[ad_1]

Note: Complete and return only the last page answer sheet.

I. Matching

Normal Distribution a. Used for small samples

Standard Normal Distribution b. Asserts a significant difference

Student t-Distribution c. Mean = 0; Standard deviation = 1

Degrees of Freedom d. Rule giving a probability interval

Standard Normal Deviate e. Me

Standard Error f. Bell curve

Confidence Interval g. Alpha error probability

Type I Error h. S/sqrt n

Research Hypothesis i. Rule providing a probability

Successful In Bus3104 j. N – 1

11. Null Hypothesis k. Rejecting a true Ho

12. Type II Error l. Asserts no significant difference

13. P-Value m. SS+/-(CC)(SE)

14. Probability Density Function o. Z-score

15. Cumulative Density Function p. Failing to reject a false Ho

II. Multiple Choice

For questions 1-5 use a normal random variable X with x-bar = 60 and s = 6.

Compute the Z-score for X = 52.

a. 1.33 b. 0.4082 c. –1.33 d. 0.0918 e. not given

Determine the value of X that is equivalent to a Z-score of 1.96.

a. 67.96 b. 61.96 c. 48.24 d. 76.71 e. not given

Find the probability that X is between forty-five and seventy-five.

a. 0.9587 b. 0.0062 c. 0.0537 d. 0.9463 e. not given

Give the probability that X is at most 78.5.

a. 0.999 b. 0.4990 c. 3.08 d. 0.001 e. not given

Calculate the value, Xo, such that P (X Xo) is 0.3023?

a. -0.85 b. 65.1 c. 52.9 d. 0.85 e. not given

What is the value for P(Z<1.96).

a. 0.9750 b. 0.4500 c. 0.0250 d. 0.4750 e. not given

Which value of Zo satisfies P(Z<Zo) = 0.67?

a. –0.44 b. 0.954 c. 0.44 d. 0.7486 e. not given

Choose the correct standard error for x-bar if s = 5.4756 and n = 36.

a. 0.5121 b. 0.39 c. 0.9126 d. 0.065 e. not given

Calculate the z-score for x = 58 if x-bar is 70 and s^2 is 5.

a. –5.37 b. –2.4 c. 2.4 d. 5.37 e. not given

Determine P(-1.23<z<-0.45).

a. 0.2171 b. 0.5643 c. 0.1736 d. 0.3907 e. not given

In a population of normally distributed aptitude scores of 1000 academy students with mean 70 and

Standard deviation 10, how many score below 95?

a. 494 b. 6 c. 25 d. 250 e. not given

The z-score for x-bar = 2.7, µ = 3, s = 1, and n = 100 is:

a. –30 b. –3 c. –0.3 d. 3 e. not given

Find a 95% confidence interval for µ if s = 5.26, x-bar = 70.1, and n = 49.

a. 70.1+/-0.2104 b. 70.1+/-1.2361 c. 70.1+/-1.4728 d. 70.1+/-1.6772 e. not given

Compute a 99% confidence interval for µ if x-bar = 16.3, s = 1.5, and n = 25.

a. 16.3+/-0.8391 b. 16.3+/-0.774 c. 16.3+/-0.8361 d. 16.3+/-0.16782 e. not given

Given Ho: µ ≤ 80 testing at alpha level 0.01 with s = 7, n = 49, and x-bar = 83. Find the p-value.

a. 0.0013 b. 0.05 c. 0.01 d. 0.17 e. not given

A company claims a mean battery life of 42 months. A sample of 36 yields a mean of 39 months

with standard deviation 7 months. Test the null hypothesis Ho: µ ≥ 42 with alpha 0.05.

a. Fail to reject Ha b. Reject Ha c. Fail to reject Ho d. Reject Ho e. not given

17. If n = 15 and = 0.01, then the critical value of t is:

a. 1.753 b. 2.602 c. 1.345 d. 2.624 e. not given

18.The standard deviation for a sampling distribution of sample means is called:

a. alpha error b. beta error c. standard error d. sample error e. not given

Given a sample of nine and assuming a two-tailed test, what percent of the area under the appropriate

t-curve will fall beyond 1.86 standard errors of the mean?

a. 10 b. 5 c. 0.10 d. 0.05 e. not given

For the standard normal distribution the total area under the curve equals:

a. 100 b. 10 c. 0.5 d. 0 e. not given

Choose the standard error for n = 2000 and p-hat = 0.113.

a. 0.014 b.0.099 c. 0.0025 d. 0.127 e. not given

Answer true (T) or false (F). (True answers are highlighted in yellow)

The total area under the bell curve equals one.

2. In a population of heights of 10,000 women, m = 63 inches and s = 5 inches. The number of

heights in the population that fall between 66 inches and 68 inches is 1156.

3. In a normal distribution, approximately 91.31% of the area under the curve is found to the

right of a point -1.36 standard deviations from the mean.

4. Approximately 92.5% of the area under the normal curve is located between the mean and

+/-1.78 standard deviations from the mean.

5. It is incorrect to say that the standard error of the mean is also the standard deviation of a

sampling distribution of means.

6. The most practical way to reduce standard error is to increase the sample size.

7. The standard error of a sampling distribution of means can be larger than the standard

deviation of the population upon which the sampling distribution is based.

8. With remaining constant, increasing the sample size yields a smaller standard error.

9. As n increases, the width of a confidence interval for m increases, assuming s is constant.

10. Expanding confidence from 95% to 99% will increase the width of a confidence interval.

11. Given n = 100 and s^2 = 400, the estimate of the standard error of the mean is 2.

12. Given a sample with n = 100, Ho: m = 500, Ha: m ¹ 500, and = 0.05, the null hypothesis

can be rejected if and only if z ³ 1.645 or z £ -1.645.

13. In problem 12 if x-bar = 505 and s = 20, the null hypothesis can be rejected with p = 0.0062.

14. Convicting an innocent defendant is a Type I Error.

If n = 16, x-bar = 43 and s = 4, then a 98% confidence interval for is (40.869, 45.131).

16. The table value of Z required to reject a null hypothesis at a given -level is larger for a

two-tailed test than for a one-tailed test.

17. According to the Central Limit Theorem, a distribution of sample means based on a sample of

n = 7 will approximate normality even if the data in the parent population are not normal.

18. When performing hypothesis tests or computing confidence intervals based on large samples,

it is necessary to assume that the data in the parent population(s) are distributed normally.

19. To estimate within two units with 95% confidence and = 10 requires n be at least 100.

20. If a two-tailed test assumes a 5% rejection region, then the total rejection region is 5%.

21. The p-value for a hypothesis test resulting in z < -2.74 is approximately 0.0310.

22. Allowing a guilty defendant to go free is a Type II Error.

23. The Student t-distribution approaches a normal distribution as the df get larger.

24. Statistical significance is the rejection of a null hypothesis at some alpha level.

25. Two samples of size 25 each yielding x-bar = 6.47, Sx = 1.14, y-bar = 6.17, and Sy =1.166

show a significant difference between means with a p-value of 0.3623.