Stats final exam; Yufei; UMD

null hypothesis

denoted by H0, is a tentative assumption about a population parameter. It is the opposite of what you're testing Begin with this when the test is challenging the assumption and is determining if there's statistical evidence to show the assumption is incorrect *Assumes whatever you're testing has no effect, there won't be a change from the original*

alternative hypothesis

denoted by Ha, is the opposite of what is stated in the null hypothesis. It is the claim you're testing Begin with this when testing involves an attempt to gather evidence in support of a research hypothesis *Assumes whatever you're testing will have an effect, will change from original*

null hypothesis

equality part of the hypotheses always appears in the

H0: p1-p2 ≤ 0 Ha: p1-p2 > 0

hypothesis tests about p̂1-p̂2--> Right-tailed

H0: p1-p2 = 0 Ha: p1-p2 ≠ 0

hypothesis tests about p̂1-p̂2--> Two-tailed

H0: p1-p2 ≥ 0 Ha: p1-p2 < 0

hypothesis tests about p̂1-p̂2--> left-tailed

p̂1-p̂2

if the sample sizes are large, the sampling distribution of ______ can be approximated by a normal probability distribution

Type II error

is accepting H0 when it is false difficult to control for probability of making Type II error.

a true alternative hypothesis is mistakenly rejected

A Type II error is committed when

H0: μ = 5.8 Ha: μ ≠ 5.8

A machine is designed to fill toothpaste tubes with 5.8 ounces of toothpaste. The manufacturer does not want any underfilling or overfilling. The correct hypotheses to be tested are

20/2= 10% find t value with 0.1 and df of 19 --> 1.328 Answer= 1.328

For a two-tailed test, a sample of 20 at 80% confidence, the critical value =

proportion: 150/300= 0.5 for p1 121/275= 0.44 for p2 Find pooled proportion: p̂= n1p̂1 + n2p̂2/ n1 + n2 300(0.5) + 275(0.44)/ 300 + 275 (150+121)/(300+275)=0.47 test statistic for 2 pop. proportions: z= (p̂1-p̂2)/√p(1-p)(1/n1+1/n2) (0.5-0.44)/√0.47(1-0.47)(1/300+1/275) = 1.44 The alternative hypothesis is p1-p2>0, increases significantly, one tailed test z0.10=1.282, since 1.440>1.282, we support the alternative and reject the null at 10% level At 10% significant level, we can reject the null hypothesis

A poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken five years ago. Results are summarized below. We are interested whether the proportion increased significantly comparing with five years ago? The conclusion is Present sample: Sample size: 300 Number considered themselves overweight: 150 Previous sample: Sample size: 275 Number considered themselves overweight: 121

proportion: 150/300= 0.5 for p1 121/275= 0.44 for p2 Find pooled proportion: p̂= n1p̂1 + n2p̂2/ n1 + n2 300(0.5) + 275(0.44)/ 300 + 275 (150+121)/(300+275)=0.47 test statistic for 2 pop. proportions: z= (p̂1-p̂2)/√p(1-p)(1/n1+1/n2) (0.5-0.44)/√0.47(1-0.47)(1/300+1/275) = 1.44 Answer= 1.44

A poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken five years ago. Results are summarized below. We are interested whether the proportion increased significantly comparing with five years ago? The test statistic is: (please keep two decimal places) Present sample: Sample size: 300 Number considered themselves overweight: 150 Previous sample: Sample size: 275 Number considered themselves overweight: 121

H0: μ ≤ 0.8 (upper) Ha: μ > 0.8 1. population proportion test statistics z= (p̂-p0) /√p0(1-p0)/n (0.85-0.8)/√0.8(1-0.8)/100= 1.25 2. t value at 95% and df of 99= 1.645 1.25 < 1.645 Upper tail: reject H0 if z > za, so do not reject null. null is correct Answer= not significantly greater than 80%

A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 80%. Refer to Exhibit 9-5. At 95% confidence, it can be concluded that the proportion of the population in favor of candidate A

population proportion test statistics z= (p̂-p0) /√p0(1-p0)/n n = 100 p̂ = 85/100 p0= 80/100 (0.85-0.8)/√0.8(1-0.8)/100= 1.25 Answer= 1.25

