STATISTICSsite.iugaza.edu.ps/mriffi/files/2018/02/ch11.pdf · Copyright © 2017, 2013, 2010 Pearson...

STATISTICSINFORMED DECISIONS USING DATAFifth Edition

Chapter 11

Inferences on Two Samples

Copyright © 2017, 2013, 2010 Pearson Education, Inc. All Rights Reserved


11.1 Inference about Two Population ProportionsLearning Objectives

1. Distinguish between independent and dependent sampling

2. Test hypotheses regarding two proportions from independent samples

3. Construct and interpret confidence intervals for the difference between two population proportions

4. Determine the sample size necessary for estimating the difference between two population proportions


11.1 Inference about Two Population Proportions11.1.1 Distinguish between Independent and Dependent Sampling (1 of 4)

A sampling method is independent when the individuals selected for one sample do not dictate which individuals are to be in a second sample. A sampling method is dependentwhen the individuals selected to be in one sample are used to determine the individuals to be in the second sample.



Dependent samples are often referred to as matched-pairssamples. It is possible for an individual to be matched against him- or herself.



Parallel Example 1: Distinguish between Independent and Dependent Sampling

For each of the following, determine whether the sampling method is independent or dependent.

A researcher wants to know whether the price of a one night stay at a Holiday Inn Express is less than the price of a one night stay at a Red Roof Inn. She randomly selects 8 towns where the location of the hotels is close to each other and determines the price of a one night stay.

A researcher wants to know whether the “state” quarters (introduced in 1999) have a mean weight that is different from “traditional” quarters. He randomly selects 18 “state” quarters and 16 “traditional” quarters and compares their weights.



Solution

The sampling method is dependent since the 8 Holiday Inn Express hotels can be matched with one of the 8 Red Roof Inn hotels by town.

The sampling method is independent since the “state” quarters which were sampled had no bearing on which “traditional”quarters were sampled.


11.1 Inference about Two Population Proportions11.1.2 Test Hypotheses Regarding Two Population Proportions from Independent Samples (1 of 30)

Sampling Distribution of the Difference between Two Proportions (Independent Sample)



Sampling Distribution of the Difference between Two Proportions (Independent Sample)



Sampling Distribution of the Difference between Two Proportions



Hypothesis Test Regarding the Difference between Two Population Proportions

To test hypotheses regarding two population proportions, p1 and p2, we can use the steps that follow, provided that:

– the samples are independently obtained using simple random sampling,



Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:

Two-Tailed Left-Tailed Right-Tailed

H0: p1 = p2 H0: p1 = p2 H0: p1 = p2

H1: p1 ≠ p2 H1: p1 < p2 H1: p1 > p2

Note: p1 is the population proportion for population 1, and p2 is the population proportion for population 2.



Step 2: Select a level of significance, α, depending on the seriousness of making a Type I error.



Classical Approach

Step 3: Compute the test statistic



Classical Approach

Use Table V to determine the critical value.



Classical Approach

Two-Tailed



Classical Approach

Left-Tailed



Classical Approach

Right-Tailed



P-value Approach

By Hand Step 3: Compute the test statistic



P-value Approach

Use Table V to determine the P-Value.



P-value Approach

Two-Tailed



P-value Approach

Left-Tailed



P-value Approach

Right-Tailed



P-value Approach

Technology Step 3: Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value. The directions for obtaining the P-value using the TI-83/84 Plus graphing calculator, Excel, MINITAB, and StatCrunch are in the Technology Step-by-Step in the text.



Classical Approach

Step 4: Compare the critical value with the test statistic:



P-value Approach

Step 4: If P-value < α, reject the null hypothesis.



P-value Approach

Step 5: State the conclusion.



Parallel Example 1: Testing Hypotheses Regarding Two Population Proportions

An economist believes that the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access. He obtains a random sample of 800 urban households and finds that 338 of them have Internet access. He obtains a random sample of 750 rural households and finds that 292 of them have Internet access. Test the economist’s claim at the α = 0.05 level of significance.



Solution

We must first verify that the requirements are satisfied:

The samples are simple random samples that were obtained independently.



