Question

Two samples are taken with the following sample means, sizes, and standard deviations x?? = 36 x?? = 23 n? = 56 n? = 45 s? = 4 s? = 3 We want to estimate the difference in population means using a 93% confidence level. What distribution does this require? z t, df = 99 NOTE: The more accurate df formula, used above, is: df = (s?²/n? + s?²/n?)² / [1/(n?-1) * (s?²/n?)² + 1/(n?-1) * (s?²/n?)²] What is ?? What is the critical value for the confidence interval? NOTE: Use the df given in the question above, 99 Use Technology The 93% confidence interval is: 11.72 < ?? ? ?? < 14.28 Does this interval include 0? What does that decision mean? It excludes 0. The two population means are different. It does include 0. The two population means seem to be equal.

          Two samples are taken with the following sample means, sizes, and standard deviations
x?? = 36 x?? = 23
n? = 56 n? = 45
s? = 4 s? = 3

We want to estimate the difference in population means using a 93% confidence level. What distribution does this require?
z
t, df = 99

NOTE: The more accurate df formula, used above, is:
df = (s?²/n? + s?²/n?)² / [1/(n?-1) * (s?²/n?)² + 1/(n?-1) * (s?²/n?)²]

What is ??

What is the critical value for the confidence interval?
NOTE: Use the df given in the question above, 99

Use Technology

The 93% confidence interval is:
11.72 < ?? ? ?? < 14.28

Does this interval include 0? What does that decision mean?
It excludes 0. The two population means are different.
It does include 0. The two population means seem to be equal.

Show more…

Two samples are taken with the following sample means, sizes, and standard deviations
x?? = 36 x?? = 23
n? = 56 n? = 45
s? = 4 s? = 3

We want to estimate the difference in population means using a 93% confidence level. What distribution does this require?
z
t, df = 99

NOTE: The more accurate df formula, used above, is:
df = (s?²/n? + s?²/n?)² / [1/(n?-1) * (s?²/n?)² + 1/(n?-1) * (s?²/n?)²]

What is ??

What is the critical value for the confidence interval?
NOTE: Use the df given in the question above, 99

Use Technology

The 93% confidence interval is:
11.72 < ?? ? ?? < 14.28

Does this interval include 0? What does that decision mean?
It excludes 0. The two population means are different.
It does include 0. The two population means seem to be equal.

Added by Jessica D.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Jon Southam

08/10/2023

Step 1

Since we are estimating the difference in population means and do not know the population standard deviations, we use the t-distribution. Show more…

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Two samples are taken with the following sample means, sizes, and standard deviations: x̄₁ = 36, x̄₂ = 23 n₁ = 56, n₂ = 45 s₁ = 4, s₂ = 3 We want to estimate the difference in population means using a 93% confidence level. What distribution does this require? t, df = 99 NOTE: The more accurate df formula, used above, is: df = (s₁²/n₁ + s₂²/n₂)² / [1/(n₁-1) * (s₁²/n₁)² + 1/(n₂-1) * (s₂²/n₂)²] What is α? What is the critical value for the confidence interval? NOTE: Use the df given in the question above, 99. Use Technology. The 93% confidence interval is: 11.72 < μ₁ − μ₂ < 14.28 Does this interval include 0? What does that decision mean? It excludes 0. The two population means are different.

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Transcript

00:01 So we have two samples taken from the following sample means and sizes and standard deviations.

00:06 Population 1, population 2, mean of 36, mean of 23, size 56, size 45, and standard deviations 4 and 3.

00:13 And we want to make a 93 % confidence interval for the difference of the means.

00:30 So it's mu1 minus mu2.

00:36 And what should we use? we should use a z or a t with 99 degrees of freedom.

00:45 The reason we're going to use a t distribution is because we don't know the population standard deviations.

00:51 These are the sample standard deviations.

00:52 These are based on the sample.

00:54 If we knew the population sigmas of each one, the population standard deviations, then we could use the z, but we don't.

01:01 So we use our t distribution with 99 degrees of freedom.

01:04 But just so you know, we're told if the variances aren't approximately equal, which these ones wouldn't be, we should use this formula for degrees of freedom.

01:14 But for simplicity's sake, we'll go with 99.

01:18 The alpha for our confidence interval, because our confidence interval is formed by taking the x bar 1 minus x bar 2.

01:28 The difference plus minus t alpha over 2 with 99 degrees of freedom, multiplied by, it's called the standard error.

01:42 What's the alpha? well, 1 minus the alpha gives us 0 .93 or 99, 93 % confidence.

01:50 So that means the alpha is 0 .07...

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