Stat 101 Exam 3 · 2020-04-27 · Stat 101 Exam 3: Important Formulas and Concepts 1 1 Sampling...

Stat 101 Exam 3:Important Formulas and Concepts 1

1 Sampling Distributions and Confidence Intervals

1.1 Definitions

1. Sampling DistributionDifferent random samples give different values of a statistic. Distribution of the statis-tics over all possible samples is called the sampling distribution. Sampling distributionmodel shows the behavior of the statistic over all the possible samples for the samesize n.

2. Sampling Distribution ModelBecause we can never see all possible samples, we often use a model as a practical wayof describing the theoretical sampling distribution.

3. Sampling Distribution Model for a ProportionIf assumptions of independence and random sampling are met, and we expect at least10 successes and 10 failures, then the sampling distribution of a proportion is modeledby a normal model with a mean equal to the true proportion value p and has a standarddeviation equal to

√p(1− p)/n.

p̂ ∼ N

(p,√

p(1−p)n

)4. Sampling Error

Sample-to-sample variation

5. Central Limit Theorem (CLT)The sampling distribution model of the sample mean (and proportion) is approximatelyNormal for large n, regardless of the distribution of the population as long as theobservations are independent. The larger the sample, the better the approximationwill be.

6. Sampling Distribution Model for a MeanIf assumptions of independence and random sampling are met, and the sample size islarge enough, the sampling distribution of the sample mean is modeled by a normalmodel with a mean equal to the population mean and has a standard deviation equalto σ/

√n.

X ∼ N(µ, σ√

n

)1This version: April 27, 2020, by Dale Embers. May not include all things that could possibly be tested

on. To be used as an additional reference to studying.

7. Confidence Interval (CI)A level C confidence interval for a model parameter is an interval of values usually ofthe form Estimate ± Margin of Error found from data in such a way that C% of allrandom samples will yield intervals that capture the true parameter value.

8. Margin of Error (MOE)In a confidence interval, the extent of the interval on either side of the observed statisticvalue. It is typically the produce of a critical value from the sampling distribution anda standard error from the data. A small MOE corresponds to a confidence interval thatpins down the parameter precisely. A large MOE corresponds to a confidence intervalthat gives relatively little information about the estimated parameter.

9. Critical ValueThe number of standard errors to move away from the mean of the sampling distribu-tion to correspond to the specified level of confidence. The critical value, for a normalsampling distribution, denoted z∗, is usually found from a table or technology. Thecritical value, for a t-distribution, denoted t∗, is also found from a table or technology.

10. Some z∗ values (Critical Values) for Confidence Intervals

CI: 90% CI 95% CI 98% CI 99% CIz 1.645 1.96 2.326 2.576

2 Chapter 16

1. One-sample z-interval for the meanThis is the confidence interval for the mean. This is given by x± z∗SD(X), SD(X) =σ√n

. The critical value z∗ depends on the particular confidence level that you specify.

2. The margin of error is z∗σ√n

.

3. Given a desired margin of error m, solve m = z∗σ√n

for n to get the desired sample

size. This will result in n = (z∗σ

m)2.

3 Chapter 17

1. HypothesisA model or proposition that we adopt in order to test.

2. Null Hypothesis (H0)The claim being assessed in a hypothesis test that states “no change from the tradi-tional value,” “no effect”, “no difference”, or “no relationship”. For a claim to be atestable null hypothesis, it must specify a value for some population parameter thatcan form the basis for assuming a sampling distribution for a test statistic.

3. Alternative Hypothesis (HA)The alternative hypothesis proposes what we should conclude if we reject the nullhypothesis.

4. P-valueThe probability of observing a value for a test statistic at least as far from the hy-pothesized value as the statistic value actually observed if the null hypothesis is true.A small p-value indicates either that the observation is improbable or that the proba-bility calculation was based on incorrect assumptions. The assumed truth of the nullhypothesis is the assumption under suspicion.

5. One-sample z-test for the meanThis is the hypothesis test. It tests the hypothesis H0 : µ = µ0 using the statisticz = (x− µ0)/( σ√

n).

6. One-sided (Tailed) AlternativeAn alternative hypothesis is one-sized ( for example HA : µ > µ0 or HA : µ < µ0)when we are interested in deviations in only one direction away from the hypothesizedparameter value.

7. Two-sided (Tailed) AlternativeAn alternative hypothesis is two-sided ( for example HA : µ 6= µ0) when we areinterested in deviations in either direction away from the hypothesized parameter value.

