ch 7 and 8

7.3 State whether the standardized test statistic t indicates that you should reject the null hypothesis. Explain. ​(a) t=−2.284 ​(b) t=2.231 ​(c) t=2.265 ​(d) t=-2.315

(a) t = -2.284 Reject H0, because t < -2.246 (b) t = 2.231 Fail to reject H0, because -2.246 < t < 2.246 (c) t = 2.265 Reject H0, because t > 2.246 (d) t = - 2.315 Reject H0, because t < -2.246

7.3 State whether the standardized test statistic t indicates that you should reject the null hypothesis. Explain. ​(a) t=1.774 ​(b) t=0 ​(c) t=1.664 ​(d) t=-1.751

(a) t = 1.774 Reject H0, because t > 1.741 (b) t = 0 Fail to reject H0, because t < 1.741 (c) t = 1.664 Fail to reject H0, because t < 1.741 (d) t = -1.751 Fail to reject H0, because t < 1.741

Test the claim about the difference between two population means μ1 and μ2 at the level of significance α. Assume the samples are random and​ independent, and the populations are normally distributed. ​Claim: μ1≤μ2​; α=0.01. Assume σ21≠σ22 Sample​ statistics: x1=2419​, s1=170​, n1=13 and x2=2299​, s2=53​, n2=10 (a) Identify the null and alternative hypotheses (b) Find the standardized test statistic t. & P-value. (c) Reject or Fail to reject

2 SAMPLE T-TEST Right tailed (a) Identify the null and alternative hypotheses H0​: μ1 ≤ μ2 Ha​: μ1 > μ2 (b) Find the standardized test statistic t. & P-value. STAT TESTS 4: 2 sampTTest use > symbol and select NO for "pooled" t = 2.4 P = .015 (c) Reject or Fail to reject Fail to reject H0. There is not enough evidence at the 1% level of significance to reject the claim.

Construct a 95% confidence interval for μ1−μ2 with the sample statistics for mean cholesterol content of a hamburger from two fast food chains and confidence interval construction formula below. Assume the populations are approximately normal with unequal variances. Stats x1=128 mg, s1=3.88 mg, n1=20 x2=119 mg, s2=2.06 mg, n2=16

2 SampTInt 2 Tailed (a) enter the endpoints of the interval STAT TESTS 0: 2-SampTInt x1: 128 S1: 3.88 n1: 20 x2: 119 S2: 2.06 n2: 16 C-Level: .95 Pooled: NO = 7 , 11

A past survey of 1,068,000 students taking a standardized test revealed that 8.4​% of the students were planning on studying engineering in college. In a recent survey of 1,476,000 students taking the​ SAT, 9.2​% of the students were planning to study engineering. Construct a 90​% confidence interval for the difference between proportions p1−p2 by using the following inequality. Assume the samples are random and independent.

2-PropZInt First figure the proportions x1 = 1068000 @ 8.4% = 89712 x2 = 1476000 @ 9.2% = 135792 STAT TESTS B: 2-PropZInt x1: 89712 n1: 1068000 x2: 135792 n2: 1476000 C-Level: .90 = -.009 < p1 -p2 < -.007

7.2 α=0.05​ H0​: μ≤1220 Ha​:μ>1220 p=.100

Fail to reject H0 not enough evidence to reject the claim

QR Find the critical​ value(s) for a​ left-tailed z-test with α=0.02. Include a graph with your answer.

ZCRITLEV Left tailed PRGM ZCRITLEV SIGLEVEL: .02 LEFT TAIL? 1 zₒ = -2.05

7.2 Match each​ P-value with the graph that displays its area without performing any calculations. Explain your reasoning. P=0.0629 and P=0.3222.

(a) displays the area for P=0.3222 (b) displays the area for P=0.0629

7.2 A scientist estimates that the mean nitrogen dioxide level in a city is greater than 35 parts per billion. To test this​ estimate, you determine the nitrogen dioxide levels for 31 randomly selected days. The results​ (in parts per​ billion) are listed to the right. Assume that the population standard deviation is 11. At α=0.03​, can you support the​ scientist's estimate? Complete parts​ (a) through​ (e). μ≥35 α=.03 σ=​11 x=30 sx=9.57 n=31 z=-2.53 ​H0: μ≤35 Ha​: μ>35 (claim) invnorm .03 = -1.88,1.88 crit values=1.88 z=-2.53 P=. normalcdf (-9999,-2.53)= .006 reject

(a) ​H0: μ≤35 Ha​: μ>35 (claim) (b) Because Ha contains a​ greater-than inequality​ symbol, this is a​ right-tailed test.

7.2 A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 740 hours. A random sample of 28 light bulbs has a mean life of 726 hours. Assume the population is normally distributed and the population standard deviation is 63 hours. At α=0.02​, do you have enough evidence to reject the​ manufacturer's claim? Complete parts​ (a) through​ (e). μ≥740 α=.02 σ=​63 x=726 n=28 ​H0: μ≥740 (claim) Ha​: μ<740 invnorm .02 = -2.05,2.05 crit values=-2.05 P=.238 normalcdf (-9999,-1.18) z=-1.18 fail to reject

(a) H0: μ≥740 (claim) Ha​: μ<740 (b) find the critical values invnorm (.02)= -2.05 (c) identify standardized test stat stat test z: z test z= -1.18 p=. (d) fail to reject not enough evidence to reject the claim

7.4 Eric randomly surveyed 150 adults from a certain city and asked which team in a contest they were rooting​ for, either North High School or South High School. Of the surveyed​ adults, 96 said they were rooting for North High while the rest said they were rooting for South High. Eric wants to determine if this is evidence that more than half the adults in this city will root for North High School. (a) Can Eric use the​ one-sample z-methods for proportions to perform a hypothesis test in this​ problem? (b) Suppose a​ p-value from the correct hypothesis test was 0.0030. Which of the following is a correct interpretation of this​ p-value?