A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 80%. Refer to Exhibit 9-5. The test statistic is

test stat. for 2 pop. means t= (x̅1- x̅2) - D0/ √s1^2/n1 + s2^2/n2 (42-38)-0/√5^2/600+6^2/700= 13.1 Answer= 13.1

A recent Time magazine reported the following information about a sample of workers in Germany and the United States. U.S.: Average length of workweek (hrs.): 42 Sample Std. dev.: 5 Sample size: 600 Germany: Average length of workweek (hrs.): 38 Sample Std. dev.: 6 Sample size: 700 We want to determine whether or not there is a significant difference between the average workweek in the United States and the average workweek in Germany. The test statistic is (please keep 1 decimal place)

alternative hypothesis: the new bonus plan increases sales null hypothesis: the new bonus plan does not increase sales

Alternative Hypothesis as a research hypothesis example: A new sales force bonus plan is developed in an attempt to increase sales.

E(x̅1- x̅2) = μ1-μ2

Expected Value

find t value with 0.1 and df of 9 --> 1.383 It's lower tail, so make it negative Answer= -1.383

For a one-tailed test (lower tail), a sample size of 10 at 90% confidence, the critical value =

find t value with 0.05 and df of 17 --> 17.40 Answer= 17.40

For a one-tailed test (upper tail), a sample size of 18 at 95% confidence, the critical value =

two-tailed

H0: p = p0 Ha: p ≠ p0

one-tailed (upper tail)

H0: p ≤ p0 Ha: p > p0

one-tailed (lower tail)

H0: p ≥ p0 Ha: p < p0

Two-tailed

H0: μ = μ0 Ha: μ ≠ μ0 For example, suppose the null hypothesis states that the mean is equal to 10. The alternative hypothesis would be that the mean is less than 10 or greater than 10.. so you need to find 2 tails

One-tailed (upper-tail)

H0: μ ≤ μ0 Ha: μ > μ0

One-tailed (lower-tail)

H0: μ ≥ μ0 Ha: μ < μ0

Region of rejection

The set of values outside the region of acceptance if test statistic falls within this region, null hypothesis is rejected

will also not be rejected at the 1% level If we can't support alternative hypothesis with 5% chance making mistake, we also cannot support alternative hypothesis with 1% b/c critical value for 1% always is bigger than the value for 5%.

If a hypothesis is not rejected at the 5% level of significance, it

it will also be rejected at 90% confidence

If a hypothesis is rejected at 95% confidence, it

alternative hypothesis should state P1 - P2 > 0

If we are interested in testing whether the proportion of items in population 1 is larger than the proportion of items in population 2, the

the alternative hypothesis is true

In hypothesis testing if the null hypothesis is rejected,

H0: μ ≤ 2 Ha: μ > 2 t= (x̅1- x̅2) - D0/ √s1^2/n1 + s2^2/n2 -1-0/ √2^2/25 + 1^2/20= -2.18 .......................................

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the following data have been gathered. Downtown store: Sample Size 25 Sample Mean $9 Sample Std. Dev. $2 North Mall Store: Sample Size 20 Sample Mean $8 Sample Std. Dev. $1 Refer to Exhibit 10-7. If we are interested in whether the average wages of employees of Downtown Store is $2 more than the average wages of employees of North Mall Store, the conclusion is:

test statistic for 2 pop. means: t= (x̅1- x̅2) - D0/ √s1^2/n1 + s2^2/n2 D0=2 9-8-2= -1 -1/ √2^2/25 + 1^2/20= -2.18 Answer= -2.18

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the following data have been gathered. Downtown store: Sample Size 25 Sample Mean $9 Sample Std. Dev. $2 North Mall Store: Sample Size 20 Sample Mean $8 Sample Std. Dev. $1 Refer to Exhibit 10-7. If we are interested in whether the average wages of employees of Downtown Store is $2 more than the average wages of employees of North Mall Store, the test statistic is: (Please keep two decimal places, and also include the negative sign, if any)

df= (s1^2/n1+ s2^2/n2)^2 / 1/n1-1(s1^2/n1)^2+1/n2-1(s2^2/n2)^2 (2^2/25 + 1^2/20)^2 = 0.0441 1/25-1(0.16)^2= 0.0010666 1/20-1(0.05)^2= 0.000131578 0.0441/0.0010666+0.000131578= 36.805 Answer= 37

