Stats Review Chapters 9-10

Created by Teri Johnson Math Coordinator, Mary Stangler Center for Academic Success

Examples are taken from Statistics 4 E by Michael Sullivan, III

And the corresponding Test Generator from Pearson

Revised 12/13

Note:

This review is composed of questions the textbook and the test generator. This review is meant to highlight basic concepts from the course. It does not cover all concepts presented by your instructor. Refer back to your notes, unit objectives, handouts, etc. to further prepare for your exam. A copy of this review can be found at www.sctcc.edu/cas.

The final answers are displayed in red and the chapter/section number is the corner.

Confidence Interval for Population Proportion

In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90% confidence interval for the population proportion, p?

– Condition 1: n(.05)≤N • The sample size is less than 5% of the population, so the

condition is met.

– Condition 2: np(1-p)≥10 • p= 2/10=.2, so np(1-p)=10(.2)(1-.2)=1.6. This is less than 10

so this condition is not met. So it would not be practical to construct the confidence interval.

9.1

Confidence Interval for Population Proportion

A poll conducted found that 944 of 1748 adults do not believe that people with tattoos are more rebellious.

If appropriate construct a 90% confidence interval.

Is it appropriate? Yes, it satisfies both conditions.

9.1

Confidence Interval for Population Proportion

What is the point estimate?

𝑝 =944

1748= .54

Find 𝑧𝛼/2. First take 1−.90

2= .05

The corresponding z-value is -1.645. We will ignore the negative and just use 1.645. Find the Margin of Error

𝐸 = 𝑧𝛼/2𝑝 (1−𝑝 )

𝑛= 1.645

.54(1−.54)

1748= .0196

Take the point estimate and add/subtract the margin of error .54-.0196=.5204, .54+.0196=.5596 The confidence interval is (.5204,.5596)

9.1

Common 𝑧𝛼/2 values

Confidence Level Z-value

90% 1.645

95% 1.96

99% 2.576

9.1

Sample Size Needed of Population Proportion

A researcher wants to estimate the proportion of Americans that have sleep deprivation. How large a sample is needed in order to be 95% confident within 5% if

a) the researcher used a previous estimate of 60%?

b) the researcher doesn’t use a previous estimate?

9.1

Sample Size Needed of Population Proportion

a) What is the sample size if the researcher used a previous estimate of 60%?

𝑁 = 𝑝 1 − 𝑝 𝑧𝛼2

𝐸

2

= .60 1 − .601.96

.05

2

= 369 b) What is the sample size if the researcher doesn’t use a previous estimate?

𝑁 = .25𝑧𝛼2

𝐸

2

= .251.96

.05

2

= 385

9.1

Find the t-value Find the t-value (using the table)

• Right tail=.1, n=6 – Degrees of Freedom (df)=n-1=5, t= 1.476

• Left tail=.05 n=16 – df=15, t=-1.753 (t is negative for left tail)

• 90% confidence, n=21

– df=20, 1−.90

2= .05, 𝑡 = 1.725

• 95% Confidence , n=83 – df= 82 (this is not in the table so choose the closest one: 80),

1−.95

2= .025, 𝑡 = 1.990

• 99% Confidence, n=1200 – df=1199, since this is more than 1000, we use the z-value

1−.99

2= .005, 𝑡 𝑜𝑟 𝑧 = 2.576

9.1,

Estimating a population mean

In a sample of 81 SARS patients, the mean incubation period was 4.6 days with a standard deviation of 15.9 days. Construct a 95% confident interval.

– We cannot use z because we do not have the population standard deviation.

– Find 𝑡𝛼/2: 1−.95

2= .025, degrees of freedom =n-1=81-80. The

corresponding t-value is 1.99 – Find the Margin of Error

𝐸 = 𝑡𝛼/2

𝑠

𝑛= 1.99

15.9

81= 3.516

– Add/subtract the margin of error from the sample mean 4.6-3.516 =1.084, 4.6+3.516=8.116 – We are 95% confident that the mean incubation period for SARS

patients is between 1.084 and 8.116.

9.2

Sample Size for Estimating Population Mean

How large must a sample be in order to be 95% confident within 2 points given a sample standard deviation of 13.67?

𝑛 =𝑧𝛼2∙ 𝑠

𝐸

2

=1.96 ∙ 13.67

2

2

= 175

9.2

What Happens to the Width of the Confidence Interval

• As the sample size increases – The width decreases

• As the level of confidence increase – The width Increases

To find out which margin of error is smaller (smaller width) with different sample sizes and level of confidence when everything else is the same calculate 𝑧𝛼/2

𝑛 (for proportion) or

𝑡𝛼/2

𝑛 (for mean) for both

cases to see which is smaller.