(a) Can Eric use the​ one-sample z-methods for proportions to perform a hypothesis test in this​ problem? Yes, because all the conditions are satisfied. The variable of interest is categorical with two​ categories, the randomization condition is​ met, and the​ "success/failure" condition is met. We can not be absolutely sure if the​ 10% condition is met and 150 is less than​ 10% of the number of adults in the​ city, because it is not known how many adults there are in this city.​ But, with two high​ schools, this is likely met.​ Also, it cannot be said for sure if the independence condition is met. Perhaps one person is rooting for North High because a friend of theirs is. It is reasonable to assume that this condition is met. As a random sample was​ taken, the chance of two people in the sample knowing each other is​ reduced, so this assumption is reasonable as well. (b) Suppose a​ p-value from the correct hypothesis test was 0.0030. Which of the following is a correct interpretation of this​ p-value? If half of all adults in this city root for North​ High, 3 out of every 1000 random samples of the same size from this population would produce the same result observed in this study or a result more unusual.

QR Match the alternative hypothesis shown below with its graph to the right. Then state the null hypothesis and sketch its graph. (a) Ha​: σ > 2 (b) State & graph Ho

(a) Ha: σ > 2 empty circle @ 2 and points right (b) State & graph Ho Ho: σ ≤ 2 solid circle on 2 pointing left

7.3 You are testing a claim and incorrectly use the normal sampling distribution instead of the​ t-sampling distribution. Does this make it more or less likely to reject the null​ hypothesis? Is this result the same no matter whether the test is​ left-tailed, right-tailed, or​ two-tailed? Explain your reasoning.

(a) Is the null hypothesis more or less likely to be​ rejected? Explain. More likely; for degrees of freedom less than​30, the tail of the curve are thicker for a t-sampling distribution.​Therefore, if you incorrectly use a standard normal sampling​ distribution, the area under the curve at the tails will be smaller than what it would be for the​t-test, meaning the critical​value(s) will lie closer to the mean. (b) Is the result the​ same? The result is the same. In each​ case, the tail thickness affects the location of the critical​ value(s).

7.3 State whether the standardized test statistic t indicates that you should reject the null hypothesis. (see pic to right) Explain. ​(a) t=1.878 ​(b) t=0 ​(c) t=−1.734 ​(d) t=-1.853

(a) t = 1.878 Fail to reject H0, because t > -1.811 (b) t = 0 Fail to reject H0, because t > -1.811 (c) t = -1.734 Fail to reject H0, because t > -1.811 (d) t = -1.853 Reject H0, because t < -1.811

QR The​ P-value for a hypothesis test is shown. Use the​ P-value to decide whether to reject H0 when the level of significance is​ (a) α=0.01​ (b) α=0.05 ​ (c) α=0.10 P=0.1175

(a) α=0.01​ Fail to reject Ho because the P-value, 0.1175, is greater than α=0.01​ (b) α=0.05 ​ Fail to reject Ho because the P-value, 0.1175, is greater than α=0.05 (c) α=0.10 Fail to reject Ho because the P-value, 0.1175, is greater than α=0.10

Test the claim about the difference between two population means μ1 and μ2 at the level of significance α. Assume the samples are random and​ independent, and the populations are normally distributed. ​Claim: μ1=μ2​; α=0.10. Assume σ21=σ22 Sample​ statistics: x1=32.9​, s1=3.5​, n1=13 and x2=35.1​, s2=2.3​, n2=15

2 SAMPLE T-TEST 2 Tailed (a) Identify the null and alternative hypotheses ​H0: μ1=μ2 (claim) Ha​: μ1≠μ2 (b) Find the standardized test statistic t. when assuming the population standard deviations are the same, select the "pooled" option. STAT TEST Stats 4: 2-sample t-test enter info Pooled: YES t = -1.99 Find the p value P = .057 (c) Reject or Fail to reject Reject H0. There is enough evidence at the 10% level of significance to reject the claim.

Construct a 90% confidence interval for μ1−μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pies and confidence interval construction formula below. Assume the populations are approximately normal with equal variances. Bakery A x1=1777 s1=158 n1=10 Barker B x2=1600 s2=187 n2=8

2 SampTInt 2 Tailed (a) enter the endpoints of the interval STAT TESTS 0: 2-SampTInt x1: 1777 S1: 158 n1: 10 x2: 1600 S2: 187 n2: 8 C-Level: .90 Pooled: YES = .35 < μ1−μ2 < 319

A past survey of 1,068,000 students taking a standardized test revealed that 7.7​% of the students were planning on studying engineering in college. In a recent survey of 1,476,000 students taking the​ SAT, 9.2​% of the students were planning to study engineering. Construct a 95​% confidence interval for the difference between proportions p1−p2 by using the following inequality. Assume the samples are random and independent.

