Ww2.justanswer.com



Short Answer/Problem Portion Chapters 6 – 10

1. For a normal distribution,

a. What z-score separates the highest 10% from the rest of the scores?

By table lookup, z = 1.28 when p = 0.9

b. What z-score separates the highest 30% from the rest of the scores?

By table lookup, z = 0.52 when p = 0.7

c. What z-score separates the lowest 40% from the rest of the scores?

By table lookup, z = -0.25 when p = 0.4

d. What z-score separates the lowest 20% from the rest of the scores?

By table lookup, z = -0.84 when p = 0.2

2. A normal distribution has a mean of µ = 100 with σ = 20. Find the following probabilities:

a. p(X > 102)

z = (x - mu)/sigma

z = (102 - 100)/20

z = 2/20

z = 0.1

Using Excel NORMSDIST, p = 0.5398

BUT since we want "greater than," we need to subtract the probability from 1

p = 1 - 0.5398 = 0.4602

c. p(X < 130)

z = (x - mu)/sigma

z = (130 - 100)/20

z = 30/20

z = 1.5

Using Excel NORMSDIST, p = 0.9332

b. p(X < 65)

z = (x - mu)/sigma

z = (65 - 100)/20

z = -35/20

z = -1.75

Using Excel NORMSDIST, p = 0.0401

d. p(95 < X < 105)

z = (x - mu)/sigma

z = (95 - 100)/20

z = -5/20

z = -0.25

Using Excel NORMSDIST, p = 0.4013

z = (x - mu)/sigma

z = (105 - 100)/20

z = 5/20

z = 0.25

p = 0.5987

To find p(95 < x < 105), subtract p(x < 95) from p(x < 105)

0.5987 - 0.4013 = 0.1974

3. Describe the shape, the mean, and the standard deviation for each of the following two distributions.

a. A population of scores with µ = 50 and σ = 6.

The shape is unknown (could be any shape). The mean is 50. The standard deviation is 6

b. The distribution of sample means based on samples of n = 36 selected from a

population with µ = 50 and σ = 6.

The shape is approximately normal, due to the central limit theorem, regardless of the shape of the original distribution. The mean is approximately 50. The stdev is 6/sqrt(36) = 6/6 = 1

4. A population has a mean of µ = 80 with σ = 20.

a. If a single score is randomly selected from this population, how much distance, on

average, should you find between the score and the population mean?

0 . On average, the sample mean will equal the population mean.

b. If a sample of n = 4 scores is randomly selected from this population, how much

distance, on average, should you find between the sample mean and the population

mean?

0, as above.

c. If a sample of n = 100 scores is randomly selected from this population, how much

distance, on average, should you find between the sample mean and the population

mean?

0, as above.

5. Each of the following samples was obtained from a population with µ = 100 and σ = 10. Find the z-score corresponding to each sample mean.

a. M = 95 for a sample of n = 4

z = (xbar - mu)/(sigma/√n)

z = (95 - 100)/(10/√4)

z = -5 / 5

z = -1

b. M = 104 for a sample of n = 25

z = (xbar - mu)/(sigma/√n)

z = (104 - 100)/(10/√25)

z = 4 / 2

z = 2

c. M = 103 for a sample of n = 100

z = (xbar - mu)/(sigma/√n)

z = (103 - 100)/(10/√100)

z = 3 / 1

z = 3

6. A researcher would like to determine whether a new tax on cigarettes has had any effect on people’s behavior. During the year before the tax was imposed, stores located in rest areas on the state thruway reported selling an average of µ = 410 packs per day with σ = 60. The distribution of daily sales was approximately normal. For a sample of n = 9 days following the new tax, the researcher found an average of M = 386 packs per day for the same stores.

a. Is the sample mean sufficient to conclude that there was a significant change in cigarette

purchases after the new tax. Use a two-tailed test with α = .05.

Critical value of z at 0.05 2-tailed is +/- 1.96

z = (xbar - µ)/(σ/√n)

z = (386 - 410 )/(60/√9)

z = -24 / (60/3)

z = -1.2

Since |z| < 1.96, there’s insufficient evidence to reject the null hypothesis. There is not a significant difference in cigarette purchases

b. If the population standard deviation was σ = 30, is the result sufficient to conclude that

there is a significant difference?

z = (xbar - µ)/(σ/√n)

z = (386 - 410 )/(30/√9)

z = -24 / (30/3)

z = -2.4

Since |z| > 1.96, there’s sufficient evidence to reject the null hypothesis. There is a significant difference in cigarette purchases.

c. Explain why the two tests lead to different outcomes.