9.1-2

Reasonable Interpretation of Confidence Intervals

A 90% confidence interval for the hours that college students sleep during the weekday is (7.8, 8.8). Which interpretations are correct? a) 90% of college students sleep between 7.8 and

8.8 hours Flawed: makes an implication about individuals rather than the mean

b) We’re 90% confident that the mean number of hours of sleep that college students get any day of the week is between 7.8 and 8.8 hours

Flawed: should be about the weekday, not any day of the week

9.2

Reasonable Interpretation of Confidence Intervals

A 90% confidence interval for the hours that college students sleep during the weekday is (7.8, 8.8). Which interpretations are correct? c) There is a 90% probability that the mean hours

of sleep that college students get during a weekday is between 7.8 and 8.8 hours.

Flawed: implies the population mean varies rather than the interval

d) We’re 90% confident that the mean number of hours of sleep that college students during a weekday is between 7.8 and 8.8 hours.

Correct!

9.2

Find the Chi-Squared Values Find 𝝌𝟏−𝜶/𝟐

𝟐 𝐚𝐧𝐝𝝌𝜶/𝟐𝟐

• 90% confidence, n=20 – α=1-.90=.1; df=n-1=19

– 𝜒1−𝛼/22 = 𝜒1−.1/2

2 = 𝜒.952 = 10.117

– 𝜒𝛼/22 = 𝜒.1/2

2 = 𝜒.052 = 30.144

• 95% confidence, n=25 – α=1-.95=.05; df=n-1=24

– 𝜒1−𝛼/22 = 𝜒1−.05/2

2 = 𝜒.9752 = 12.401

– 𝜒𝛼/22 = 𝜒..05/2

2 = 𝜒.0252 = 39.364

• If the degrees of freedom is not on the table, use the closest degrees of freedom

• If the degrees of freedom is directly between 2 values, find the mean of the values. Ex: For 65, take the χ2 values for both 60 and 70 and averge their χ2 values.

9.3

Estimating a Population Standard Deviation A student randomly selects 10 paperbacks at a store. The mean price is $8.75 with a standard deviation of $1.50. Construct a 95% confidence interval for the population standard deviation, σ. Assume the data are normally distributed

1. Find 𝜒1−𝛼/22 and𝜒𝛼/2

2 for 95% confidence level and df=9

𝜒1−.05/22 = 2.7

𝜒.05/22 = 19.023

2. Find the lower and upper bounds of σ2

Lower: (𝑛−1)𝑠2

𝜒𝛼/22 =

(10−1)1.52

19.023=1.065

Upper: (𝑛−1)𝑠2

𝜒1−𝛼/22 =

(10−1)1.52

2.7=7.5

3. Since we want the confidence interval for σ and we have the lower/upper bounds of σ2, we need to take the square roots of the lower and upper bound.

Lower: 1.065=1.03

Upper: 7.5=2.74

The confidence interval is (1.03,2.74)

9.3

Which Procedure to Use?

9.4 Copyright © 2013 Pearson Higher Ed

Hypothesis Testing Determine if it is a right-, left-, or two-tailed test

𝐻𝑜: 𝜇 = 6 𝐻1: 𝜇 > 6

𝐻𝑜: 𝜇 = 6 𝐻1: 𝜇 < 6

𝐻𝑜: 𝜇 = 6 𝐻1: 𝜇 ≠ 6

Right -tailed Left -tailed Two -tailed

The mean age of lawyers in New York is 50.7 years. Two-tailed The mean annual return for an employee's IRA is at most 3.4 percent. Left-Tailed A popular referendum on the ballot is favored by more than half of the voters. Right-Tailed

10.1

Type I and Type II errors

Determine the type I and type II error. Find the probability of making a type I error. A referendum for an upcoming election is favored by more than half of the voters. Level of Significance, α, is 0.01. • Type I error: rejecting that p=.5, when in reality p≤5. • Type II error: not rejecting p=.5, when p>.5. • Probability(type I error)=α=.01

10.1

Proper Conclusion for Hypothesis Test

A candidate for state representative of a certain state claims to be favored by at least half of the voters. If a hypothesis test is preformed, how should you interpret a decision that fails to reject the null hypothesis?

A) There is not sufficient evidence to reject the claim p≥.5.