Solution

Step 1: We want to determine whether the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access. So,

H0: p1 = p2 versus H1: p1 > p2

or, equivalently,

H0: p1 − p2 = 0 versus H1: p1 − p2 > 0

Step 2: The level of significance is α = 0.05.



Solution



Solution: Classical Approach

This is a right-tailed test with α = 0.05.The critical value is z0.05 = 1.645.



Step 4: Since the test statistic, z0 = 1.33 is less than the critical value z.05 = 1.645, we fail to reject the null hypothesis.



Solution: P-value Approach

Because this is a right-tailed test, the P-value is the area under the normal to the right of the test statistic z0 = 1.33.That is, P-value = P(Z > 1.33) ≈ 0.09.




Step 4: Since the P-value is greater than the level of significance α = 0.05, we fail to reject the null hypothesis.



Solution

Step 5: There is insufficient evidence at the α = 0.05 level to conclude that the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access.


11.1 Inference about Two Population Proportions11.1.3 Construct and Interpret Confidence Intervals for the

Difference between Two Population Proportions (1 of 6)

Constructing a (1 − α) • 100% Confidence Interval for the Difference between Two Population Proportions

To construct a (1 − α) • 100% confidence interval for the difference between two population proportions, the following requirements must be satisfied:

1. the samples are obtained independently using simple random sampling,




Constructing a (1 − α) • 100% Confidence Interval for the Difference between Two Population Proportions

Provided that these requirements are met, a (1 − α) • 100% confidence interval for p1 − p2 is given by




Parallel Example 3: Constructing a Confidence Interval for the Difference between Two Population Proportions

An economist obtains a random sample of 800 urban households and finds that 338 of them have Internet access. He obtains a random sample of 750 rural households and finds that 292 of them have Internet access. Find a 99% confidence interval for the difference between the proportion of urban households that have Internet access and the proportion of rural households that have Internet access.




Solution

We have already verified the requirements for constructing a confidence interval for the difference between two population proportions in the previous example.




Solution

Thus,




Solution

We are 99% confident that the difference between the proportion of urban households that have Internet access and the proportion of rural households that have Internet access is between −0.03 and 0.10. Since the confidence interval contains 0, we are unable to conclude that the proportion of urban households with Internet access is greater than the proportion of rural households with Internet access.


11.1 Inference about Two Population Proportions11.1.4 Determine the Sample Size Necessary for Estimating the Difference between Two Population Proportions (1 of 4)

Sample Size for Estimating p1 − p2

The sample size required to obtain a (1 − α)•100% confidence interval with a margin of error, E, is given by



Parallel Example 5: Determining Sample Size

A doctor wants to estimate the difference in the proportion of 15-19 year old mothers that received prenatal care and the proportion of 30-34 year old mothers that received prenatal care. What sample size should be obtained if she wished the estimate to be within 2 percentage points with 95% confidence assuming:

A. she uses the results of the National Vital Statistics Report results in which 98% of the 15-19 year old mothers received prenatal care and 99.2% of 30-34 year old mothers received prenatal care.

B. she does not use any prior estimates.



The doctor must sample 265 randomly selected 15-19 year old mothers and 265 randomly selected 30-34 year old mothers.



The doctor must sample 4802 randomly selected 15-19 year old mothers and 4802 randomly selected 30-34 year old mothers. Note that having prior estimates of p1 and p2 reduces the number of mothers that need to be surveyed.


11.2 Inference about Two Means: Dependent SamplesLearning Objectives

1. Test hypotheses for a population mean from matched-pairs data

2. Construct and interpret confidence intervals about the population mean difference of matched-pairs data


11.2 Inference about Two Means: Dependent Samples11.2.1 Test Hypotheses for a Population Mean from Matched-Pairs Data (1 of 30)

“In Other Words”

Statistical inference methods on matched-pairs data use the same methods as inference on a single population mean, except that the differences are analyzed.



Testing Hypotheses Regarding the Difference of Two Means Using a Matched-Pairs Design

To test hypotheses regarding the mean difference of data obtained form a dependent sample (matched-pairs data), use the following steps. provided that

• the sample is obtained using simple random sampling,

• the sample data are matched pairs,

• the differences are normally distributed with no outliers or the sample size, n, is large (n ≥ 30),

• the sampled values are independent



Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways, where μd is the population mean difference of the matched-pairs data.