8. One rejects H0 and accepts Ha and calls the results statistically significant if the P -value is sufficiently small (less than α).

4 Chapter 18

1. Statistically significantWhen the p-value falls below the alpha level, we say that the test is “statisticallysignificant” at that alpha level.

2. Alpha levelThe threshold p-value that determines when we reject a null hypothesis. If we observea statistic whose p-value based on the null hypothesis is less than α, we reject thatnull hypothesis.

3. Significance levelThe alpha level is also called the significance level, most often in a phrase such as aconclusion that a particular test is “significant at the 5% significance level”

4. Type I ErrorThe error of rejecting a null hypothesis when in fact it is true (also called a falsepositive). The probability of a Type I Error is α.

5. Type II ErrorThe error of failing to reject a null hypothesis when in fact it is false (also called a falsenegative). The probability of a Type II Error is β.

6. βThe probability of a Type II Error is commonly denoted β and depends on the effectsize.

7. PowerThe probability that a hypothesis test will correctly reject a false null hypothesis is thepower of the test. For any specific value in the alternative, the power is 1− β.

5 Chapter 20

1. Student’s t distributionA family of distributions indexed by its degrees of freedom. The t-models are unimodal,symmetric, and bell shaped, but have fatter tails and a narrower center than theNormal model. As the degrees of freedom increase, t-distributions approah the Normaldistribution.

2. Degrees of Freedom for Student’s t distribution (df)For the t-distribution, the degrees of freedom are equal to n− 1, where n is the samplesize.

3. One-sample t-interval for the meanThis is the confidence interval for the mean. This is given by y ± t∗n−1SE(y), SE(y) =s/√n. The critical value t∗n−1 depends on the particular confidence level that you

specify and on the number of degrees of freedom n− 1.

4. One-sample t-test for the meanThis is the hypothesis test. It tests the hypothesis H0 : µ = µ0 using the statistict = (x− µ0)/( s√

n), which has a t-distribution with n− 1 degrees of freedom.

5. One-sided (Tailed) AlternativeAn alternative hypothesis is one-sized ( for example HA : µ > µ0 or HA : µ < µ0)when we are interested in deviations in only one direction away from the hypothesizedparameter value.

6. Two-sided (Tailed) AlternativeAn alternative hypothesis is two-sided ( for example HA : µ 6= µ0) when we areinterested in deviations in either direction away from the hypothesized parameter value.

6 Chapter 22

1. One-sample z-interval for the population proportion

This is the confidence interval for the proportion. This is given by p̂± z∗√p̂ · (1− p̂)

n.

The critical value z∗ depends on the particular confidence level that you specify.

2. The margin of error is z∗√p̂ · (1− p̂)

n.

3. Given a desired margin of error m, solve m = z∗√p∗ · (1− p∗)

nfor n to get the desired

sample size. This will result in n = (z∗

m)2 · p∗ · (1 − p∗). Use an estimate for p∗ or

p∗ = .5 if no estimate exists.

4. One-proportion Z-testA test of the null hypothesis that the proportion of a single sample equals a specified

value H0 : p = p0 by computing the test statistic z = (p̂− p0)/

√p0 · (1− p0)

n.

5. One-sided (Tailed) AlternativeAn alternative hypothesis is one-sized ( for example HA : p > p0 or HA : p < p0)when we are interested in deviations in only one direction away from the hypothesizedparameter value.

6. Two-sided (Tailed) AlternativeAn alternative hypothesis is two-sided ( for example HA : p 6= p0) when we are inter-ested in deviations in either direction away from the hypothesized parameter value.

7 Confidence Interval Creation and Hypothesis Test-

ing Summary

7.1 One-Proportion

Proportion - always use p

• Confidence Interval Creation

CI: p̂± z∗√p̂(1− p̂)

n︸︷︷︸MOE

z∗ = criticalvalue

Table of critical values for z∗ for Confidence Intervals:

CI: 90% 95% 96% 98% 99%z∗ 1.645 1.96 2.054 2.326 2.576

• Hypothesis TestingStep 1: Write down your hypothesis

H0 : p = p0

HA : p <or>or6= p0

Step 2: Calculate your test statistic

z = p̂−p0√p0(1−p0)

n

Step 3: Calculate the p-valueStep 4: State your conclusion. If p-value ≤ α, (usually α = 0.05), then Reject H0. Ifp-value is > α, then Do Not Reject H0.