2-PropZInt First figure the proportions x1 = 1068000 @ 7.7% = 82236 x2 = 1476000 @ 9.2% = 135792 STAT TESTS B: 2-PropZInt x1: 82236 n1: 1068000 x2: 135792 n2: 1476000 C-Level: .95 = -.016 < p1 -p2 < -.014

Use the figure to the​ right, which shows the percentages of adults from several countries who favor building new nuclear power plants in their country. The survey included random samples of 1009 adults from Country​ A, 1069 adults from Country​ B, 1127 adults from Country​ C, and 1023 adults from Country D. At α=0.06​, can you reject the claim that the proportion of adults in Country A who favor building new nuclear power plants in their country is the same as the proportion of adults from Country B who favor building new nuclear power plants in their​ country? Assume the random samples are independent. A = 1009 = 49% P = 494.41 B = 1069 = 47% P = 502.43 C = 1127 = 43% D = 1023 = 34%

2-PropZTest First figure the proportions A = 1009*.49 = 494 B = 1069*.47 = 502 (a) Identify the claim and state Ho and Ha The claim is​ "the proportion of adults in Country A who favor building new nuclear power plants in their country is the same as the proportion of adults from Country B who favor building new nuclear power plants in their​ country." Ho and Ha ​Ho: p1 = p2 (CLAIM) Ha​: p1 ≠ p2 (b) Find the standardized test statistic STAT TESTS 6: 2-PropZTest x1: 494 n1: 1009 x2: 502 n2: 1069 z = .91 P-Value = .362 (c) Conclusions Fail to reject H0 because the​ P-value is greater than the significance level α. (d) Can you reject the original claim? ​No, at the 6​% significance ​level, there is insufficient evidence to reject the claim.

QR *******State whether each standardized test statistic X2 allows you to reject the null hypothesis. Explain. (a) X2 = 22.756 (b) X2 = 0 (c) X2 = 1.899 (d) X2 = 27.287

???

QR ************A snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.18. A dietician is asked to test this claim and finds that a random sample of 19 servings has a variance of 1.07. At α=0.10​, is there enough evidence to reject the​ manufacturer's claim? Assume the population is normally distributed. Complete parts​ (a) through​ (e) below (a) Write the claim mathematically (b) Find the critical values and rejection regions Choose the correct statement (c) Find the standardized test Stat (d) Reject / Fail to reject µ = 1.18 sx2 = 1.07 n = 19 α = .10

??? 2 tailed (a) Write the claim mathematically µ = 1.18 (CLAIM) µ ≠ 1.18 (b) Find the critical values/rejection regions Choose the correct statement (c) standard test Stat first convert variance into standard deviation sx2 = 1.07 √1.07 = 1.034 s = 1.034 (d) Reject / Fail to reject

7.2 A company that makes cola drinks states that the mean caffeine content per​ 12-ounce bottle of cola is 35 milligrams. You want to test this claim. During your​ tests, you find that a random sample of thirty​ 12-ounce bottles of cola has a mean caffeine content of 33.4 milligrams. Assume the population is normally distributed and the population standard deviation is 6.3 milligrams. At α=0.06​, can you reject the​ company's claim? Complete parts​ (a) through​ (e). μ=35 α=0.06 / 2 = .03 σ=6.3​ x=33.4 n=30 H0: μ=35 (claim) Ha​: μ≠35 ZCRIT .03 = -1.88, 1.88 z=-1.39 P=.082 since z is not in the rejection region, fail to reject the null hypothesis (e) At the 6​% significance​ level, there is not enough evidence to reject the​ company's claim that the mean caffeine content per​ 12-ounce bottle of cola is equal to 3535 milligrams

A hypothesis test is​ right-tailed if the alternative hypothesis contains the​ greater-than inequality symbol ​(>​), ​left-tailed if it contains the​ less-than inequality symbol ​(<​), or​ two-tailed if it contains the​ not-equal-to symbol ​(≠​).

Elmo likes music. He wondered if listening to music while studying will improve scores on an exam. Fifty students who were to take the midterm in a week agreed to be part of a study. Half were randomly assigned to listen to classical music while studying for the exam. The other half were told not to listen to any music while studying for the exam. A hypothesis test is to be performed to determine if the average scores of those listening to music while studying for the exam were higher than those who did not listen to any music while studying for the exam. Which of the following hypothesis tests should be​ used?

A two-sample t-test Two groups are being compared and the response variable​ (exam scores) is quantitative. This leads to performing a​ two-sample t-test.

Determine whether a normal sampling distribution can be used for the following sample statistics. If it can be​ used, test the claim about the difference between two population proportions p1 and p2 at the level of significance α. Assume that the samples are random and independent. ​Claim: p1≠p2​, α=0.01 Sample​ Statistics: x1=37​, n1=72​, x2=36​, n2=62

CKp̂ (custom program) 2-PropZTest (a) Determine whether a normal sampling distribution can be used PRGM CKPEEHAT The samples are random and independent. A normal sampling distribution can be used because n1p=39.22, n1q=32.78, n2p=33.78​, and n2q=28.22 State the null & alternative hypotheses ​Ho: p1 = p2 Ha​: p1 ≠ p2​ (claim) (b) Calculate the standardized test stat STAT TESTS 6: 2-PropZTest z = -.77 (c) Conclusions Since P>α​, fail to reject H0. There is not enough evidence at the α=0.01 level of significance to support the claim

Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the difference between two population proportions p1 and p2 at the given level of significance α using the given sample statistics. Assume the sample statistics are from independent random samples. ​Claim: p1=p2​, α=0.10 Sample​ statistics: x1=43​, n1=105 x2=49​, n2=206

CKp̂ (custom program) InvT 2-PropZTest 2 Tailed (a) Determine whether a normal sampling distribution can be used PRGM CKPEEHAT YES (b) Find the critical values PRGM InvT AREA LEFT: .05 DF: 309 (N1 + N2 - 2) = -1.65 , 1.65 (c) Conclusion Reject H0. There is sufficient evidence that there is a difference between p1 and p2.