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the following data have been gathered. Downtown store: Sample Size 25 Sample Mean $9 Sample Std. Dev. $2 North Mall Store: Sample Size 20 Sample Mean $8 Sample Std. Dev. $1 Refer to Exhibit 10-7. The degrees of freedom for for the t-distribution are

≥ za This is upper tail, so reject H0 if z > za

In order to test the following hypotheses at an α level of significance H0: μ ≤ 800 Ha: μ > 800 the null hypothesis will be rejected if the test statistic Z is

null hypothesis: we trust the company and label is correct. μ ≥ 67.6 fl oz alternative hypothesis: label is not correct, company needs to make the change. μ < 67.6 fl oz

Null Hypothesis as an assumption to be challenged example: label on a soft drink bottle states it contains 67.6 fl oz

p̂= n1p̂1 + n2p̂2/ n1 + n2

Pooled estimator of p

Lower tail: reject H0 if z ≤ -za Upper tail: reject H0 if z > za For two tailed: reject H0 if -z ≥ za or z ≤ za

Rejection rule:

σx̅1- x̅2 = √σ1^2/n1+ σ2^2/n2

Std. Dev. (Std. error)

H0: μ ≤ 40,000 Ha: μ > 40,000

The average life expectancy of tires produced by the Whitney Tire Company has been 40,000 miles. Management believes that due to a new production process, the life expectancy of their tires has increased. In order to test the validity of their belief, the correct set of hypotheses is

μ1-μ2

The difference between the two population means is

df= (s1^2/n1 + s2^2/n2)^2 / 1/n1-1(s1^2/n1)^2+1/n2-1(s2^2/n2)^2 top part is same as σx̅1- x̅2 from previous problem--> 16 16^2 / 1/10-1(8.5)^2 + 1/12-1(7.5)^2 256/ 0.111(72.25) + 0.0909(56.25) 256/ (8.0277 + 5.1136)= 19.4805 Answer= 19

The following information was obtained from independent random samples. Assume normally distributed populations. Sample 1: Sample Mean 45 Sample Variance 85 Sample Size 10 Sample 2: Sample Mean 42 Sample Variance 90 Sample Size 12 Refer to Exhibit 10-4. The degrees of freedom for the t-distribution are

Point estimator = x̅1- x̅2 x̅1= 45 x̅2= 42 45-42=3 Answer= 3

The following information was obtained from independent random samples. Assume normally distributed populations. Sample 1: Sample Mean 45 Sample Variance 85 Sample Size 10 Sample 2: Sample Mean 42 Sample Variance 90 Sample Size 12 Refer to Exhibit 10-4. The point estimate for the difference between the means of the two populations is

Std. error (dev.) of 2 pop. means: σx̅1- x̅2 = √σ1^2/n1+ σ2^2/n2 σ1 and σ2 = std. dev. of pop. 1 and 2 sample variance= 85 and 90 √85= 9.2195 √90= 9.4868 (but then you square these two anyway so just leave them as the variance) n1 and n2= sample size from pop. 1 and 2 σx̅1- x̅2= √85/10+90/12 Answer= 4

The following information was obtained from independent random samples. Assume normally distributed populations. Sample 1: Sample Mean 45 Sample Variance 85 Sample Size 10 Sample 2: Sample Mean 42 Sample Variance 90 Sample Size 12 Refer to Exhibit 10-4. The standard error of x̅1 -x̅2 is

test statistic for 2 pop. means: t= (x̅1- x̅2) - D0/ √s1^2/n1 + s2^2/n2 x̅1- x̅2= 45-42= 3 D0= 0 3-0/ 4 = 0.75 Answer= 0.75

