Two-Sample t Procedures

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Question 1

Question 1.

True or false: If σ₁ and σ₂ were known, then $\overline{x}$ ₁ – $\overline{x}$ ₂ would have a Normal distribution.

True

False

Correct. If σ₁ and σ₂ were known, then

$\overline{x}$ ₁ –

$\overline{x}$ ₂ would have a Normal distribution with mean µ₁ – µ₂ and standard deviation

$\sqrt{ \frac{ \sigma _{1} ^{2} }{ n_{1} } + \frac{ \sigma _{2} ^{2} }{ n_{2} } }$ .

Incorrect. Correct. If σ₁ and σ₂ were known, then

$\overline{x}$ ₁ –

$\overline{x}$ ₂ would have a Normal distribution with mean µ₁ – µ₂ and standard deviation

$\sqrt{ \frac{ \sigma _{1} ^{2} }{ n_{1} } + \frac{ \sigma _{2} ^{2} }{ n_{2} } }$ .

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Question 2

111

Question 2.

What should we do when we don’t know σ₁ and σ₂?

Estimate σ₁ with s₁ and σ₂ with s₂, the sample standard deviations for the two samples.

Use the average of s₁ and s₂ to estimate both σ₁ and σ₂.

Combine the two sets of data and use the standard deviation of the combined data to estimate both σ₁ and σ₂.

Correct. Just as with a one-sample t, we use s to estimate σ. Specifically, we use s₁ to estimate σ₁ and s₂to estimate σ₂.

Incorrect. Just as with a one-sample t, we use s to estimate σ. Specifically, we use s₁ to estimate σ₁ and s₂to estimate σ₂.

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Question 3

156

Question 3.

With statistical software, which option is the best for degrees of freedom?

Option 1: Approximate two-sample t

Option 2: Conservative two-sample t

Correct. Using software and the approximate two-sample t option is the best option for degrees of freedom as those degrees of freedom are higher than for the conservative two-sample t.

Incorrect. Using software and the approximate two-sample t option is the best option for degrees of freedom as those degrees of freedom are higher than for the conservative two-sample t.

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Question 4

184

Question 4.

True or false: The null hypothesis is a statement of no difference between the two population means.

False

True

Correct. H₀: µ₁ – µ₂ = 0 which is equivalent to H₀: µ₁ = µ₂ is a statement of zero difference or no difference.

Incorrect. H₀: µ₁ – µ₂ = 0 which is equivalent to H₀: µ₁ = µ₂ is a statement of zero difference or no difference.

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Question 5

265

Question 5.

What value is used as the claimed parameter value for µ₁ – µ₂ in the test statistic for testing H₀: µ₁ = µ₂?

Depends on the situation.

One

One hundred

Zero

Correct. H₀: µ₁ = µ₂ is equivalent to H₀: µ₁ – µ₂ = 0, so the value used for µ₁ – µ₂ is zero.

Incorrect. H₀: µ₁ = µ₂ is equivalent to H₀: µ₁ – µ₂ = 0, so the value used for µ₁ – µ₂ is zero.

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Question 6

356

Question 6.

What do we use as a point estimate for µ₁ – µ₂ in a confidence interval?

Zero

$\overline{x} _{1}$ -

$\overline{x} _{2}$

One

$\overline{x} _{2}$

$\overline{x} _{1}$

Correct. We use

$\overline{x} _{1}$ -

$\overline{x} _{2}$ as a point estimate of µ₁ – µ₂.

Incorrect. We use

$\overline{x} _{1}$ -

$\overline{x} _{2}$ as a point estimate of µ₁ – µ₂.

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Question 7

412

Question 7.

What table do we use to find the P-value for $t = \frac{ \overline{x} _{1} - \overline{x} _{2} }{\sqrt{ \frac{ s _{1} ^{2} }{ n_{1} } + \frac{ s _{2} ^{2} }{ n_{2} } } }$ with df = smaller of (n₁ – 1, n₂ – 1)?

Standard Normal table

t table

Correct. Since the formula begins with “t =”, we use the t table to find P-value.

Incorrect. Since the formula begins with “t =”, we use the t table to find P-value.

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Question 8

430

Question 8.

What table do we use to find the multiplier in the formula $\overline{x} _{1} - \overline{x} _{2} \pm t^{*} \sqrt{ \frac{ s _{1} ^{2} }{ n_{1} } + \frac{ s _{2} ^{2} }{ n_{2} }}$ with df = smaller of (n₁ – 1, n₂ – 1)?

t table

Standard Normal table

Correct. Since the formula includes “t*," we use the t table to find the multiplier.

Incorrect. Since the formula includes “t*," we use the t table to find the multiplier.

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Question 9

534

Question 9.

Why do we only need to check for outliers if n₁ + n₂ < 40?

Because if n₁ + n₂ ≥ 40, we can apply the Central Limit Theorem.

Because we need to check for strong skewness rather than outliers if n₁ + n₂ ≥ 40.

Because there are never any outliers if n₁ + n₂ ≥ 40.