QR Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the population proportion p at the given level of significance α using the given sample statistics. ​Claim: p<0.14 α=0.05​ Sample​ statistics: p̂=0.12​ n=25

Can normal sampling be used? No because np is less than 5

QR Determine whether the claim stated below represents the null hypothesis or the alternative hypothesis. If a hypothesis test is​ performed, how should you interpret a decision that​ (a) rejects the null hypothesis or​ (b) fails to reject the null​ hypothesis? A scientist claims that the mean incubation period for the eggs of a species of bird is at least 56 days. Ho ≤ 56 (CLAIM)

Does the claim represent the null hypothesis or alternative hypothesis? Since the claim contains a statement of equality, it represents the null hypothesis. (a) How should you interpret a decision that rejects the null hypothesis? There is sufficient evidence to reject the claim that the mean incubation period for the eggs of a species of bird is at least 56 days. (b) How should you interpret a decision that fails to reject the null hypothesis? There is insufficient evidence to reject the claim that the mean incubation period for the eggs of a species of a bird is at least 56 days.

What conditions are necessary in order to use the z​-test to test the difference between two population​ proportions?

Each sample must be randomly​ selected, independent, and n1p1, n1q1, n2p2, and n2q2 must be at least five.

7.2 Claim: μ≤1240​ α=0.01 σ=210.96 x=1263.09 n=200 H0​: μ≤1240 Ha​:μ>1240 Z=1.55 p=.061

Fail to reject H0 there is not enough evidence to reject

7.3 Find the critical​ value(s) and rejection​ region(s) for the indicated​ t-test, level of significance α​, and sample size n. Two​-tailed test α=0.02​ n=11

INVT (custom program) 2 tailed test Since the test is two-tailed, divide the level of significance by 2. PRGM TCINTVAL AREA LEFT: .01 DF: 10 ENTER = -2.764 (a) Critical values = -2.764, 2.764 (b) Rejection regions = t < -2.764 and t > 2.764

7.3 Find the critical​ value(s) and rejection​ region(s) for the indicated​ t-test, level of significance α​, and sample size n. Right​-tailed test α=0.10​ n=16

INVT (custom program) Right tailed test Since the test is right-tailed, you can either: 1) subtract 1-0.1 = .90 for "AREA LEFT" or 2) use .10 as "AREA LEFT", and leave off the negative sign in the answer. PRGM TCINTVAL AREA LEFT: .90 DF: 10 ENTER = 1.341 (a) Critical value = 1.341 (b) Rejection regions = t > 1.341

7.3 Find the critical​ value(s) and rejection​ region(s) for the indicated​ t-test, level of significance α​, and sample size n. Left​-tailed test, α=0.10​, n=11

INVT(custom program) Left tailed test PRGM TCINTVAL AREA LEFT: .10 DF: 10 ENTER = -1.372 (a) Critical value = -1.372 (b) Rejection region = t < -1.372

7.3 A used car dealer says that the mean price of a​ three-year-old sports utility vehicle is $23,000. You suspect this claim is incorrect and find that a random sample of 22 similar vehicles has a mean price of $23,838 and a standard deviation of $1968. Is there enough evidence to reject the claim at α=0.10​? Complete parts​ (a) through​ (e) below. Assume the population is normally distributed α= 0.10 μ= 23000 x= 23838 s= 1968 n= 22

INVT(custom program) STAT T-Test 2 tailed test (a) Identify H0 and Ha ​H0: μ =​ $23,000 (CLAIM) Ha​: μ ≠​ $23,000 (b)Critical values PRGM TCINTVAL AREA LEFT: .05 DF: 21 ENTER = -1.721 =-1.721, 1.721 (c) standardized test statistic STAT TESTS 2: T-Test Stats μ: 23000​ x: 23838 Sx: 1968 n: 22 ​μ: ≠​μ Calculate ENTER = 2 (d) reject or fail to reject Reject H0 because the test statistic is in the rejection regions (e) Interpret At the 10% level of significance, there is sufficient evidence to reject the claim that the mean price is $23,000.

7.2 Find the critical​ value(s) for a​ left-tailed z-test with α=0.11. Include a graph with your answer.

InvNorm Left tailed 2nd vars 3: invnorm (.11) = -1.23

A pet association claims that the mean annual cost of food for dogs and cats are the same. The results for samples for the two types of pets are shown below. At α=0.05​, can you reject the pet​ association's claim? Assume the population variances are not equal. Assume the samples are random and​ independent, and the populations are normally distributed. Complete parts​ (a) through​ (e) below. Dogs x1 = 201 s1 = 43 n1 = 17 CATS x2 = 166 s2 = 39 n2 = 14

InvT 2 SAMPLE T-TEST 2 Tailed test (a) identify the claim and state Ho & Ha. The correct claim is: ​"The mean annual costs of food for dogs and cats are​ equal." What are Ho & Ha: Ho: μ1 = μ2 Ha: μ1 ≠ μ2 Which hypothesis is the claim? Ho (b) Find the critical values & identify the rejection regions Population variance are not equal so use the lower sample# minus 1. PRGM invT area to left: .025 DF: 13 = -2.16,2.16 Rejection regions: t < -to, t > to (c) Find the standardized test statistic STAT TEST 4: 2 sample ttest "POOLED" NO t = 2.374 P= .025 (d) Reject the null? Reject the null At the 5% significance level, there is enough evidence to reject the claim that the mean annual costs of food for dogs and cats are the same.

A researcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of a sample of 13 blue crabs from Location A contain a mean of 191 milligrams of fish and a standard deviation of 36 milligrams. The stomach contents of a sample of 9 blue crabs from Location B contain a mean of 189 milligrams of fish and a standard deviation of 43 milligrams. At α=0.10​, can you support the​ researcher's claim? Assume the population variances are equal. (will use n1+n2-2 for DF) Complete parts​ (a) through​ (d) below. Location A x1=191 s1=36 n1=13 Location B x2=189 s2=43 n2=9

InvT 2 SAMPLE T-TEST Right tailed (a) Identify H0 &Ha ​Ho: μ1−μ2 ≤ 0 Ha​: μ1−μ2 > 0 (claim) (b) Find the standardized test statistic STAT TESTS 4: 2sample t-test Stats select > Yes "Pooled" t = .118 (c) Find the P-value P = .4535 (d) State conclusion Fail to reject H0. There is not enough evidence at the 10​% level of significance to support the​researcher's claim.

A magazine claims that the mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry World. The results for samples of customer transactions for the two fast food restaurants are shown below. At α=0.01​, can you support the​ magazine's claim? Assume the population variances are equal. Assume the samples are random and​ independent, and the populations are normally distributed. Complete parts​ (a) through​ (e) below. Burger Stop x1=8.83 s1=.83 n1=12 Fry World x2=8.20 s2=.72 n2=10 *when assuming the population variances are equal: 1: use n1+n2 -2 2: select yes for "POOLED"

InvT 2 SAMPLE T-TEST Right tailed (a) identify the claim and state Ho & Ha ​"The mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry​ World." state Ho & Ha Ho: μ1 ≤ μ2 Ha: μ1 ≥ μ2 (claim) which hypothesis is the claim? Ha (b) Find the critical values PRGM invT AREA LEFT: .01 DF: 20 (since variances are equal, n1+n2 - 2) t = 2.528 rejection regions: t > to (c) Find the standardized test statistic STAT TESTS 4: 2sample t-test Stats select > Yes "Pooled" t = 1.881 (d) Support or reject null Fail to reject the null (e) Interpret At the 1% significance​ level, there is not enough evidence to support the claim of the magazine that the mean amount spent by customers at Burger World is greater than the mean amount spent by customers at Fry World.

An engineer wants to know if producing metal bars using a new experimental treatment rather than the conventional treatment makes a difference in the tensile strength of the bars​ (the ability to resist tearing when pulled​ lengthwise). At α=0.02​, answer parts​ (a) through​ (e). Assume the population variances are equal and the samples are random. If​ convenient, use technology to solve the problem. Experimental 388 392 354 446 444 359 446 Conventional 387 354 434 416 402 416 445 364 379 439

InvT 2 sample t-test 2 Tailed (a) identify the claim & state Ho & Ha The claim is​ "The new treatment makes a difference in the tensile strength of the​ bars." state Ho & Ha Ho: μ1 = μ2 Ha: μ1 ≠ μ2 (claim) Which hypothesis is the claim? Ha (b) Find the critical values & rejection regions PRGM invT AREA LEFT: .01 DF: 15 (n1+n2 -2 bc variances are equal) t = -2.602,2.602 Rejection regions t < -to, t > to (c) Find standardized test stat STAT TESTS 4: 2sample t-test Data select = Yes "Pooled" t = .031 (d) reject or fail to reject Fail to reject the null hypothesis (e) Interpret At the 2% significance​ level, there is not enough evidence to support the claim.

A​ state-by-state survey found that the proportions of adults who are smokers in state A and state B were 22.9​% and 17.6​%, respectively.​ (Suppose the number of respondents from each state was 3000​.) At α=0.10​, can you support the claim that the proportion of adults who are smokers is lower in state A than in state​ B? Assume the random samples are independent. Complete parts​ (a) through​ (e). A = 3000 22.9% = 687 B = 3000 17.6% = 528

InvT 2-PropZTest Left Tailed Figure the proportions A = 3000*.229 = 687 B = 3000*.176 = 528 (a) Identify the claim and state Ho and Ha The claim is​ "the proportion of adults who are smokers in state A is lower than the proportion of adults who are smokers in state​ B." Ho and Ha H0​: p1 ≥ p2 Ha​: p1 < p2 (claim) (b) Critical values and rejection regions PRGM InvT AREA LEFT: .1 DF: 5998 zₒ = -1.28 Identify rejection regions z < -1.28 (c) Find the standardized test statistic STAT TESTS 6: 2-PropZTest x1: 687 n1: 3000 x2: 528 n2: 3000 z = 5.11 P-Value = .999 (d) Conclusions Fail to reject Ho (e) Interpret the decision At the 10​% significance​ level, there is insufficient evidence to support the claim.

Use the figure to the​ right, which shows the percentages of adults from several countries who favor building new nuclear power plants in their country. The survey included random samples of 1115 adults from Country​ A, 1025 adults from Country​ B, 1000 adults from Country​ C, and 1055 adults from Country D. At α=0.05​, can you reject the claim that the proportion of adults in Country C who favor building new nuclear power plants in their country is greater than the proportion of adults from Country D who favor building new nuclear power plants in their​ country? Assume the random samples are independent. x1 = 460 (.46 * 1000) n1 = 1000 @ 46% x2 = 422 (.4 * 1055) n2 = 1055 @ 40%

InvT 2-PropZTest Right Tailed Figure the proportions x1 = 1000*.46 = 460 B = 1055*.4 = 422 (a) Identify the claim and state Ho and Ha The claim is​ "the proportion of adults in Country C who favor building new nuclear power plants in their country is greater than the proportion of adults from Country D who favor building new nuclear power plants in their​ country." Ho and Ha H0​: p1 ≤ p2 Ha​: p1 > p2 (claim) (b) Critical values and rejection regions PRGM InvT AREA LEFT: .05 DF: 2053 (1000+1055-2) zₒ = 2.75 P-value = .003 (c) Find the standardized test statistic STAT TESTS 6: 2-PropZTest x1: 687 n1: 3000 x2: 528 n2: 3000 z = 5.11 P-Value = .999 (d) Conclusions Reject H0 because the​P-value is less than the significance level α. (e) Can you support the original claim? ​Yes, at the 5​% significance​ level, there is sufficient evidence to support the claim.

