 400

71.22+ 350

66.42√25.26

CI: ( x ) = ((1120.9 - 1070.5) (1.81)(5.03)) = (50.4 9.10) 1 − x2 ± z2σ σx −x 1 2± ± CI = (41.3, 59.2)

The 93% CI for μ is (41.3, 59.2) 1 − μ2

︿

212 p2︿600 120

16. n1 = 800 n2 = 600 x1 = 212 x2 = 120 p1 = = 0.265 = = 0.20

800

p q1︿1︿+ n2

︿ √ n1

︿

︿√ 800

p q2

(.265)(.735) + 600 (.2)(.8)

σ = = = 0.023 p −p 1︿2 2

2σ 21−α2.94

z = = = 0.47 → 1.88

CI: ( p ) = ( (1.88)(0.023)) = ( 0.043)= (0.022, 0.108) 1︿− p2︿± z2σ σp −p 1︿2

︿ 0.265 − 0.2 ± 0.065 ±

The 94% CI for p is (0.022, 0.108) 1︿− p2︿

17. n=250 x = 13.6 s= 4.3 α = 0.065

Step 1:

H μ 4 0: = 1

Ha : μ < 14

Step 2:

Level α = 0.065

Step 3: Test statistic

x−μ

zobs = s√n

Step 4: Critical region

We reject H0in favor of Haif zobs <- zαis called a level α

Step 5:

zα = z.065 = 0.435 → 1.51 -zα = -1.51

x−μ

13.6−140.27 −0.4

z = = = -1.47 obs = s√n

4.3

√250

Given that = -1.47 is not less than = -1.51 we fail to reject H0

zobs − zα in favor of Ha at level α = 0.065. So the data does not provide significant evidence that the population mean hours spent in class is less than 14.

18. H0: μ= 16

Ha: μ< 16

Level α = 0.03

(x−μ )0

Test statistic: zobs = s = √n

Critical Region:

(x−16) σ

√n

Reject H0in favor of Haif zobs < -zα at level 3% -zα = -z.03→ 0.5 - 0.03 = 0.47 → zα = 1.88 -zα = -1.88 (15.9−16) −0.1

zobs = 0.2 = = -4

√64

0.025

Since zobs = -4 and is not less than -zα = -1.88 we do not reject H0 and conclude that the data does not provide significant evidence at the 3% level that the net population mean weight of contents of cereal box is less than 16 ounces.

19. H0: p= 0.25

Ha: p< 0.25,

where p is the population proportion of households that tune in to the show. Level α= 0.03

︿

Test statistic: zobs =(p−p )

0

p q0 0

√ n

︿

=(p−.25)

√ n

(.25)(.75)

Critical region (or rejection region):

The test rejects H0 at level α in favor of Haif zobs < - z

Calculations and conclusion:

210

= = 0.21; n=1000

1000

α= 0.03; so 0.5- α= 0.47\ zα = z.03 = 1.88.

- zα = -1.88

(.21−.25)

zobs = = -2.921

√ 1000

(.25)(.75)

Since zobs = -2.921 < -1.88= -zα

Reject H0in favor of Ha at the 3% level.

So, the data do provide significant evidence, at the 3% level, that the population proportion of households that tune in to the show is less than 0.25.

20. From the last sentence, the alternative is Ha: p ≠ 0.25.

So in step 1 of the solution,

H0: p= 0.25

Ha: p≠ 0.25.

(...change in step 4)

Critical region (or rejection region):

The test rejects at H0 at level α in favor of Ha

if zobs > or if z z obs < - .

α z

α

2

2

(change in calculations and conclusion, step 5)

α = 0.03 ; = =0.015 2α2.03

0.5- = 0.5-0.015 = 0.485 2α

= z z .015 = 2.17.

α

2

Since zobs =-2.921 < -2.17= - (= -z.015

z )

α

2

Reject H0in favor of Ha at the 3% level. The data do provide significant evidence, at the 3% level, that the population proportion of households that tune in to the show is different from 0.25

21. Assumption: the population is assumed to be normal

H0: μ= 3.0

Ha: μ> 3.0

Where μ is the population mean GPA

Level α= 0.05

(x−μ )0s

Test statistic: tobs = s = √n

(x−3.0) √n

Critical region (or rejection region):

The test rejects H0 at level α in favor of Haif tobs > tα, where tαis a critical value from the t-distribution with df=n-1= 10-1=9

Calculations and conclusion:

x = 3.2; n=10 ; s = 0.41;

α= 0.05; df=10-1=9

look for tα = t.05for 9 df, get 1.833

(3.2−3.0)

tobs = = 1.5426 . .41

√10

Since tobs =1.5426 is not greater than (not >) 1.833= t.05(for 9 df),

do not reject H0in favor of Ha at the 5% level.