The following information was obtained from independent random samples. Assume normally distributed populations. Sample 1: Sample Mean 45 Sample Variance 85 Sample Size 10 Sample 2: Sample Mean 42 Sample Variance 90 Sample Size 12 Refer to Exhibit 10-4. The test statistic is

t0.05/2= t0.025 df= 5-1= 4 find on table--> 2.776 test statistic for difference btwn 2 pop. means = -1 -1 < 2.776 (not bigger than) -1 > -2.776 (not smaller than) so alternative is incorrect Answer= should not be rejected

The following information was obtained from matched samples. Individual 1: Method 1: 7 Method 2: 5 Individual 2: Method 1: 5 Method 2: 9 Individual 3: Method 1: 6 Method 2: 8 Individual 4: Method 1: 7 Method 2: 7 Individual 5: Method 1: 5 Method 2: 6 Refer to Exhibit 10-5. If the null hypothesis is tested at the 5% level, the null hypothesis

test statistic for matched samples: t= d (bar) - μd/ sd/√n d (bar)= (mean of sample)= -1 μd=0 sd= (std. dev. of columns)= 2.236 Find std. dev. for the two columns: 7-5, 5-9, 6-8, 7-7, 5-6 2, -4, -2, 0, -1 ---> mean= -1 subtract mean from each number 2--1, -4--1, -2--1, 0--1, -1--1 3, -3, -1, 1, 0---> square results 9, 9, 1, 1, 0 --> mean of results= 20 20/n-1---> 20/5-1= 5---> √5= 2.236 -1-0/ 2.236/√5= -1 Answer= -1

The following information was obtained from matched samples. Individual 1: Method 1: 7 Method 2: 5 Individual 2: Method 1: 5 Method 2: 9 Individual 3: Method 1: 6 Method 2: 8 Individual 4: Method 1: 7 Method 2: 7 Individual 5: Method 1: 5 Method 2: 6 Refer to Exhibit 10-5. The null hypothesis tested is H0: μd = 0. The test statistic for the difference between the two population means is

point estimator= x̅1- x̅2 mean of method 1: 7+5+6+7+5= 6 mean of method 2: 5+9+8+7+6= 7 6-7= -1 Answer= -1

The following information was obtained from matched samples. Individual 1: Method 1: 7 Method 2: 5 Individual 2: Method 1: 5 Method 2: 9 Individual 3: Method 1: 6 Method 2: 8 Individual 4: Method 1: 7 Method 2: 7 Individual 5: Method 1: 5 Method 2: 6 Refer to Exhibit 10-5. The point estimate for the difference between the means of the two populations (Method 1 - Method 2) is

is (1 - confidence level)

The level of significance

rejecting a true null hypothesis

The level of significance in hypothesis testing is the probability of

H0: μ ≤ 3 (upper) Ha: μ > 3 1. sampling distribution of z = (x̅ - μ0)/ (σ/√n) n = 100 x̅ = 3.1 S = 0.5 μ0= 3 (3.1-3)/ (0.5/√100)= 2 3. t value of 95% confidence= 0.05, degrees of freedom= 99 --> t value= has to be in the 1.660 region 2 > 1.660 Upper tail: reject H0 if z > za, so Ha is correct --> μ > 3 Answer = significantly greater than 3

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. Refer to Exhibit 9-4. At 95% confidence, it can be concluded that the mean of the population is

sampling distribution of z = (x̅ - μ0)/ (σ/√n) n = 100 x̅ = 3.1 S = 0.5 μ0= 3 (3.1-3)/ (0.5/√100)= 2 Answer= 2

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. Refer to Exhibit 9-4. The test statistic is

x̅1-x̅2 these are the means of sample 1 and 2 taken from the population

The point estimator of the difference between the means of the populations 1 and 2 (μ1-μ2) is

H0: μ ≤ 8,000 (upper) Ha: μ > 8,000

The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200. Refer to Exhibit 9-9. At the 5% significant level, what is your conclusion?