7.2 One-Sample Mean

Sample Mean - always use x or yThe degrees of freedom is given by df = n− 1.

• Confidence Interval Creation

CI : x± t∗n−1

s√n︸︷︷︸

MOE

t∗n−1 = criticalvalue

Use Appendix D Table T to determine the critical values of t.


H0 : µ = µ0

HA : µ <or>or6= µ0


tn−1 = x−µ0s√n


7.3 Paired Differences of Means

The degrees of freedom is given by df = n− 1.

• Confidence Interval Creation (n = number of pairs)

CI: d̄± t∗n−1

sd√n︸︷︷︸

MOE

t∗n−1 = criticalvalue


H0 : µd = 0HA : µd <or>or 6= ∆0


tn−1 = d̄sd√n

d̄ = averageofthedifferencessd = standarddeviationofthedifferences


8 Extra Information

Review any and all notes and supplementary materials. It may be the case that somethingwas accidentally omitted from this study guide. Also, review any problems that may havebeen discussed in class as not all example problems may have been provided here.

9 Example Problems

1. The 95% confidence interval for the number of teens who reported that they hadmisrepresented their age online is from 45.6% to 52.5%. There were 799 teens in thisstudy.

(a) Interpret the interval in this context.

(b) Explain the meaning of “95% confident” in this context.

2. A study found that 16 of 40 peanut candy bars in fact did not contain peanuts.

(a) Construct a 90% confidence interval.

(b) Interpret your 90% confidence interval.

(c) Construct a 95% confidence interval.

(d) Interpret your 95% confidence interval.

3. Several factors are involved in the creation of a confidence interval. Among them arethe sample size, the level of confidence, and the margin of error. Which statements aretrue?

(a) For a given sample size, higher confidence means a smaller margin of error.

(b) For a given confidence level, halving the margin of error requires a sample twiceas large.

(c) For a certain confidence level, you can get a smaller margin of error by selectinga bigger sample.

(d) For a fixed margin of error, larger samples provide greater confidence.

4. I sample 600 people and 432 of them like cats. Construct a 95% confidence intervalfor the population proportion.

5. I think the proportion of people that eat candy is around 0.75. I am going to constructa 90% confidence interval and want the margin of error to be ±0.025. How large shouldthe sample size be?

6. Jimmy samples 930 people and 234 took public transportation. Construct a 99%confidence interval for the population proportion.

7. I am going to construct a 95% confidence interval for the proportion of people thatwear eyeglasses and want the margin of error to be ±0.2. I have no idea what toestimate for the population proportion. How large should the sample size be?

8. A researcher believes that more than 50% of all people voted in the last election. Shesamples 800 people and 420 of them voted. Test her claim at a significance level of0.05 (i.e. compare the P-value to 0.05).

(a) State the hypotheses to be tested.

(b) Compute the test statistics (z-value). You must show your computation to receivecredit.

(c) Compute the P-value associated with your test statistic.

(d) Make a conclusion about the hypotheses.

9. A researcher believes that fewer than 75% of all mollusks are tasty. He samples 1200mollusks and 865 of them are tasty. Test his claim at a significance level of 0.05 (i.e.compare the P-value to 0.05).





10. A researcher believes that the percentage of people that watch Game of Thrones isdifferent than 27%. He samples 900 people and 220 of them watch. Test his claim ata significance level of 0.05 (i.e. compare the P-value to 0.05).





11. A butcher wants to estimate the mean weight of a ham. She samples 33 hams andcomputes a sample mean weight of 8.2 pounds and a sample standard deviation of 3.3pounds. What is a 90% confidence interval for the population mean weight of ham?Please indicate the value you used for z∗ or t∗.

12. A professor is interested in the mean length of a letter of recommendation. He samples51 letters and finds a sample mean length of 620 words with a sample standard deviationof 90 words. What is a 95% confidence interval for the population mean length of aletter? Please indicate the value you used for z∗ or t∗.

13. A computer professional wants to know the mean number of emails people receive eachday. She is going to compute a 95% confidence interval and wants a margin of error of±2 emails. She believes the standard deviation to be 18 emails. How large should thesample size be to ensure this margin of error?

14. A researcher believes that the mean age at which a person first votes is greater than 22years. He samples 27 people and computes a sample mean of 24.3 years and a samplestandard deviation of 8 years.


(b) What is the value of your test statistic (t or z value)?