H0: µd = 0 H0: µd = 0 H0: µd = 0

H1: µd ≠ 0 H1: µd < 0 H1: µd > 0



Classical Approach




Classical Approach

Use Table VII to determine the critical value using n − 1 degrees of freedom.



Classical Approach

Two-Tailed



Classical Approach

Left-Tailed



Classical Approach

Right-Tailed



Classical Approach




P-value Approach




P-value Approach

Use Table VII to approximate the P-value using n − 1 degrees of freedom.



P-value Approach

Two-Tailed



P-value Approach

Left-Tailed



P-value Approach

Right-Tailed



P-value Approach

Technology Step 3: Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value. The directions for obtaining the P-value using the TI-83/84 Plus graphing calculator, MINITAB, Excel and StatCrunch, are in the Technology Step-by-Step in the text.



P-value Approach




These procedures are robust, which means that minor departures from normality will not adversely affect the results. However, if the data have outliers, the procedure should not be used.



Parallel Example 2: Testing a Claim Regarding Matched-Pairs Data

The following data represent the cost of a one night stay in Hampton Inn Hotels and La Quinta Inn Hotels for a random sample of 10 cities. Test the claim that Hampton Inn Hotels are priced differently than La Quinta Hotels at the α = 0.05 level of significance.



City Hampton Inn La Quinta

Dallas 129 105

Tampa Bay 149 96

St. Louis 149 49

Seattle 189 149

San Diego 109 119

Chicago 160 89

New Orleans 149 72

Phoenix 129 59

Atlanta 129 90

Orlando 119 69



Solution

This is a matched-pairs design since the hotel prices come from the same ten cities. To test the hypothesis, we first compute the differences and then verify that the differences come from a population that is approximately normally distributed with no outliers because the sample size is small.



Solution



Solution

Step 1: We want to determine if the prices differ:

H0: μd = 0 versus H1: μd ≠ 0


Step 3: The test statistic is




This is a two-tailed test so the critical values at the α = 0.05 level of significance with n − 1 = 10 − 1 = 9 degrees of freedom are −t0.025 = −2.262 and t0.025 = 2.262.




Step 4: Since the test statistic, t0 = 5.27 is greater than the critical value t.025 = 2.262, we reject the null hypothesis.




Because this is a two-tailed test, the P-value is two times the area under the t-distribution with n − 1 = 10 − 1 = 9 degrees of freedom to the right of the test statistic t0 = 5.27.

That is, P-value = 2P(t > 5.27) ≈ 2(0.00026) = 0.00052 (using technology). Approximately 5 samples in 10,000 will yield results at least as extreme as we obtained if the null hypothesis is true.




Step 4: Since the P-value is less than the level of significance α = 0.05, we reject the null hypothesis.



Solution

Step 5: There is sufficient evidence to conclude that Hampton Inn hotels and La Quinta hotels are priced differently at the α = 0.05 level of significance.


11.2 Inference about Two Means: Dependent Samples11.2.2 Construct and Interpret Confidence Intervals for the Population Mean Difference of Matched-Pairs Data (1 of 5)

Confidence Interval for Matched-Pairs Data

A (1 − α) • 100% confidence interval for μd is given by



Confidence Interval for Matched-Pairs Data

Note: The interval is exact when the population is normally distributed and approximately correct for nonnormal populations, provided that n is large.



Parallel Example 4: Constructing a Confidence Interval for Matched-Pairs Data

Construct a 90% confidence interval for the mean difference in price of Hampton Inn versus La Quinta hotel rooms.



Solution

We have already verified that the differenced data come from a population that is approximately normal with no outliers.



We are 90% confident that the mean difference in hotel room price for Hampton Inn versus La Quinta Inn is between $33.53 and $69.27.