Correct. We can apply the Central Limit Theorem whenever the combined sample size is large and data are collected appropriately.

Incorrect. We can apply the Central Limit Theorem whenever the combined sample size is large and data are collected appropriately.

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Question 10

569

Question 10.

If we want to test whether the mean water intake for rats receiving the anti-depressant is greater than the mean water intake for rats receiving the placebo, what alternative hypotheses should we test? Note: D = anti-depressant drug and P = placebo

$\mathrm{H_{a}: \mu_{D} \neq \mu_{P}}$

$\mathrm{H_{a}: \mu_{D} < \mu_{P}}$

$\mathrm{H_{a}: \mu_{D} > \mu_{P}}$

Correct. If we want to show that the mean water intake of rats receiving the drug is greater than the mean water intake of rats receiving the placebo, then we want

$\mathrm{H_{a}: \mu_{D} > \mu_{P}}$ .

Incorrect. If we want to show that the mean water intake of rats receiving the drug is greater than the mean water intake of rats receiving the placebo, then we want

$\mathrm{H_{a}: \mu_{D} > \mu_{P}}$ .

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Question 11

641

Question 11.

Why do we need a two-sample t test for means?

Because we have two treatment groups with a categorical response variable.

Because we have two treatment groups with a quantitative response variable.

Because we have two populations of rats with a categorical response variable.

Because we have two populations of rats with a quantitative response variable.

Correct. When we have two separate sets of quantitative data, we compare two means. In this case, the response variable is water intake which is quantitative and we have a data set for the drug group of rats and a data set for the placebo group of rats. We don’t have two populations of rats for this example.

Incorrect.When we have two separate sets of quantitative data, we compare two means. In this case, the response variable is water intake which is quantitative and we have a data set for the drug group of rats and a data set for the placebo group of rats. We don’t have two populations of rats for this example.

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Question 12

650

Question 12.

How should these rats be allocated to the two treatments?

By randomly selecting twenty rats from all rats

By random allocation of the twenty rats to two treatment groups

By selecting the fastest ten rats to go in the drug group and the slowest ten rats to go in the placebo group.

Correct. Since this is an experiment, the rats should be assigned to treatments with random allocation.

Incorrect. Since this is an experiment, the rats should be assigned to treatments with random allocation.

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Question 13

673

Question 13.

What is the response variable?

Drug versus placebo

Water output

Water intake

Weight of rat

Correct. Water intake was measured on each rat; this is the response variable.

Incorrect. Water intake was measured on each rat; this is the response variable.

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Question 14

685

Question 14.

Are there any outliers in either data set?

Yes

Correct. We observe no observation for which you might say, “Oh, wow!!! That’s an outlier!”

Incorrect. We observe no observation for which you might say, “Oh, wow!!! That’s an outlier!”

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Question 15

728

Question 15.

For the conservative two-sample t procedure, degrees of freedom are the smaller of (n₁ – 1) and (n₂ – 1). What are the degrees of freedom for this example?

Correct. n₁ – 1 = 10 – 1 = 9 and n₂ – 1 = 10 – 1 = 9 so df = 9.

Incorrect. n₁ – 1 = 10 – 1 = 9 and n₂ – 1 = 10 – 1 = 9 so df = 9.

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Question 16

788

Question 16.

With 0.025 < P-value < 0.05 and α = 0.05, should we reject H₀?

Yes

Correct. 0.025 < P-value < 0.05 is less than α = 0.05 so we reject H₀.

Incorrect. 0.025 < P-value < 0.05 is less than α = 0.05 so we reject H₀.

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Question 17

798

Question 17.

Can we conclude that the anti-depressant drug caused an increase in water intake?

Yes

Correct. Because this was a valid experiment, we can conclude that the drug caused an increase in water intake.

Incorrect. Because this was a valid experiment, we can conclude that the drug caused an increase in water intake.

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Question 18

883

Question 18.

In symbols, the parameter being estimated with this confidence interval is µ₁ – µ₂. What is this parameter in context?

The mean water intake of rats receiving the anti-depressant drug.

The mean water intake of rats receiving the placebo.

Both answers A and B.

The difference between the mean water intake of rats receiving the anti-depressant drug and the mean water intake of rats receiving a placebo.

Correct. µ₁ – µ₂ is the difference between the two treatment means. In this example, it is the difference between the mean water intake of rats receiving the anti-depressant drug and the mean water intake of rats receiving a placebo.

Incorrect. µ₁ – µ₂ is the difference between the two treatment means. In this example, it is the difference between the mean water intake of rats receiving the anti-depressant drug and the mean water intake of rats receiving a placebo.

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Question 19

890

Question 19.

This confidence interval estimates µ₁ – µ₂. On the basis of this interval, can we say that µ₁ – µ₂ ≠ 0?

Yes

Correct. Since the interval does not include the value of zero, we can say that µ₁ – µ₂ does not equal zero.

Incorrect. Since the interval does not include the value of zero, we can say that µ₁ – µ₂ does not equal zero.

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StatTutor Lesson - Two-Sample t Procedures

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Question 5

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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