(c) What is the P-value?

(d) What conclusion should be drawn (compare p-value to 0.05).

15. A researcher believes that the mean age at which a person first tries chocolate is lessthan 3 years. He samples 24 people and computes a sample mean of 2.3 years and asample standard deviation of 1.5 years.





16. A researcher believes that the mean height of a prairie dog is different than 14 inches.She samples 31 prairie dogs and computes a sample mean of 15.8 inches and a samplestandard deviation of 3.6 inches.





17. Which of the following are true? If false, explain briefly.

(a) A very low P-value provides evidence against the null hypothesis.

(b) A high P-value is strong evidence in favor of the null hypothesis.

(c) A P-value above 0.10 shows that the null hypothesis is true.

(d) If the null hypothesis is true, you can’t get a p-value below 0.01.

18. Which of the following statements are true? If false, explain briefly.

(a) Using an alpha level of 0.05, a p-value of 0.04 results in rejecting the null hypoth-esis.

(b) The alpha level depends on the sample size.

(c) With an alpha level of 0.01, a p-value of 0.10 results in rejecting the null hypoth-esis.

(d) Using an alpha level of 0.05, a p-value of 0.06 means the null hypothesis is true.

19. For each of the following situations, state whether a Type I or Type II, or neither errorhas been made. Explain briefly.

(a) A bank wants to know if the enrollment on their website is above 30% based ona small sample of customers. they test H0 : p = 0.3 versus HA : p > 0.3 andreject the null hypothesis. Later they find out that actually 28% of all customersenrolled.

(b) A student tests 100 students to determine whether other students on her campusprefer Coke or Pepsi and finds no evidence that preference for Coke is not 0.5.Later, a marketing company tests all students on campus and finds no difference.

(c) A human resource analyst wants to know if the applicants this year score, onaverage, higher on their placement exam than the 52.5 points the candidatesaveraged last year. She samples 50 recent tests and finds the average to be 54.1points. She fails to reject the null hypothesis that the mean is 52.5 points. Atthe end of the year, they find that the candidates this year had a mean of 55.3points.

(d) A pharmaceutical company tests whether a drug lifts the headache relief ratefrom the 25% achieved by the placebo. They fail to reject the null hypothesisbecause the p-value is 0.465. Further testing shows that the drug actually relievesheadaches in 38% of people.

20. We want to estimate the healing rate for a wound. A sample of size 17 is collectedand the sample mean is computed to be 24.3 micrometers per hour, with a samplestandard deviation of s= 8 micrometers per hour. What is a 95% confidence intervalfor the population mean?

21. A sample of size n=150 people is collected and the sample proportion of people who areilliterate is computed to be .20. Compute a 95% confidence interval for the populationproportion of illiterate people.

22. You believe that the proportion of people that like cheese is .80. You are going toconstruct a 95% confidence interval and want the margin of error to be plus or minus.03. What should the sample size be?

23. Teresa knows that appointment times are approximately normally distributed. Shebelieves the mean wait time is longer than 25 minutes. She conducts a test with α= 0.05 and the appropriate hypotheses. She selects 25 random appointments and thesample mean was found to be 25.66 minutes and a sample standard deviation of 10minutes.


(b) Compute the value of the test score.

(c) Give the P-value or range of P-values.

(d) Do the results seem significant?

24. You claim that the proportion of people who watch American Idol is greater than .50.You sample n=200 people and compute a sample proportion of .53. Assume α = 0.05.


(b) Compute the value of the test score.

(c) Give the P-value or range of P-values.

(d) Do the results seem significant?

10 Example Solutions

1. (a) We are 95% confident that, if we were to ask all teens whether they have mis-represented their age online, between 45.6% and 52.5% of them would say theyhave.

(b) If we were to collect many random samples of 799 teens, about 95% of the con-fidence intervals would contain the true proportion of all teens who admit tomisrepresenting their age online.

2. This problem tells us that p̂ = 16/40, n = 40.

(a) For a 90% confidence interval, z∗ = 1.645. The 90% confidence interval wouldthen be

p̂± z∗√

p̂(1−p̂)n

= 1640± 1.645

√( 1640)(1− 16

40)40

= 1640± 1.645

√0.006

= (0.2726, 0.5274)

(b) We are 90% confidence that between 27% and 53% of all peanut candy bars didnot contain peanuts.