The data do not provide significant evidence at the 5% level, that the population mean GPA is greater than 3.0.

22. H0: μ1 − μ2 = 10

Ha: μ1 − μ2 > 10

Where μ1is the population mean amount spent by females and μ2is the population mean amount spent by males

Level α = 0.017

x1−x2−D0 σx1−x2= √s21n1+s22

Test statistic: zobs = where σx −x 1 2

n2

Critical region: test rejects H0in favor of Ha at the 1.7% level if zobs > zα zα = z.017 = 0.5-0.017=0.483 → zα = 2.12

= = = 0.5902 √s21n1+s22

n2 √ 480

(9,9)2+ 390

(7.5)2√0.348

x1−x2−D0 zobs = σx −x 1 2

38.75−26.30−10 2.45

zobs = = = = 4.15 σx −x 1 2

0.5902

Given that z =4.15 > =2.12 we reject in favor of at the 1.7% level and conclude that obszα H0 Ha the data do provide significant evidence that the population mean amount spent by female shoppers exceeds the mean amount spent by males by more than $10.

23. H0: p1 − p2 = 0

Ha : p1 − p2 > 0

Where p1is the population proportion of year 2011 students who had a scholastic test score above twelve hundred and p2is the population proportion of year 2016 students who had a scholastic test score above twelve hundred.

Level α = 0.03

︿

︿

Test statistic: zobs =p −p −0

1

2

√+ n1

︿︿ pq

︿︿pq n2

Critical region: The test rejects H0in favor of Ha at level α = 0.03 if zobs > zα

︿=x +x 1 2

212+120 332

︿

︿

p = = = 0.237

212 = 0 p .20

n +n 1 2 800+600

1400p .265 1 = 800

120 = 0

2 = 600

︿︿√ 800

︿︿+ n2

2 √n1

pq

pq

(.237)(.763) + 600 (.237)(/763)

σ︿p −p = = = 0.023

︿

1

︿

︿

z = = = 2.83 obs =p −p −0 0.265−0.20−00.023

1

2

︿︿ 0.023

0.065

√+ n1 ︿︿

pq

pq n2

zα = z.03 → 0.5-0.03=0.47 → zα = 1.88

Given that z = 2.83 > = 1.88 we reject in favor of at level = 0.03 and conclude obszα H0 Ha α that the data do provide significant evidence that the population proportion of year 2011 students who had a scholastic test score above twelve hundred is greater than the population proportion of year 2016 students who had a scholastic test score above twelve hundred.

24. We assume the regression model

︿

H0: β 1 =0

︿β︿

Ha: 1 >0 where 1 β is the slope of the model equation

Level α= 0.01

Test statistic: tobs =

Critical region (or rejection region): The test rejects H0 at level α in favor of Haif tobs > tα, where tα is a critical value from the t-distribution with df=n-2= 5-2=3

Calculations and conclusion:

tα = t.01 where df= n-2= 3 t.01 =4.541

t.01 where df= n-2= 3

Xi Yi Xi Yi Xi2 Yi2

3.3 3.2 10.56 10.89 10.24

3.5 3.3 11.55 12.25 10.89

3.6 3.6 12.96 12.96 12.96

3.7 3.6 13.32 13.69 12.96

3.4 3.3 11.22 11.56 10.89

_____________________________

Totals:

17.5 17 59.61 61.33 57.94

︿

(β – 0 )

1 σβ︿1s

tobs = where = σβ︿1

√SSxx

︿

β =

SSxy

1 SSxx

(Σ x ) (Σ y )

SSxy = Σ xi yi- = 0.11 n

i i

SSxx = Σ xi2- (Σ xi)2/n = 0.10

︿

1 β︿0

β = 1.1 & = -0.45

1.1

tobs = = 4.369

0.0796

√.10

Given that tobs = 4.369 < tα =4.541 the test fails to reject H0in favor of Ha at level α = 0.01 and cannot conclude that the data provide significant evidence to indicate that as HS GPA increases, on the average college GPA increases.