Refer to the figure at the right. At α=0.05​, can you support the claim that the proportion of men ages 18 to 24 living in​ parents' homes was greater in 1998 than in 2006​? Assume the survey included 12,000 men and 11,900 women in 1998 and 14,200 men and 14,000 women in 2006​, and assume the samples are random and independent. SAMPLE SIZE MEN 1998 = 12000 @ 60.9% 2006 = 14200 @ 56.3% WOMEN 1998 = 11900 @ 46.3% 2006 = 14000 @ 50.8% PROPORTIONS MEN 1998 = 12000 * .609 = 7308 2006 = 14200 *.563 = 7995

InvT 2-PropZTest Right Tailed Figure the proportions x1 = 12000*.609 = 7308 B = 14200*.563 = 7995 (a) Identify Ho and Ha H0​: p1≤p2 Ha​: p1 > p2 (claim) (b) Find the critical value PRGM InvT AREA LEFT: .05 DF: 26198 (12000+14200 - 2) zₒ = 1.64 (because it is right tailed the answer is positive) (c) Find the standardized test stat STAT TESTS 6: 2-PropZTest x1: 7308 n1: 12000 x2: 7995 n2: 14200 z = 7.52 (d) Can you support the claim? Yes, at the 5​% significance​level, there is sufficient evidence to support the claim.

Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the difference between two population proportions p1 and p2 at the given level of significance α using the given sample statistics. Assume the sample statistics are from independent random samples. ​Claim: p1=p2​, α=0.10 Sample​ statistics: x1=43​, n1=105 x2=49​, n2=206

InvT Can normal sampling distribution be used? Yes Identify Ho & Ha H0​: p1=p2 Ha​: p1≠p2 Find the critical values PRGM InvT AREA LEFT: .05 DF: 309 = -1.65, 1.65

Use the​ t-distribution table to find the critical​ value(s) for the indicated alternative​ hypotheses, level of significance α​, and sample sizes n1 and n2. Assume that the samples are​ independent, normal, and random. Answer parts​ (a) and​ (b). Ha​: μ1<μ2​, α=0.10​, n1=6​, n2=8 (a) Find the critical values assuming population variances are equal (b) Find the critical values assuming population variances are not equal

InvT Left tailed (a) Find the critical values assuming population variances are equal PRGM InvT AREA LEFT: .1 DF: 12 (n1 + n2 - 2) t0 = -1.356 (b) Find the critical values assuming population variances are not equal RGM InvT AREA LEFT: .1 DF: 12 (n1 - 1) n1 bc its lowest t0 = -1.476

Use the​ t-distribution table to find the critical​ value(s) for the indicated alternative​ hypotheses, level of significance α​, and sample sizes n1 and n2. Assume that the samples are​ independent, normal, and random. Answer parts​ (a) and​ (b). Ha​:μ1>μ2​, α=0.025​, n1=13​, n2=11 (a) Find the critical values assuming population variances are equal (b) Find the critical values assuming population variances are not equal

InvT Right tailed (a) Find the critical values assuming population variances are equal PRGM InvT AREA LEFT: .025 DF: 22 (n1 + n2 - 2) t0 = 2.074 (b) Find the critical values assuming population variances are not equal RGM InvT AREA LEFT: .025 DF: 10 (n2 - 1) n1 bc its lowest t0 = 2.228

An instructor for a test preparation course claims that the course will improve the test scores of students. The table shows the critical reading scores for 14 students the first two times they took the test. Before taking the test for the second​ time, the students took the​ instructor's course to try to improve their critical reading test scores. At a​ = 0.01, is there enough evidence to support the​ instructor's claim? Complete parts​ (a) through​ (f).

InvT STAT EDIT STAT T-TEST Left Tailed Enter data into L1 and L2 at the top of L3 L1-L2 ENTER (a) identify the claim and state Ho and Ha The claim is​ "The students' critical reading test scores improved the second time they took the​ test." Ho and Ha H0​: μd ≥ 0 Ha​: μd < 0 (claim) (b) Find the critical values & rejection regions PRGM InvT AREA LEFT: .01 DF: 13 tₒ = -2.65 Identify the rejection regions t < -2.650 (c) Calculate d̅ and sd STAT CALC 1: 1 VAR STATS 2nd 3 (L3) ENTER d̅ = x̅= -38.643 sd = 46.256 (d) Use the t-test to find standard test stat t. STAT TEST 2: T-Test Data 2nd 3 (L3) µ < t = -3.126 (e) Decision Reject the null hypothesis (f) Interpret decision At the​ 1% significance​ level, there is enough evidence that the​ students' critical reading scores improved the second time they took the test.

A physical therapist claims that one​ 600-milligram dose of Vitamin C will increase muscular endurance. The table available below shows the numbers of repetitions 15 males made on a hand dynamometer​ (measures grip​ strength) until the grip strengths in three consecutive trials were​ 50% of their maximum grip strength. At α=0.01​, is there enough evidence to support the ​ therapist's claim? Assume the samples are random and​ dependent, and the population is normally distributed. Complete parts​ (a) through​ (e) below.