(c) For a 95% confidence interval, z∗ = 1.96. The 95% confidence interval would thenbe

p̂± z∗√

p̂(1−p̂)n

= 1640± 1.96

√( 1640)(1− 16

40)40

= 1640± 1.96

√0.006

= (0.2482, 0.5518)

(d) We are 95% confident that between 25% and 55% of all peanut candy bars didnot contain peanuts.

3. (a) False. Higher confidence means a larger margin of error.Suppose n = 10. Suppose p̂ = 0.5. Start with 90% Confidence (z∗ = 1.645.)Calculate MOE. Now change to 95% Confidence (z∗ = 1.96). Calculate MOE.Compare the two results.

MOE90 = 1.645√

0.5(1−0.5)10

= 1.645√

0.2510

= 1.645√

0.025= 0.26,

MOE95 = 1.96√

0.5(1−0.5)10

= 1.96√

0.2510

= 1.96√

0.025= 0.31.

From this, we can see that MOE increases when confidence increases.

(b) False. The margin of error decreases as the square root of the sample size increases.Halving the margin of error requires a sample four times as large as the original.Suppose p̂ = 0.5. Suppose 95% Confidence (z∗ = 1.96). Start with MOE = 0.6.Then compare with MOE = 0.3.

0.6 = 1.96√

0.5(1−0.5)n

⇒ 0.61.96

=√

0.25n

⇒ 0.306 =√

0.25n

⇒ 0.0937 = 0.25n

⇒ n = 0.250.0937

⇒ n = 2.668,

0.3 = 1.96√

0.5(1−0.5)n

⇒ 0.31.96

=√

0.25n

⇒ 0.153 =√

0.25n

⇒ 0.0234 = 0.25n

⇒ n = 0.250.0234

⇒ n = 10.68.

So our original n = 2.668 and the new n = 10.68, which is approximately 4 timesthe original value of n.

(c) True. Larger samples are less variable, which translates to a smaller margin oferror. We can be more precise at the same level of confidence.Suppose p̂ = 0.5. Suppose 90% Confidence. Start with n = 2 and compare ton = 18.

MOE2 = 1.645√

0.5(1−0.5)2

= 1.645√

0.252

= 1.645√

0.125= 0.582,

MOE18 = 1.645√

0.5(1−0.5)18

= 1.645√

0.2518

= 1.645√

0.139= 0.194.

Our MOE decreased when n increased.

(d) True. Larger samples are less variable, which makes us more confident that agiven confidence interval succeeds in catching the population proportion.Suppose MOE = 0.4. Suppose p̂ = 0.5. Compare the confidence of n = 5 ton = 8.

0.4 = z∗5

√0.5(1−0.5)

5⇒ 0.4 = z∗5

√0.05

⇒ 0.4√0.05

= z∗5⇒ 1.789 = z∗5 ,

0.4 = z∗8

√0.5(1−0.5)

8⇒ 0.4 = z∗8

√0.03125

⇒ 0.4√0.03125

= z∗8⇒ 2.263 = z∗8 .

As the sample sizes increases, z∗ increases, which means that the confidence levelincreases.

4. I sample 600 people and 432 of them like cats. Construct a 95% confidence intervalfor the population proportion.

p̂ = 432600

= 0.72z∗ = 1.96n = 600

CI : p̂± z∗√

p̂(1−p̂)n

⇒ 0.72± 1.96√

0.72(1−0.72)600

⇒ (0.684, 0.756)

5. I think the proportion of people that eat candy is around 0.75. I am going to constructa 90% confidence interval and want the margin of error to be ±0.025. How large shouldthe sample size be?

p̂ = 0.75z∗ = 1.645MOE = 0.025

MOE = z∗√

p̂(1−p̂)n

⇒ 0.025 = 1.645√

0.75(1−0.75)n

⇒ 0.0251.645

=√

0.1875n

⇒(

0.0251.645

)2= 0.1875

n

⇒ n = 0.1875

( 0.0251.645)

2

⇒ n = 811.8075⇒ n ≈ 812

6. Jimmy samples 930 people and 234 took public transportation. Construct a 99%confidence interval for the population proportion.

p̂ = 234930

z∗ = 2.576n = 930

CI : p̂± z∗√

p̂(1−p̂)n

⇒ 234930± 2.576

√(234/930)(1−234/930)

930

⇒ 0.252± 2.576√

0.188930

⇒ (0.215, 0.289)