11.3 Inference about Two Means: Independent SamplesLearning Objectives

1. Test hypotheses regarding the difference of two independent means

2. Construct and interpret confidence intervals regarding the difference of two independent means


11.3 Inference about Two Means: Independent SamplesIntroduction (1 of 2)

Sampling Distribution of the Difference of Two Means: Independent Samples with Population

Standard Deviations Unknown (Welch’s t)

Suppose that a simple random sample of size n1 is taken from a population with unknown mean μ1 and unknown standard deviation σ1. In addition, a simple random sample of size n2 is taken from a population with unknown mean μ2 and unknown standard deviation σ2. If the two populations are normally distributed or the sample sizes are sufficiently large (n1 ≥ 30, n2 ≥ 30), then


11.3 Inference about Two Means: Independent SamplesIntroduction (2 of 2)

Sampling Distribution of the Difference of Two Means: Independent Samples with Population

Standard Deviations Unknown (Welch’s t)


11.3 Inference about Two Means: Independent Samples11.3.1 Test Hypotheses Regarding the Difference of Two Independent Means (1 of 29)

Testing Hypotheses Regarding the Difference of Two Means

To test hypotheses regarding two population means, μ1 and μ2, with unknown population standard deviations, we can use the following steps, provided that:

– the samples are obtained using simple random sampling;

– the samples are independent;

– the populations from which the samples are drawn are normally distributed or the sample sizes are large (n1 ≥ 30, n2 ≥ 30);

– for each sample, the sample size is no more than 5% of the population size.



Step 1: Determine the null and alternative hypotheses. The hypotheses are structured in one of three ways:


H0: µ1 = µ2 H0: µ1 = µ2 H0: µ1 = µ2

H1: µ1 ≠ µ2 H1: µ1 < µ2 H1: µ1 > µ2

Note: µ1 is the population mean for population 1, and µ2 is the population mean for population 2.



Classical Approach




Classical Approach

Use Table VII to determine the critical value using the smaller of n1 − 1 or n2 − 1 degrees of freedom.



Classical Approach

Two-Tailed



Classical Approach

Left-Tailed



Classical Approach

Right-Tailed



Classical Approach




P-value Approach




P-value Approach

Use Table VII to determine the P-value using the smaller of n1 − 1 or n2 − 1 degrees of freedom.



P-value Approach

Two-Tailed



P-value Approach

Left-Tailed



P-value Approach

Right-Tailed



P-value Approach

Technology Step 3: Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value. The directions for obtaining the P-value using the TI-83/84 Plus graphing calculator, Excel, MINITAB, and StatCrunch are in the Technology Step-by-Step in the text.



P-value Approach




These procedures are robust, which means that minor departures from normality will not adversely affect the results. However, if the data have outliers, the procedure should not be used.



Parallel Example 1: Testing Hypotheses Regarding Two Means

A researcher wanted to know whether “state” quarters had a weight that is more than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the following data.



STATE(grams)

blank TRADITIONAL(grams)

blank

5.70 5.67 5.67 5.55

5.73 5.61 5.70 5.61

5.70 5.67 5.72 5.58

5.65 5.62 5.66 5.74

5.73 5.65 5.70 5.68

5.79 5.73 5.68 5.53

5.77 5.71 5.67 5.55

5.70 5.76 5.61 5.74

5.73 5.72 blank blank



Test the claim that “state” quarters have a mean weight that is more than “traditional” quarters at the α = 0.05 level of significance.

NOTE: A normal probability plot of “state” quarters indicates the population could be normal. A normal probability plot of “traditional” quarters indicates the population could be normal



No outliers.



Solution

Step 1: We want to determine whether state quarters weigh more than traditional quarters:

H0: μ1 = μ2 versus H1: μ1 > μ2


Step 3: The test statistic is




This is a right-tailed test with α = 0.05. Since n1 − 1 = 17 and n2 − 1 = 15, we will use 15 degrees of freedom. The corresponding critical value is t0.05 = 1.753.




Step 4: Since the test statistic, t0 = 2.53 is greater than the critical value t.05 = 1.753, we reject the null hypothesis.




Because this is a right-tailed test, the P-value is the area under the t-distribution to the right of the test statistict0 = 2.53. That is, P-value = P(t > 2.53) ≈ 0.01.




Step 4: Since the P-value is less than the level of significance α = 0.05, we reject the null hypothesis.



Solution

Step 5: There is sufficient evidence at the α = 0.05 level to conclude that the state quarters weigh more than the traditional quarters.