In the second test, the variation in sales from day to day is much less. Therefore, there’s much less of a chance that the low sample mean is due to random chance. In the first test, there’s so much variation that there’s more than a 5% chance of seeing a sample mean this far from the population mean, even if the population mean hasn’t changed.

7. A sample of n = 16 individuals is selected from a population with µ = 30. After a treatment is administered to the individuals, the sample mean is found to be M = 33.

a. If the sample variance is s2 = 16, then calculate the estimated standard error and

determine whether the sample is sufficient to conclude that the treatment has a

significant effect? Use a two-tailed test with α = .05.

Sample stdev s = sqrt(16) = 4

Estimated standard error if 4/sqrt(16) = 4/4 = 1

Critical value of t at alpha = 0.05 2-tailed and 15 df is 2.131

t = (xbar - µ)/(s/√n)

t = (33 - 30 )/(4/√16)

t = 3 / (4/4)

t = 3

|t| > 2.131, so there is sufficient evidence to reject the null hypothesis. The treatment has had a significant effect.

b. If the sample variance is s2 = 64, then calculate the estimated standard error and

determine whether the sample is sufficient to conclude that the treatment has a

significant effect? Use a two-tailed test with α = .05.

Sample standard deviation s = sqrt(64) = 8

Standard error is 8/sqrt(16) = 8/4 = 2

t = (xbar - µ)/(s/√n)

t = (33 - 30 )/(8/√16)

t = 3 / (8/4)

t = 1.5

Since |t| < 2.131, there is insufficient evidence to reject the null hypothesis. The treatment has not had a significant effect.

c. Describe how increasing variance affects the standard error and the likelihood of

rejecting the null hypothesis.

The standard error increases as the variance goes up. That’s because the more the variance, the more likely any difference is due to sampling error. Therefore, we’re less likely to reject the null hypothesis. We’d need a bigger difference to get the same probability with a larger standard error.

8. Although you can compute a z-score for a single score (a sample of n = 1), it is impossible to compute a t statistic for a sample that has only one score. Explain why.

That’s because the t distribution is a family of distributions based on degrees of freedom, which is n-1. When n = 1, there are no degrees of freedom.

9. An educational psychologist studies the effect of frequent testing on retention of class material. In one section of an introductory course, students are given quizzes each week. A second section of the same course receives only two tests during the semester. At the end of the semester, both sections receive the same final exam, and the scores are summarized below.

Frequent Quizzes Two Exams

n = 20 n = 20

M = 73 M = 68

a. If the first sample variance is s2 = 38 and the second sample has s2 = 42, do the data

indicate that testing frequency has a significant effect on performance? Use a two-

tailed test at the .05 level of significance. (Note: Because the two sample are the same

size, the pooled variance is simply the average of the two sample variances.)

df: n1 + n2 - 2 = 38

Critical value 2-tailed at alpha = 0.05, 38 df, is 2.02

Find the pooled variance and standard deviation

s^2 = (38 + 42)/ 2 = 40

s = sqrt(40) = 6.3246

Then find the t statistic

t = (m1 - m2) / [s*(sqrt((1/n1) + (1/n2)))]

t = (73 - 68) /[ 6.3246*(SQRT((1/20) + (1/20)))]

t = 5 / [6.3246*(SQRT(0.1))]

t = 5 / (6.3246*0.3162)

t = 2.5

|t| > 2.02, so there is sufficient evidence to reject the null hypothesis. There is a significant difference between the two groups.

b. If the first sample variance is s2 = 84 and the second sample has s2 = 96, do the data

indicate that testing frequency has a significant effect? Again, use a two-tailed test with

α = .05.

Find the pooled variance

s^2 = (84 + 96)/ 2 = 90

s = sqrt(90) = 9.4868

Then find the t statistic

t = (m1 - m2) / [s*(sqrt((1/n1) + (1/n2)))]

t = (73 - 68) /[ 9.4868*(SQRT((1/20) + (1/20)))]

t = 5 / [9.4868*(SQRT(0.1))]

t = 5 / (9.4868*0.3162)

t = 1.667

|t| < 2.02, so there is insufficient evidence to reject the null hypothesis. There is no significant difference between the two groups.

c. Describe how the size of the variance affects the outcome of the hypothesis test?

When the variance is larger, there’s more probability that any difference between the two groups is due to chance. Therefore, it’s harder to reject the null hypothesis, and more likely that no significant difference will be found.

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download