7. I am going to construct a 95% confidence interval for the proportion of people thatwear eyeglasses and want the margin of error to be ±0.2. I have no idea what toestimate for the population proportion. How large should the sample size be?

p̂ = 0.5whenwedon′thaveanyideaforthepopulationproportionz∗ = 1.96MOE = 0.2

MOE = z∗√

p̂(1−p̂)n

⇒ 0.2 = 1.96√

0.5(1−0.5)n

⇒ 0.21.96

=√

0.25n

⇒(

0.21.96

)2= 0.25

n

⇒ n = 0.25

( 0.21.96)

2

⇒ n = 24.01⇒ n ≈ 25

8. A researcher believes that more than 50% of all people voted in the last election. Shesamples 800 people and 420 of them voted. Test her claim at a significance level of0.05 (i.e. compare the P-value to 0.05).


H0 : p = 0.5HA : p > 0.5

(b) Compute the test statistics (z-value). You must show your computation to receivecredit.p̂ = 420/800 = 0.525. n = 800.

z = p̂−p0√p0(1−p0)

n

= 0.525−0.5√0.5(1−0.5)

800

= 0.025√0.25800

= 1.41


P (Z > 1.41) = normalcdf(1.41, 999) = 0.0793

(d) Make a conclusion about the hypotheses.Since the p-value is “large” (0.0793 > 0.05), Do Not Reject H0. The results arenot significant.

9. A researcher believes that fewer than 75% of all mollusks are tasty. He samples 1200mollusks and 865 of them are tasty. Test his claim at a significance level of 0.05 (i.e.compare the P-value to 0.05).


H0 : p = 0.75HA : p < 0.75


z = p̂−p0√p0(1−p0)

n

= 0.721−0.75√0.75(1−0.75)

1200

= −0.029√0.18751200

= −2.32


P (Z < −2.32) = normalcdf(−999,−2.32) = 0.0102

(d) Make a conclusion about the hypotheses.Since the p-value is “small” (0.0102 < 0.05), Reject H0. The results are significant.

10. A researcher believes that the percentage of people that watch Game of Thrones isdifferent than 27%. He samples 900 people and 220 of them watch. Test his claim ata significance level of 0.05 (i.e. compare the P-value to 0.05).


H0 : p = 0.27HA : p 6= 0.27


z = p̂−p0√p0(1−p0)

n

= 0.244−0.27√0.27(1−0.27)

900

= −0.026√0.1971900

= −1.76

(c) Compute the P-value associated with your test statistic.Note that this is a 2-sided test.

p-value= 2P (Z < −1.76)= 2 (normalcdf(−999,−1.76))= 2(0.039)= 0.078

(d) Make a conclusion about the hypotheses.Since the p-value is “large” (0.078 > 0.05), Do Not Reject H0. The results arenot significant.

11. A butcher wants to estimate the mean weight of a ham. She samples 33 hams andcomputes a sample mean weight of 8.2 pounds and a sample standard deviation of 3.3pounds. What is a 90% confidence interval for the population mean weight of ham?Please indicate the value you used for z∗ or t∗.Summary of what is given:

n = 33y = 8.2s = 3.3.

For confidence intervals for the mean, we use t∗, with n − 1 degrees of freedom and90% confidence (for this case). Thus, t∗32 = 1.694.

CI : y ± t∗n−1s√n

⇒ 8.2± 1.694 3.3√33

⇒ (7.227, 9.173)

12. A professor is interested in the mean length of a letter of recommendation. He samples51 letters and finds a sample mean length of 620 words with a sample standard deviationof 90 words. What is a 95% confidence interval for the population mean length of aletter? Please indicate the value you used for z∗ or t∗.Summary of what is given:

n = 51y = 620s = 90.

For confidence intervals for the mean, we use t∗, with n − 1 degrees of freedom and95% confidence (for this case). Thus, t∗50 = 2.009.


⇒ 620± 2.009 90√51

⇒ (594.682, 645.318)

13. A computer professional wants to know the mean number of emails people receive eachday. She is going to compute a 95% confidence interval and wants a margin of error of±2 emails. She believes the standard deviation to be 18 emails. How large should thesample size be to ensure this margin of error? Summary of what is given:

MOE = 2s = 18

For sample size calculation, since this is based on the mean, use t∗, with n− 1 degreesof freedom. Note that as n becomes really large, the t-distribution becomes more likethe normal distribution. Therefore, use the 95% confidence interval critical value fromthe normal distribution instead. z∗ = 1.96. Sample size can be calculated as follows:

MOE = t∗n−1s√n⇒ 2 = 1.96 18√

n

⇒ 21.96×18

= 1√n

⇒√n = 1.96×18

2

⇒ n =(

1.96×182

)2

⇒ n = 311.1696⇒ n ≈ 312

14. A researcher believes that the mean age at which a person first votes is greater than 22years. He samples 27 people and computes a sample mean of 24.3 years and a samplestandard deviation of 8 years.