NOTE: The degrees of freedom used to determine the critical value in the last example are conservative. Results that are more accurate can be obtained by using the following degrees of freedom:


11.3 Inference about Two Means: Independent Samples11.3.2 Construct and Interpret Confidence Intervals Regarding the Difference of Two Independent Means (1 of 8)

Constructing a (1 − α) • 100% Confidence Interval for the Difference of Two Means

A simple random sample of size n1 is taken from a population with unknown mean μ1 and unknown standard deviation σ1. Also, a simple random sample of size n2 is taken from a population with unknown mean μ2 and unknown standard deviation σ2. If the two populations are normally distributed or the sample sizes are sufficiently large (n1 ≥ 30 and n2 ≥ 30), a(1 − α) • 100% confidence interval about μ1 − μ2 is given by . . .



Constructing a (1 − α) • 100% Confidence Interval for the Difference of Two Means



Parallel Example 3: Constructing a Confidence Interval for the Difference of Two Means

Construct a 95% confidence interval about the difference between the population mean weight of a “state” quarter versus the population mean weight of a “traditional” quarter.



Solution

We have already verified that the populations are approximately normal and that there are no outliers.



Solution

Thus,



Solution

We are 95% confident that the mean weight of the “state” quarters is between 0.0086 and 0.0974 ounces more than the mean weight of the “traditional” quarters. Since the confidence interval does not contain 0, we conclude that the “state” quarters weigh more than the “traditional” quarters.



When the population variances are assumed to be equal, the pooled t-statistic can be used to test for a difference in means for two independent samples. The pooled t-statistic is computed by finding a weighted average of the sample variances and uses this average in the computation of the test statistic.

The advantage of this test statistic is that it exactly follows Student’s t-distribution with n1 + n2 − 2 degrees of freedom.

The disadvantage to this test statistic is that it requires that the population variances be equal.



CAUTION!

We would use the pooled two-sample t-test when the two samples come from populations that have the same variance. Pooling refers to finding a weighted average of the two sample variances from the independent samples. It is difficult to verify that two population variances might be equal based on sample data, so we will always use Welch’s t when comparing two means.


11.4 Inference about Two Population Standard DeviationsLearning Objectives

1. Find critical values of the F-distribution

2. Test hypotheses regarding two population standard deviations


11.4 Inference about Two Population Standard Deviations11.4.1 Find Critical Values of the F-distribution (1 of 11)

Requirements for Testing Claims Regarding Two Population Standard Deviations

1. The samples are independent simple random samples or from a completely randomized design.

2. The populations from which the samples are drawn are normally distributed.



CAUTION!

If the populations from which the samples are drawn are not normal, do not use the inferential procedures discussed in this section.



Notation Used When Comparing Two Population Standard Deviations



follows the F-distribution with n1 − 1 degrees of freedom in the numerator and n2 − 1 degrees of freedom in the denominator.



Characteristics of the F-distribution

1. It is not symmetric. The F-distribution is skewed right.

2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the χ2 distribution and Student’s t-distribution, whose shape depends upon their degrees of freedom.

3. The total area under the curve is 1.

4. The values of F are always greater than or equal to zero.



is the critical F with n1 − 1 degrees of freedom in the numerator and n2 − 1 degrees of freedom in the denominator and an area of α to the right of the critical F.



To determine the critical value that has an area of α to the left, use the following property:



Parallel Example 1: Finding Critical Values for the F-distribution

Find the critical F-value:

a) for a right-tailed test with α = 0.1, degrees of freedom in the numerator = 8 and degrees of freedom in the denominator = 4.

b) for a two-tailed test with α = 0.05, degrees of freedom in the numerator = 20 and degrees of freedom in the denominator = 15.



Solution

a) F0.1, 8, 4 = 3.95

b) F.025, 20, 15 = 2.76



NOTE:

If the number of degrees of freedom is not found in the table, we follow the practice of choosing the degrees of freedom closest to that desired. If the degrees of freedom is exactly between two values, find the mean of the values.


11.4 Inference about Two Population Standard Deviations11.4.2 Test Hypotheses Regarding Two Population Standard Deviations (1 of 22)

Test Hypotheses Regarding Two Population Standard Deviations

To test hypotheses regarding two population standard deviations, σ1 and σ2, we can use the following steps, provided that

1. the samples are obtained using simple random sampling.

2. the sample data are independent.

3. the populations from which the samples are drawn are normally distributed.



Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:


H0: σ1 = σ2 H0: σ1 = σ2 H0: σ1 = σ2

H1: σ1 ≠ σ2 H1: σ1 < σ2 H1: σ1 > σ2



which follows Fisher’s F-distribution with n1 − 1 degrees of freedom in the numerator and n2 − 1 degrees of freedom in the denominator.



Classical Approach

Use Table IX to determine the critical value(s) using n1 − 1 degrees of freedom in the numerator and n2 − 1 degrees of freedom in the denominator. The shaded regions in the graphs on the following slides represent the critical regions.



Classical Approach

Two-Tailed



Classical Approach

Left-Tailed



Classical Approach

Right-Tailed



Classical Approach




P-value Approach

Step 3: Use technology to determine the P-value.



P-value Approach




CAUTION!

The test for equality of population standard deviations is not robust. Thus, any departures from normality make the results of the inference useless.



Parallel Example 2: Testing Hypotheses Regarding Two Population Standard Deviations

A researcher wanted to know whether “state” quarters had a standard deviation weight that is less than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional”quarters, weighed each of them and obtained the data on the next slide. A normal probability plot indicates that the sample data could come from a population that is normal. Test the researcher’s claim at the α = 0.05 level of significance.



STATE(grams)

blank TRADITIONAL(grams)

blank

5.70 5.67 5.67 5.55

5.73 5.61 5.70 5.61

5.70 5.67 5.72 5.58

5.65 5.62 5.66 5.74

5.73 5.65 5.70 5.68

5.79 5.73 5.68 5.53

5.77 5.71 5.67 5.55

5.70 5.76 5.61 5.74

5.73 5.72 blank blank



Solution

Step 1: The researcher wants to know if “state” quarters have a standard deviation weight that is less than “traditional”quarters. Thus

H0: σ1 = σ2 versus H1: σ1 < σ2

This is a left-tailed test.




Solution

Step 3: The standard deviation of “state” quarters was found to be 0.0497 and the standard deviation of “traditional” quarters was found to be 0.0689. The test statistic is then




Since this is a left-tailed test, we determine the critical value at the 1 − α = 1 − 0.05 = 0.95 level of significance with n1 − 1 = 18 − 1 = 17 degrees of freedom in the numerator and n2 − 1 = 16 − 1 = 15 degrees of freedom in the denominator. Thus,




Step 4: Since the test statistic F0 = 0.52 is greater than the critical value F0.95,17,15 = 0.42, we fail to reject the null hypothesis.




Step 3: Using technology, we find that the P-value is 0.097. If the statement in the null hypothesis were true, we would expect to get results at least as extreme as obtained about 10 out of 100 times. This is not very unusual.




Step 4: Since the P-value is greater than the level of significance, α = 0.05, we fail to reject the null hypothesis.



Solution

Step 5: There is not enough evidence to conclude that the standard deviation of weight is less for “state” quarters than it is for “traditional” quarters at the α = 0.05 level of significance.


11.5 Putting It Together: Which Method Do I Use?Learning Objectives

1. Determine the appropriate hypothesis test to perform


11.5 Putting It Together: Which Method Do I Use?11.5.1 Determine the Appropriate Hypothesis Test to

Perform (1 of 9)

What parameter is addressed in the hypothesis?

– Proportion, p

– σ or σ2

– Mean, μ



Perform (2 of 9)

Proportion, p

Is the sampling Dependent or Independent?



Perform (3 of 9)

Proportion, p

Dependent samples:

Provided the samples are obtained randomly and the total number of observations where the outcomes differ is at least 10, use the normal distribution with



Perform (4 of 9)

Proportion, p

Independent samples:



Perform (5 of 9)

σ or σ2

Provided the data are normally distributed, use the F-distribution with



Perform (6 of 9)

Mean, μ

Is the sampling Dependent or Independent?



Perform (7 of 9)

Mean, μ

Independent samples:

Provided each sample size is greater than 30 or each population is normally distributed, use Student’s t-distribution



Perform (8 of 9)

Mean, μ

Dependent samples:

Provided each sample size is greater than 30 or the differences come from a population that is normally distributed, use Student’s t-distribution with n − 1 degrees of freedom with



Perform (9 of 9)

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