μ ≤ 8000 the hypothesis we are trying to find is H1: μ > 8000 the null is what will happen if the hypothesis doesn't come true.. can either stay the same, or be less than the original amount

The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200. Refer to Exhibit 9-9. The correct null hypothesis for this problem is

1. sampling distribution of z = (x̅ - μ0)/ (σ/√n) n = 64 x̅ = 8,300 S= 1,200 μ0= 8,000 (8,300-8,000)/ (1,200/√64)= 2 Answer= 2

The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200. Refer to Exhibit 9-9. The value of the test statistic is

standard deviation of the sampling distribution of x̅1 -x̅2

The standard error of x̅1 -x̅2 is the (Hint page 395 on textbook)

test statistic= 2.256 since μd=0, it's a two-tailed test t= 0.10/2= 0.05 df= 8-1= 7 t value= 1.895 and -1.895 2.256 > 1.895 and 2.256 > -1.895 2.256 exceeds 1.895 on the bell curve to the right, so we can support alternative and reject null Answer= should be rejected

Two major automobile manufacturers have produced compact cars with the same size engines. We are interested in determining whether or not there is a significant difference in the MPG (miles per gallon) of the two brands of automobiles. A random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance (matched samples). The following data show the results of the test. Driver 1: Manufacturer A: 32 Manufacturer B: 28 Driver 2: Manufacturer A: 27 Manufacturer B: 22 Driver 3: Manufacturer A: 26 Manufacturer B: 27 Driver 4: Manufacturer A: 26 Manufacturer B: 24 Driver 5: Manufacturer A: 25 Manufacturer B: 24 Driver 6: Manufacturer A: 29 Manufacturer B: 25 Driver 7: Manufacturer A: 31 Manufacturer B: 28 Driver 8: Manufacturer A: 25 Manufacturer B: 27 Refer to Exhibit 10-9. At 90% confidence the null hypothesis (please feel free to use excel)

32-28, 27-22, 26-27, 26-24, 25-24, 29-25, 31-28, 25-27 4, 5, -1, 2, 1, 4, 3, -2= 16/8= 2 Answer= 2

Two major automobile manufacturers have produced compact cars with the same size engines. We are interested in determining whether or not there is a significant difference in the MPG (miles per gallon) of the two brands of automobiles. A random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance (matched samples). The following data show the results of the test. Driver 1: Manufacturer A: 32 Manufacturer B: 28 Driver 2: Manufacturer A: 27 Manufacturer B: 22 Driver 3: Manufacturer A: 26 Manufacturer B: 27 Driver 4: Manufacturer A: 26 Manufacturer B: 24 Driver 5: Manufacturer A: 25 Manufacturer B: 24 Driver 6: Manufacturer A: 29 Manufacturer B: 25 Driver 7: Manufacturer A: 31 Manufacturer B: 28 Driver 8: Manufacturer A: 25 Manufacturer B: 27 Refer to Exhibit 10-9. The mean for the differences is (please feel free to use excel)

test statistic for matched samples: t= d (bar) - μd/ sd/√n d (bar)= (mean of sample)= 2 μd=0 sd= (std. dev. of columns)= 2.5071 4, 5, -1, 2, 1, 4, 3, -2 subtract mean from each number 4-2, 5-2, -1-2, 2-2, 1-2, 4-2, 3-2, -2-2 2, 3, -3, 0, -1, 2, 1, -4 ---> square results 4, 9, 9, 0, 1, 4, 1, 16--> mean of results= 44 44/n-1---> 44/8-1= ---> √6.2857 = 2.5071 2-0/ 2.507/√8= 2.256 Answer= 2.256

Two major automobile manufacturers have produced compact cars with the same size engines. We are interested in determining whether or not there is a significant difference in the MPG (miles per gallon) of the two brands of automobiles. A random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance (matched samples). The following data show the results of the test. Driver 1: Manufacturer A: 32 Manufacturer B: 28 Driver 2: Manufacturer A: 27 Manufacturer B: 22 Driver 3: Manufacturer A: 26 Manufacturer B: 27 Driver 4: Manufacturer A: 26 Manufacturer B: 24 Driver 5: Manufacturer A: 25 Manufacturer B: 24 Driver 6: Manufacturer A: 29 Manufacturer B: 25 Driver 7: Manufacturer A: 31 Manufacturer B: 28 Driver 8: Manufacturer A: 25 Manufacturer B: 27 Refer to Exhibit 10-9. The test statistic is (please feel free to use excel)