InvT STAT EDIT STAT T-TEST LeftTailed Enter data into L1 and L2 at the top of L3 L1-L2 ENTER (a) identify the claim and state Ho and Ha Using Vitamin C increases muscular endurance. Ho and Ha H0​: μd ≥ 0 Ha​: μd < 0 (claim) (b) Find the critical values & rejection regions PRGM InvT AREA LEFT: .01 DF: 14 tₒ = -2.624 Identify the rejection regions t > -2.624 (c) Calculate d̅ and sd STAT CALC 1: 1 VAR STATS 2nd 3 (L3) ENTER d̅ = x̅= 75.8 sd = 276.2 (d) Use the t-test to find standard test stat t. STAT TEST 2: T-Test Data 2nd 3 (L3) µ < t = 1.063 (e) Decision Fail to reject the null hypothesis. There is not enough evidence to support the claim that using Vitamin C increases muscular endurance.

A scientist claims that pneumonia causes weight loss in mice. The table shows the weights​ (in grams) of six mice before infection and two days after infection. At α=0.05​, is there enough evidence to support the​ scientist's claim? Assume the samples are random and​ dependent, and the population is normally distributed. Complete parts​ (a) through​ (e) below.

InvT STAT EDIT STAT T-TEST Right Tailed Enter data into L1 and L2 at the top of L3 L1-L2 ENTER (a) Identify the claim Pneumonia causes weight loss in mice. What are Ho and Ha? ​Ho: μd ≤ 0 Ha​: μd > 0 (claim) (b) Find the critical values & rejection regions PRGM InvT AREA LEFT: .05 DF: 5 = tₒ > 2.02 (c) Calculate d and sd STAT CALC 1: 1 VAR STATS 2nd 3 (L3) ENTER d̅ = x̅= .95 sd = .345 (d) Find the standard statistic t STAT TEST 2: T-Test µ: 0 x̅: .95 sd: .345 n: 6 µ > t = 6.746 P=5.43 (e) Inference Reject the null hypothesis. There is enough evidence to support the claim that pnuemonia causes weight loss in mice.

A researcher claims that a​ post-lunch nap decreases the amount of time it takes males to sprint 20 meters after a night with only 4 hours of sleep. The table shows the amounts of time​ (in seconds) it took for 10 males to sprint 20 meters after a night with only 4 hours of sleep when they did not take a​ post-lunch nap and when they did take a​ post-lunch nap. At α=0.05​, is there enough evidence to support the​ researcher's claim? Assume the samples are random and​ dependent, and the population is normally distributed. Complete parts​ (a) through​ (e) below.

InvT STAT EDIT STAT T-TEST RightTailed Enter data into L1 and L2 at the top of L3 L1-L2 ENTER (a) identify the claim and state Ho and Ha A​ post-lunch nap decreases the amount of time it takes males to sprint 20 meters. Ho and Ha H0​: μd ≤ 0 Ha​: μd > 0 (b) Find the critical values & rejection regions PRGM InvT AREA LEFT: .05 DF: 9 tₒ = 1.833 Identify the rejection regions t > 1.833 (c) Calculate d̅ and sd STAT CALC 1: 1 VAR STATS 2nd 3 (L3) ENTER d̅ = x̅= .006 sd = .018 (d) Use the t-test to find standard test stat t. STAT TEST 2: T-Test Data 2nd 3 (L3) µ > t = 1.068 (e) Decision Fail to reject the null hypothesis. There is not enough evidence to support the claim that a post dash lunch nappost-lunch nap decreases the amount of time it takes males to sprint 20 meters.

Use the​ t-distribution table to find the critical​ value(s) for the indicated alternative​ hypotheses, level of significance α​, and sample sizes n1 and n2. Assume that the samples are​ independent, normal, and random. Answer parts​ (a) and​ (b). Ha​: μ1≠μ2​, α=0.02​, n1=15​, n2=8 (a) Find the critical values assuming population variances are equal (b) Find the critical values assuming population variances are not equal

InvT (custom program) 2 tailed (a)Find the critical values assuming the population variances are equal 2 tailed test so divide .02 / 2 = .01 DF = N1 +N2 - 2 = 21 PRGM InvT AREA LEFT: .01 DF: 21 T0 = -2.518,2.518 (b) critical values assuming the population variances are not equal DF = lower of n1 or n2 numbers minus 1 PRGM InvT AREA LEFT: .01 DF: 7 t0 = -2.998,2.998

7.2 Find the critical​ value(s) and rejection​ region(s) for the type of​ z-test with level of significance α. Include a graph with your answer. Right​-tailed test, α=0.03

Invnorm (a) 2nd vars 3: invnorm (.03) = 1.88 use + # when right tailed (b) The rejection region is z>1.88 (c)see image

7.4 A humane society claims that less than 69​% of households in a certain country own a pet. In a random sample of 400 households in that​ country, 264 say they own a pet. At α=0.10​, is there enough evidence to support the​ society's claim? Complete parts​ (a) through​ (c) below. H0​: p ≥ .69 Ha​: p < .69 (claim) Right tailed z = -1.31 P = .097 α=0.10 f

PRGM PROPZCLM normalcdf Left tailed test (a) claim scenario Less than 69​% of households in the country own a pet. hypotheses H0​: p ≥ .69 Ha​: p < .69 (claim) (b) standardized test stat z phat = 264/400 = .66 PRGM PROPZCLM p: 69, phat: .66, n:400 Z = -1.31 2nd Vars 2: normalcdf (-9999,-1.31) = .097 P = .097 (c) Reject or fail to reject H0​: p ≥ .69 Ha​: p < .69 (claim) P value is less than α=0.10 = reject the null Reject the null hypothesis. There is enough evidence to support the​society's claim.

