Math Problem Statement

  1. Left sided, = 0.05 = 5%, n = 50, NORMAL DISTRIBUTION

NEED: InvNorm(area to the left) = InvNorm(0.05) =

  1. Right sided, = 0.05 = 5%, n = 50, NORMAL DISTRIBUTION. (5% area to the right means 1-.05 = .95 (95%) are to the left)

NEED: InvNorm(area to the left) = InvNorm(1-0.05) =

  1. Two-sided, = 0.05 = 5%, n = 50, NORMAL DISTRIBUTION (5% area split in 2 tails means half in each tail).

NEED: InvNorm(area to the left) = InvNorm(0.05/2) = AND InvNorm(1-0.05/2) =

USING T-DISTRIBUTION: (Note df = n - 1 = 49 for 1 sample means)

  1. Left sided, = 0.05 = 5%, n = 50, t-DISTRIBUTION.

NEED: InvT(area to the left,df) = InvT(0.05,49) =

  1. Right sided, = 0.05 = 5%, n = 50, t-DISTRIBUTION. (5% area to the right means 1-.05 = .95 (95%) are to the left)

NEED: InvT(area to the left, df) = InvT(1-0.05, 49) =

  1. Two-sided, = 0.05 = 5%, n = 50, t-DISTRIBUTION (5% area split in 2 tails means half in each tail).

NEED: InvT(area to the left, df) = InvT(0.05/2, 49) = AND InvNorm(1-0.05/2) =

Solution

Let's solve each of these scenarios using the normal and t-distributions.


Normal Distribution

For a standard normal distribution, we use the ZZ-score:

  1. Left-sided: InvNorm(0.05)=1.645\text{InvNorm}(0.05) = -1.645 This is the ZZ-score that leaves 5% of the area to the left.

  2. Right-sided: InvNorm(10.05)=InvNorm(0.95)=1.645\text{InvNorm}(1 - 0.05) = \text{InvNorm}(0.95) = 1.645 This is the ZZ-score that leaves 5% of the area to the right, or 95% of the area to the left.

  3. Two-sided:

    • Left tail: InvNorm(0.052)=InvNorm(0.025)=1.96\text{InvNorm}\left(\frac{0.05}{2}\right) = \text{InvNorm}(0.025) = -1.96
    • Right tail: InvNorm(10.052)=InvNorm(0.975)=1.96\text{InvNorm}\left(1 - \frac{0.05}{2}\right) = \text{InvNorm}(0.975) = 1.96 These are the ZZ-scores that leave 2.5% in each tail (5% total).

t-Distribution

For the tt-distribution, with degrees of freedom df=n1=49\text{df} = n - 1 = 49, we use the tt-score:

  1. Left-sided: InvT(0.05,49)=1.677\text{InvT}(0.05, 49) = -1.677 This is the tt-score that leaves 5% of the area to the left with 49 degrees of freedom.

  2. Right-sided: InvT(10.05,49)=InvT(0.95,49)=1.677\text{InvT}(1 - 0.05, 49) = \text{InvT}(0.95, 49) = 1.677 This is the tt-score that leaves 5% of the area to the right, or 95% to the left, with 49 degrees of freedom.

  3. Two-sided:

    • Left tail: InvT(0.052,49)=InvT(0.025,49)=2.009\text{InvT}\left(\frac{0.05}{2}, 49\right) = \text{InvT}(0.025, 49) = -2.009
    • Right tail: InvT(10.052,49)=InvT(0.975,49)=2.009\text{InvT}\left(1 - \frac{0.05}{2}, 49\right) = \text{InvT}(0.975, 49) = 2.009 These are the tt-scores that leave 2.5% in each tail (5% total) with 49 degrees of freedom.

Would you like additional explanations on these calculations or more details about interpreting the results?

Related Questions

  1. How do you interpret the values obtained from InvNorm and InvT in hypothesis testing?
  2. Why do we use the t-distribution instead of the normal distribution when the sample size is small?
  3. How would these critical values change if we increased the sample size nn?
  4. What is the significance of using a two-sided test versus a one-sided test in statistical inference?
  5. Can you explain the difference between a ZZ-score and a tt-score in terms of their applications?

Tip

Remember, the t-distribution is generally wider (has heavier tails) than the normal distribution, especially for smaller sample sizes, because it accounts for additional uncertainty due to estimating the population standard deviation.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
t-Distribution
Critical Values
Hypothesis Testing
Statistical Inference

Formulas

InvNorm(area to the left)
InvT(area to the left, df)
Z-Score Formula
t-Score Formula

Theorems

Z-Score Theorem
t-Distribution Theorem

Suitable Grade Level

Grades 11-12