At 5% significant level, we cannot reject the null hypothesis because at 5% significance t=0.05. Value is 1.684 test statistic 2 > 1.684 **********************

We are interested in testing the following hypotheses. H0: μ1- μ2 ≥ 0 Ha: μ1- μ2 < 0 Based on 40 degrees of freedom, the test statistic t is computed to be 2. The conclusion is (Hint: the sign of the test statistic)

reject H0 (conclude μ >12)

When H0 is false (μ >12)

Accept H0 (conclude μ ≤ 12)

When H0 is true (μ ≤ 12)

use the sample std. devs s1 and s2 as estimates of σ1 and σ2 use tα/2 instead of zα/2.

When σ1 and σ2 are unknown, we will

Region of acceptance defined so that the change of making a type 1 error is equal to the significance level

a range of values If the test statistic falls within this, the null hypothesis is not rejected

significance tests

applications of hypothesis testing that only control the Type I error are often called

Hypothesis testing

can be used to determine whether a statement about the value of a population parameter should or should not be rejected

Type I error

is rejecting H0 when it is true

(Upper tail) 1. sampling distribution of z = (x̅ - μ0)/ (σ/√n) (24.6-20)/ (12/√36)= 2.3 2. t value of 95% and df of 35= 1.69 2.3 > 1.69 Upper tail: reject H0 if z > za (t value) Answer= be rejected

n = 36 x̅ = 24.6 S = 12 H0: μ ≤ 20 Ha: μ > 20 Refer to Exhibit 9-1 (hint: S is the standard deviation of the sample). If the test is done at 95% confidence, the null hypothesis should

sampling distribution of z = (x̅ - μ0)/ (σ/√n) (24.6-20)/ (12/√36)= 2.3 Answer= 2.3

n = 36 x̅ = 24.6 S = 12 H0: μ ≤ 20 Ha: μ > 20 Refer to Exhibit 9-1(hint: S is the standard deviation of the sample). The test statistic is

(Lower tail) 1. sampling distribution of z = (x̅ - μ0)/ (σ/√n) (50-54)/ (16/√64)= -2 2. 95% confidence, so t value is 0.05 with degrees of freedom of 63 --> find on table --> 1.669 -2< -1.669 Lower tail: reject H0 if z ≤ -za (t value) Answer= be rejected

n = 64 x̅ = 50 S = 16 H0: μ ≥ 54 Ha: μ < 54 Refer to Exhibit 9-2 (hint: S is the standard deviation of the sample). If the test is done at 95% confidence, the null hypothesis should

sampling distribution of z = (x̅ - μ0)/ (σ/√n) (50-54)/ (16/√64)= -2 Answer= -2

n = 64 x̅ = 50 S = 16 H0: μ ≥ 54 Ha: μ < 54 Refer to Exhibit 9-2(hint: S is the standard deviation of the sample). The test statistic equals

level of significance (α)

probability of making a Type I error when the null hypothesis is true as an equality is called the

n1p1≥5 n2p2≥5 n1(1-p1)≥5 n2(1-p2)≥5

sample sizes are sufficiently large if all of these conditions are met

z= (p̂1-p̂2)/ √p̂(1-p̂)(1/n1+1/n2)

test statistic for p1-p2

critical value

the value of the test statistic that established the boundary of the rejection regions called the

matched-sample (aka dependent sample) design this design often leads to a smaller sampling error than independent-sample design

with a ________ each sampled item provides a pair of data values

H0: μ1-μ2 = D0 H0: μ1-μ2 ≠ D0

μ1-μ2: σ1 and σ2 unknown--> Two-tailed

H0: μ1-μ2 ≥ D0 H0: μ1-μ2 < D0

μ1-μ2: σ1 and σ2 unknown--> left-tailed

H0: μ1-μ2 ≤ D0 H0: μ1-μ2 > D0

μ1-μ2: σ1 and σ2 unknown--> right-tailed