B) There is sufficient evidence to support the claim p≥.5.

C) There is sufficient evidence to reject the claim p≥.5. D) There is not sufficient evidence to support the claim

p≥.5.

10.1

Correct!

Hypothesis Test for Population Proportion: Classical Approach

A survey of 1000 adults, 542 found that they could not eat just one m&m. Does this sample evidence find that more than half of adults can not eat just one m&m. Use the α=.05 level of significance.

1. State the null and alternative Hypothesis 𝐻0: 𝜇 = .5 𝐻1: 𝜇 > .5

2. Find 𝑝 : 𝑝 =𝑥

𝑛=

542

1000= .542

10.2

Hypothesis Test for Population Proportion: Classical Approach

3. Compute the test statistic, z0

𝑧0 =𝑝 − 𝑝

𝑝(1 − 𝑝)𝑛

=.542 − .5

.5(1 − .5)1000

= 2.66

4. Determine the critical value For right tailed, zα=z.05=1.64

5. Compare with the test statistic with the critical value For a right-tailed, we reject the null hypothesis if z0 > zα

2.66>1.64 Therefore we reject the null hypothesis.

6. Conclusion: We reject the null hypothesis. There is sufficient evidence at the α=.05 level of significance to conclude more than half of adults cannot eat just 1 m&m.

Hypothesis Test for Population Mean: P-Value Approach

A local retailer claims that the mean waiting time is less than 9 minutes. A random sample of 20 waiting times has a mean of 7.4 minutes with a standard deviation of 2.1 minutes. At α = 0.01, test the retailer's claim. Assume the distribution is normally distributed. Test the retailer’s claim using α=.01.

10.3

Hypothesis Test for Population Mean

1) State the null and alternative hypothesis 𝐻𝑜: 𝜇 = 9 𝐻1: 𝜇 < 9

2) Find the t-value

𝑡 =𝑥 −𝜇0𝑠

𝑛 =

7.4−92.1

20 = −3.407 with 19 degrees of freedom

3) Find the p-value (use the table or excel) P-value =.00148 4) Conclusion: Since the p-value is less than .01 (α), we reject the hypothesis. There is significant evidence that the mean waiting time is less than 9 minutes.

10.3

Hypothesis Test for Population Mean

In 2002, the mean age of inmate on death-row was 40.7 years. A researcher wonders if the mean age has changed since then. She randomly selects 32 death-row inmates and finds that their mean age is 38.9 with a standard deviation of 9.6. Test the researchers claim.

10.3

Hypothesis Test for Population Mean

1) State the Null and Alternative hypothesis ℎ𝑜: 𝜇 = 40.7 ℎ1: 𝜇 ≠ 40.7

2) Calculate the test statistic

𝑡 =𝑥 −𝜇0𝑠

𝑛 =

38.9−40.79.6

32 = −1.06 with 31 degrees of freedom

3) Find the p-value (use the table or excel) If using the table or a non 2 tailed excel function, you need to divide the p-value by 2 P-value =.2970 We are not given a significance level (α), we use 5%. The p-value is greater than 5%, so we do not reject the null hypothesis. Conclusion: There is not sufficient evidence to reject the claim that the mean has changed.

10.3

Hypothesis Test for Population Standard Deviation

A professor at an all-men's college determined that the standard deviation of men's heights is 2.5 inches. The professor then randomly selected 41 female students from a nearby all-female college and found the standard deviation to be 2.9 inches. Test the professor's claim that the standard deviation of female heights is greater than 2.5 inches. Use α = 0.01.

10.4

Hypothesis Test for Population Standard Deviation

1. State the Null and Alternative hypothesis ℎ𝑜: 𝜎 = 2.5 ℎ1: 𝜎 > 2.5

2. Calculate the test statistic

𝜒02 =

(𝑛 − 1)𝑠2

𝜎02 =

(41 − 1)2.92

2.52= 53.824

3. Determine the P-value Look up value: .929118 Subtract from 1: 1-.929118=.075

4. Conclusion: p-value> α (.01) Fail to reject the null hypothesis. There is not statistical evidence to support the claim that standard deviation is greater than 2.5.

10.4

When to Use t and When to Use z

• Use Z – Inference (confidence interval/ hypothesis testing)

about population proportion

– Looking for sample sizes

– Have population standard deviation σ

• Use T – Inference (confidence interval/ hypothesis testing)

about population mean

– Sample standard deviation s (but not looking for sample size n)

10.5

10.5

Which Procedure to Use?