Handout 4

CONFIDENCE INTERVALS: ONE POPULATION PARAMETER

A (1 - a) Confidence Interval is an interval which has a probability of 1 - a of containing the given population parameter based on a sample. There is a probability of a that the population parameter will be outside the interval.

Determine the parameter (m, p, or s) for which you are doing the interval. Follow the directions on the chart.

1. Find a 99% confidence interval for the mean of a population if a sample of 50 values has a mean of 22 with a standard deviation of 8.

Zinterval (19.086,24.914)

2. Find a 99% confidence interval for the mean of a population if a sample of 20 values has a mean of 22 with a standard deviation of 8.

Tinterval (16.882,27.118)

3. Find a 95% confidence interval for the proportion of the population that will support Gore for president if, in a sample of 500 individuals, 220 say they will support him.

1 prop zinterval (.396,.484)

4. If 28% a sample of 200 individuals has blue eyes, estimate the proportion of the population with blue eyes using a 90% confidence interval.

1 prop z interval (.228,.332)

5. A sample of 20 schnauzers had a mean height of 11 .4 inches with a standard deviation of 1.21 inches. Determine a 95% confidence interval for the true standard deviation for schnauzer heights.

.92<s<1.76

Finding sample size to estimate a parameter: Determine the parameter and follow formula on chart. To find use INVNORM( , m , s )

6. The standard deviation of weights for 10 year old girls is known to be 5.6 pounds. A researcher wants to estimate the mean weight of 10 year old girls within 1 pound. If he wishes to be 99% confident in his estimate, how large a sample is necessary? 209

7. A politician wants to estimate the proportion of voters that will vote for her. She wants to be 90% certain that his estimate is within .03 of the true value. How large a sample should she take? 752

HYPOTHESIS TESTING

P-value Approach and Classical Method

STEP 1: State a null (H_o) and alternate (H₁) hypothesis.

When writing null and alternate hypothesis, ignore all information coming from the sample. If there are no words indicating direction (is better, improves, decreases, etc) than the alternate will be a not equal. Read all statements about the population parameter before deciding on a null and alternate hypothesis.

If the alternate hypothesis is a not equal to (¹) we call it a two-tail test.

If the alternate is a greater than (>) it is a one-tail upper tail test.

If the alternate is a less than (<) it is a one-tail lower tail test.

P-value Approach: TI83

STEP 2: Determine the appropriate test and follow directions on the hypothesis testing chart.

STEP 3: Look at the p-value and alpha.

If the p-value is less than or equal to alpha (a) reject H_o.

If the p-value is greater than alpha, accept H_o .

Classical Approach

STEP 2: Calculate the test statistic as on the hypothesis testing chart.

STEP 3: Draw the rejection region based on alpha and the alternate hypothesis.

STEP 4: If the test statistic is in the rejection region, reject the null hypothesis.

The last steps of both of the approaches are:

· State whether you are accepting or rejection the null hypothesis.

· State conclusion or answer question in words.

STATING NULL AND ALTERNATE HYPOTHESES

· The null hypothesis will always be the statement with the equal in it.

· The alternate will be the complement to it.

1. Write a null and alternate hypothesis for each of the following:

a) The mean number of errors on this machine is 2.

H₀: m = 2 H_a: m ¹ 2

b) The mean number of errors on this machine is less than 2

H₀: m = 2 H_a: m < 2

c) The mean grade for this exam is not 75

H₀: m = 75 H_a: m ¹ 75

d) The mean number of errors is more than 2

H₀: m = 2 H_a: m > 2

e) At least 50% of the voters will support Bush

H₀: p = .5 H_a: p £ .5

f) More than 44% of registered voters will vote.

H₀: p = .44 H_a: p > .44

g) The standard deviation of this test is at most 10.

H₀: s = 10 H_a: s > 10

h) The standard deviation is no more than 10.

H₀: s = 10 H_a: s > 10

One Population: Hypothesis Tests on Means and Proportions Variance and Standard Deviation

1. Packages of broccoli are supposed to weigh 12 ounces. The mean for a sample of 24 packages produced by a certain machine was 11.92 . If the standard deviation of all packages is known to be 1.2 ounces, can you conclude the machine is producing packages of the correct weight?

H₀: m = 12 H_a: m ¹ 12

Z test p=.744

Accept H₀

Yes. The packages are the correct weight.

2. Scores on the SAT exam have a mean of 483. A College Board Prep course promises that it will improve on the average score. If a selected group of 38 students who completed this course had a mean of 495 with a standard deviation of 75, do the data support the College Board promise?

H₀: m = 483 H_a: m >483

Z test p=.162

Accept H₀

No. The college board promise is not supported.

3. An environmental group collects a liter of water from each of 25 locations and measures the amount of dissolved oxygen in each liter. The mean of the sample is 4.62 mg and the standard deviation is .92 mg. Is there evidence that the stream has a mean oxygen content less than 5 mg per liter?

H₀: m = 5 H_a: m < 5

Z test p=.025

Reject H₀

Yes. The mean oxygen content is less than 5 mg. Per liter.

4. A pre-election poll 0f 1500 registered voters finds that 781 would vote for Senator Fillibuster. Can we conclude that a majority (more than 50%) of all voters will vote for him?

H₀: p = ..5 H_a: p >.5

1 prop z test pval=.055

Accept H₀

No. We cannot conclude a majority will vote for him.

5. A company claims that more than 80% of its employees are satisfied with their jobs. The Union feels that this is incorrect and takes a sample of 40 employees. If 78% of the employees in the sample are satisfied, does this support the company or the Union?

H₀: p = ..8 H_a: p >.8

1 prop z test pval=.653

Accept H₀

The Union is supported.

6. The GPA's of 22 randomly selected female students had a mean of 2.85 with a standard deviation of .56. If the population standard deviation for all students is .5, can we conclude that female students GPA's vary more than the average?

H₀: s = .5 H_a: s >.5

Chi Square test: C² = 26.34

Rejection region C²>32.671

Accept H₀

No. Female students do not vary more than the average.

P-VALUE, ALPHA AND TYPE I AND TYPE II ERRORS

Type I Error: Rejecting the null hypothesis when it is true or

Claim H₁ is true when H₀ is.

Alpha (a) or significance level of a test is the largest acceptable probability of making a Type I error.

p-value (also called the observed significance level) is the actual probability of making a type I error if you reject H_o for the given data.

Type II Error: Accepting the null hypothesis when it is false or

Claim H₀ is true when H₁ is.

Beta (b) or the power of a test is the probability of making a type II error.

7. Write a null and alternate hypothesis for each of the following then describe a Type I and Type II error for each. Determine which (if either) would be more serious in your viewpoint and state why.

a) H₀: The defendant is innocent

H_a: The defendant is guilty

Type I error: Conclude the defendant is guilty when (s)he is innocent.

Type II error: Conclude the defendant is innocent when (s)he is guilty.

b) H₀: The parachute will open

H_a: The parachute will not open.

Type I error: Conclude the parachute will not open when it will.

Type II error: Conclude the parachute will open when it will not.

c) The mean salary for this company is at least 25,000.

H₀: m³25000 H_a: m<25000

Type I error: Conclude the mean is less than 25,000 dollars when it is at least 25,000.

Type II error: Conclude the mean is at least 25,000 dollars when it isn’t.

d) Fewer than 5% of all high school graduates in Chicago cannot read.

H₀: p³.05 H_a: p< .05

Type I error: Conclude that fewer than 5% of Chicago grads are unable to read when that is not the case.

Type II error: Conclude the literacy rate for Chicago grads is at least 5% when it is really less than that.