H0 : µ = 22HA : µ > 22

(b) What is the value of your test statistic (t or z value)?Use the t-test statistic because we are dealing with means.

tn−1 = y−µ0s√n

t27−1 = t26 = 24.3−228√27

= 2.31.54

= 1.49

(c) What is the P-value?On your calculator: tcdf(1.49, 999, 26) = 0.0741On the table: Go to degrees of freedom 26, find where 1.49 is in the row, and thenlook at the one-tail probability values. The probability is between 0.05 and 0.10.

(d) What conclusion should be drawn (compare p-value to 0.05).Since the p-value is “large” (0.0741 > 0.05), Do Not Reject H0. The results arenot significant.

15. A researcher believes that the mean age at which a person first tries chocolate is lessthan 3 years. He samples 24 people and computes a sample mean of 2.3 years and asample standard deviation of 1.5 years.


H0 : µ = 3HA : µ < 3

(b) What is the value of your test statistic (t or z value)?Use the t-test statistic because we are dealing with means.


t24−1 = t23 = 2.3−31.5√24

= −0.70.3062

= −2.286

(c) What is the P-value?On your calculator: tcdf(−999,−2.286, 23) = 0.0159.On the table: Go to degrees of freedom 23, find where 2.286 is in the row, andthen look at the one-tail probability values. The probability is between 0.01 and0.025.

(d) What conclusion should be drawn (compare p-value to 0.05).Since the p-value is “small” (0.0159 < 0.05), Reject H0. The results are significant.

16. A researcher believes that the mean height of a prairie dog is different than 14 inches.She samples 31 prairie dogs and computes a sample mean of 15.8 inches and a samplestandard deviation of 3.6 inches.


H0 : µ = 14HA : µ 6= 14

(b) What is the value of your test statistic (t or z value)? Use the t-test statisticbecause we are dealing with means.


t31−1 = t30 = 15.8−143.6√31

= 1.80.6466

= 2.784

(c) What is the P-value?On your calculator: 2tcdf(2.784, 999, 30) = 2(0.0046) = 0.0092.On the table: go to degrees of freedom 30, find where 2.784 is in the row, andthen look at the two-tail probability values. The probability is lower than 0.01.

(d) What conclusion should be drawn (compare p-value to 0.05).Since the p-value is “small” (0.0092 < 0.05), Reject H0. The results are significant.

17. Which of the following are true? If false, explain briefly.

(a) A very low P-value provides evidence against the null hypothesis.True.

(b) A high P-value is strong evidence in favor of the null hypothesis.False. A high p-value shows that the data are consistent with the null hypothesisbut does not prove that the null hypothesis is true.

(c) A P-value above 0.10 shows that the null hypothesis is true.False. No p-value ever shows that the null hypothesis is true (or false).

(d) If the null hypothesis is true, you can’t get a p-value below 0.01.False. If the null hypothesis is true, you will get a p-value below 0.01 about oncein a hundred hypothesis tests.

18. Which of the following statements are true? If false, explain briefly.

(a) Using an alpha level of 0.05, a p-value of 0.04 results in rejecting the null hypoth-esis.True.

(b) The alpha level depends on the sample size.False. The alpha level is set independently and does not depend on the samplesize.

(c) With an alpha level of 0.01, a p-value of 0.10 results in rejecting the null hypoth-esis.False. The p-value would have to be less than 0.01 to reject the null hypothesis.

(d) Using an alpha level of 0.05, a p-value of 0.06 means the null hypothesis is true.False. It means that we do not have enough evidence at that alpha level to rejectthe null hypothesis.

19. For each of the following situations, state whether a Type I or Type II, or neither errorhas been made. Explain briefly.

(a) A bank wants to know if the enrollment on their website is above 30% based ona small sample of customers. they test H0 : p = 0.3 versus HA : p > 0.3 andreject the null hypothesis. Later they find out that actually 28% of all customersenrolled.Type I Error. The actual value is not greater than 0.3, but they rejected the nullhypothesis.

