Stats Final

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)

At 5% significant level, we cannot reject the null hypothesis

Exhibit 10-7 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 North Mall Sample size 25 20 Sample mean $9 $8 Sample standard deviation $2 $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:

At 5% significant level, we cannot reject the null hypothesis Feedback The alternative hypothesis is: m1-m2>2, wage 1 is $2 more than wage 2. (9-8-2)/sqrt(2*2/25+1*1/20)=-2.18. The critical value is t0.05 with df37=1.687. Since -2.18 isn't bigger than 1.687, we cannot support alternative, we cannot reject the null.

The level of significance

is (1 - confidence level)

Exhibit 9-2 n = 64 = 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

(50-54)/(16/sqrt(64))=-2.

Exhibit 9-9 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

(8300-8000)/(1200/sqrt(64))=2

Exhibit 10-5 The following information was obtained from matched samples. Individual Method 1 Method 2 1 7 5 2 5 9 3 6 8 4 7 7 5 5 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

-1

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

-1

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

-1.383 check t table, t10% is 1.383 with df 9. Since it's lower tail, ie. alternative is x<#, the critical value is -t.10

Exhibit 10-7 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 North Mall Sample size 25 20 Sample mean $9 $8 Sample standard deviation $2 $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)

-2.18

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

0.75 (45-42-0)/sqrt((85/10)+(90/12))=0.75. Hint, here 85 is the variance.

Exhibit 9-5 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

1.25

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

1.328

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

1.383

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

1.44

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

1.740

A recent Time magazine reported the following information about a sample of workers in Germany and the United States. United States Germany Average length of workweek (hours) 42 38 Sample Standard Deviation 5 6 Sample Size 600 700 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)

13.1 (42-38-0)/sqrt((5*5/600+6*6/700))=13.1

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

19 Please use the long equation on handout page 2.

Exhibit 10-9 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 Manufacturer A Manufacturer B 1 32 28 2 27 22 3 26 27 4 26 24 5 25 24 6 29 25 7 31 28 8 25 27 Refer to Exhibit 10-9. The mean for the differences is (please feel free to use excel)

2.0

Exhibit 9-4 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

2.00

Exhibit 10-9 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 Manufacturer A Manufacturer B 1 32 28 2 27 22 3 26 27 4 26 24 5 25 24 6 29 25 7 31 28 8 25 27 Refer to Exhibit 10-9. The test statistic is (please feel free to use excel)

2.256

Exhibit 9-1 n = 36 = 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

2.3 (24.6-20)/(12/sqrt(36))=2.27

We are interested in testing the following hypotheses. H0: P1- P2 ≤ 0 Ha: P1- P2 > 0 The test statistic Z is computed to be 1.96. The conclusion is

At 5% significant level, we can reject the null hypothesis

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

3 The point estimate is the difference between the sample mean.

Exhibit 10-7 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 North Mall Sample size 25 20 Sample mean $9 $8 Sample standard deviation $2 $1 Refer to Exhibit 10-7. The degrees of freedom for for the t-distribution are

37

Exhibit 10-4 The following information was obtained from independent random samples. Assume normally distributed populations. Sample Mean Sample 1: 45 Sample 2: 42 Sample Variance Sample 1: 85 Sample 2: 90 Sample Size Sample 1: 10 Sample 2: 12 Refer to Exhibit 10-4. The standard error of x-bar1 — x-bar2 is

4.0 The standard error of ? is the standard deviation of the sample distribution of ? , which is the denominator of the equation.

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

At 10% significant level, we can reject the null hypothesis The sample proportion is : 150/300=0.5 for p1 and 121/275=0.44 for p2. The pooled proportion is (150+121)/(300+275)=0.47. Test statistic: (0.5-0.44)/sqrt(0.47*(1-0.47)*(1/300+1/275))=1.440 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. z0.05=1.645, since 1.440 isn't bigger than 1.645, we cannot reject the null at 5% level.

We are interested in testing the following hypotheses. H0: P1- P2 = 0 Ha: P1- P2 ≠ 0 The test statistic Z is computed to be -2.0. The conclusion is

At 5% significant level, we can reject the null hypothesis

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

H0: μ = 5.8 Ha: μ ≠ 5.8

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

H0: ≤ 40,000 Ha: μ > 40,000

Exhibit 9-9 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?

Reject the null hypothesis.

A Type II error is committed when

a true alternative hypothesis is mistakenly rejected

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

alternative hypothesis should state P1 - P2 > 0

Exhibit 9-1 n = 36 = 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

be rejected

Exhibit 9-2 n = 64 = 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

be rejected (50-54)/(16/sqrt(64))=-2. t0.05with df 63=1.669. -2<-1.669, we support alternative and reject null.

Exhibit 9-5 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

is not significantly greater than 80% (0.85-0.8)/sqrt(0.8*(1-0.8)/100)=1.25, which is smaller than z0.05=1.645, we cannot choose alternative, we cannot reject null.

The level of significance in hypothesis testing is the probability of

rejecting a true null hypothesis

Exhibit 10-9 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 Manufacturer A Manufacturer B 1 32 28 2 27 22 3 26 27 4 26 24 5 25 24 6 29 25 7 31 28 8 25 27 Refer to Exhibit 10-9. At 90% confidence the null hypothesis (please feel free to use excel)

should be rejected Using excel to build a new variable: 32-28=4 5 -1 2 1 4 3 -2 the average is 2, and sample standard deviation is 2.507. Test statistic is (2-0)/(2.507/sqrt(8))=2.256, which is bigger than t0.05 with df 7=1.895. So we can support the alternative and reject the null.

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

should not be rejected We need to build a new variable: difference between Method 1 and 2. which is: 7-5=2 -4 -2 0 -1 The sample average is -1, the sample standard deviation is 2.236. The test statistic (-1-0)/(2.236/sqrt(5))=-1.000 t0.025 with df=5-1=4 is 2.766. Since -1 isn't bigger than 2.766 nor smaller than -2.766, we can not support the alternative, we cannot reject the null.

Exhibit 9-4 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

significantly greater than 3 (3.1-3)/(0.5/sqrt(100))=2, which is bigger than 1.660, t0.05 with df=99. So it's significantly bigger than 3.

The standard error of x-bar1 - x-bar2 is the (Hint page 395 on textbook)

standard deviation of the sampling distribution of x-bar1 - x-bar2

In hypothesis testing if the null hypothesis is rejected,

the alternative hypothesis is true

Exhibit 9-9 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. At the 5% significance level, the critical value is (Specify to 3 decimal places): At the 10% significance level, the critical value is (Specify to 3 decimal places):

using z value since this is population standard deviation known case. It's one tailed test, z0.05=1.645, z 0.10=1.282

If a hypothesis is rejected at 95% confidence, it

will also be rejected at 90% confidence

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

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

Exhibit 9-9 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

μ ≤ 8000

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

≥Zα