(b) A student tests 100 students to determine whether other students on her campusprefer Coke or Pepsi and finds no evidence that preference for Coke is not 0.5.Later, a marketing company tests all students on campus and finds no difference.No error. The actual value is 0.5 which was not rejected.

(c) A human resource analyst wants to know if the applicants this year score, onaverage, higher on their placement exam than the 52.5 points the candidatesaveraged last year. She samples 50 recent tests and finds the average to be 54.1points. She fails to reject the null hypothesis that the mean is 52.5 points. Atthe end of the year, they find that the candidates this year had a mean of 55.3points.Type II Error. The actual value was 55.3 points, which is greater than 52.5, whichwas not rejected.

(d) A pharmaceutical company tests whether a drug lifts the headache relief ratefrom the 25% achieved by the placebo. They fail to reject the null hypothesisbecause the p-value is 0.465. Further testing shows that the drug actually relievesheadaches in 38% of people.Type II Error. The null hypothesis was not rejected, but it was false. The truerelief rate was greater than 0.25.

20. We want to estimate the healing rate for a wound. A sample of size 17 is collectedand the sample mean is computed to be 24.3 micrometers per hour, with a samplestandard deviation of s= 8 micrometers per hour. What is a 95% confidence intervalfor the population mean?What we are given:

y = 24.3s = 8n = 17

Because we want the confidence interval for the population mean, use the formula


⇒ 24.3± t∗17−18√17

⇒ 24.3± 2.120 8√17

⇒ (20.187, 28.413)

21. A sample of size n=150 people is collected and the sample proportion of people who areilliterate is computed to be .20. Compute a 95% confidence interval for the populationproportion of illiterate people.What we are given:

n = 150p̂ = 0.2

Because we want the confidence interval for the population proportion, use the formula

CI: p̂± z∗√

p̂(1−p̂)n

⇒ 0.2± 1.96√

0.2(1−0.2)150

⇒ 0.2± 1.96√

0.16150

⇒ (0.136, 0.264)

22. You believe that the proportion of people that like cheese is .80. You are going toconstruct a 95% confidence interval and want the margin of error to be plus or minus.03. What should the sample size be?What we are given:

p̂ = 0.8MOE = 0.03

Since this is dealing with one proportion, use the formula

MOE = z∗√

p̂(1−p̂)n⇒ 0.03 = 1.96

√0.8(1−0.8)

n

⇒ 0.031.96

=√

0.16n

⇒(

0.031.96

)2= 0.16

n

⇒ n = 0.16

( 0.031.96)

2

⇒ n = 682.95⇒ n ≈ 683

23. Teresa knows that appointment times are approximately normally distributed. Shebelieves the mean wait time is longer than 25 minutes. She conducts a test with α= 0.05 and the appropriate hypotheses. She selects 25 random appointments and thesample mean was found to be 25.66 minutes and a sample standard deviation of 10minutes.


H0 : µ = 25HA : µ > 25

(b) Compute the value of the test score.Since this is asking for us to test the population mean, we need the formula


⇒ t25−1 = t24 = 25.66−2510√25

= 0.662

= 0.33

(c) Give the P-value or range of P-values.On your calculator: tcdf(0.33, 999, 24) = 0.372.On the table: go to degrees of freedom 24, find where 0.33 is on the row, and thenlook at the one-tail probability values. The probability is greater than 0.10.

(d) Do the results seem significant?Since the p-value is “large” (0.372 > α = 0.05), Do Not Reject H0. The resultsare not significant.

24. You claim that the proportion of people who watch American Idol is greater than .50.You sample n=200 people and compute a sample proportion of .53. Assume α = 0.05.


H0 : p = 0.5HA : p > 0.5

(b) Compute the value of the test score. Since this is asking for us to test the popu-lation proportion, we need the formula

z = p̂−p0√p0(1−p0)

n

= 0.53−0.5√0.5(1−0.5)

200

= 0.03√0.25200

= 0.03√0.00125

= 0.849

(c) Give the P-value or range of P-values.On your calculator: normalcdf(0.849, 999) = 0.198.

(d) Do the results seem significant?Since the p-value is “large” (0.198 > α = 0.05), Do Not Reject H0. The resultsare not significant.

Stat 101 Exam 3 · 2020-04-27 · Stat 101 Exam 3: Important Formulas and Concepts 1 1 Sampling...

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