QR A research center claims that 26​% of adults in a certain country would travel into space on a commercial flight if they could afford it. In a random sample of 1200 adults in that​ country, 30​% say that they would travel into space on a commercial flight if they could afford it. At α=0.10​, is there enough evidence to reject the research​ center's claim? Complete parts​ (a) through​ (d) below. (a) Identify the claim and state Ho and Ha (b) Identify the standardized test statistic and P-value (c) Reject / Fail to reject P = .26 p̂ = .30 n = 1200 α =.10

PropZclm (a) Identify the claim 25% of adults in the country would travel into space on a commercial flight if they could afford it. State Ho and Ha P = .26 (CLAIM) P ≠ .26 (b) Identify the standardized test statistic PRGM PropZclm P: .26 p̂: .30 N: 1200 z = 3.16 Identify the P-value manually until I learn a better way Look up 3.16 on z chart then subtract # from 1 1-.9992 = .0008 each side add L & R .0008 + .0008 P = .002 (c) Reject / Fail to reject Reject the null hypothesis. There is enough evidence to reject the claim.

7.2​ Claim: μ>1250​; α=0.06​; σ=211.78. Sample​ statistics: x=1273.62​, n=250 ​H0: μ≤1250 Ha​: μ>1250 z=1.76 p=0.39

Reject H0 there is enough evidence to support the claim

7.2 α=0.08​ H0​: μ≤1200 Ha​:μ>1200 p=.040

Reject H0 there is enough evidence to reject the claim

7.2 A nutritionist claims that the mean tuna consumption by a person is 3.2 pounds per year. A sample of 50 people shows that the mean tuna consumption by a person is 2.9 pounds per year. Assume the population standard deviation is 1.17 pounds. At α=0.03​, can you reject the​ claim? α=0.09 x=3.3 σ=1.04 μ=3.5 n=90 ​H0: μ=3.5 Ha​: μ≠3.5 z=-1.82 p=.034 * 2 = .069

Reject H0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5pounds.

Construct a​ 90% confidence​ interval, using the inequality To test the effectiveness of a new drug that is reported to increase the number of hours of sleep patients get during the​ night, researchers randomly select 13 patients and record the number of hours of sleep each gets with and without the new drug.

STAT EDIT STAT TEST T-INTERVAL Enter data into L1 and L2 at the top of L3 L1-L2 ENTER STAT TEST 8: TInterval Data List: L3 Freq: 1 C-Level: .9 = -1.73 <ud <-1.13

Construct a​ 90% confidence​ interval, using the inequality [formula] To test the effectiveness of a new drug that is reported to increase the number of hours of sleep patients get during the​ night, researchers randomly select 13 patients and record the number of hours of sleep each gets with and without the new drug.

STAT EDIT STAT TEST T-INTERVAL Enter data into L1 and L2 at the top of L3 L1-L2 ENTER STAT TEST 8: TInterval Data List: L3 Freq: 1 C-Level: .9 = -1.84 <ud <−1.44.

7.3 The dean of a university estimates that the mean number of classroom hours per week for​ full-time faculty is 11.0. As a member of the student​ council, you want to test this claim. A random sample of the number of classroom hours for eight​ full-time faculty for one week is shown in the table below. At α=0.01​, can you reject the​ dean's claim? Complete parts​ (a) through​ (d) below. Assume the population is normally distributed. 11.8 9.7 11.8 6.6 7.4 10.9 13.5 9.2 α= 0.01 μ = 11 2 tailed test n= 8

STAT T-Test 2 tailed test (a) Identify the claim H0​: μ = 11 Ha​: μ ≠ 11 (b) P-value STAT TESTS 2: T-Test Data μ: 11​ List: L1 (or whichever list you put data in) ​μ: ≠ μ Calculate ENTER p = .32 (c) reject or fail to reject Fail to reject H0 because the​P-value is greater than the significance level. (d) Interpret At the 1​% level of​ significance, there is not sufficient evidence to reject the claim that the mean number of classroom hours per week for​ full-time faculty is 11.0.

7.3 Use a​ t-test to test the claim about the population mean μ at the given level of significance α using the given sample statistics. Assume the population is normally distributed. ​Claim: μ≠24​ α=0.01 Sample​ statistics: x=23.1​ s=5.1​ n=12

STAT T-Test 2 tailed test (a) null and alternative hypotheses H0​: μ = 24 Ha​: μ ≠ 24 (b) VAlue of standardized test statistic STAT TESTS 2: T-Test Stats μ: 24​ x: 23.1 Sx: 5.1 n: 12 ​μ: ≠​μ Calculate ENTER = -.61 (c) P value = .553 ? do you not subtract from 1 when using = sign? (d) Reject or fail to reject Fail to reject H0. There is not enough evidence to support the claim.

7.3 A car company says that the mean gas mileage for its luxury sedan is at least 25 miles per gallon​ (mpg). You believe the claim is incorrect and find that a random sample of 6 cars has a mean gas mileage of 23 mpg and a standard deviation of 5 mpg. At α=0.10​, test the​ company's claim. Assume the population is normally distributed. α= 0.1 μ < 25 left tailed x= 23 s= 5 n= 6

STAT T-Test INVT Left tailed test (a) Which sampling distribution should be used and​ why? Use a​ t-sampling distribution because the population is​ normal, and σ is unknown. (b) State the appropriate hypotheses to test. H0​: μ ≥ 25 Ha​: μ < 25 (c) What is the value of the standardized test​ statistic? STAT TESTS 2: T-Test Stats μ: 25​ x: 23 Sx: 5 n: 6 ​μ: > μ Calculate ENTER = -.98 (d) What is the critical​ value? PRGM INVT AREA LEFT: .1 DF: 5 ENTER = -1.476 (e) What is the outcome and the conclusion of this​ test? For a​ left-tailed test, if the test statistic is less than the critical value reject the null hypothesis.​ Otherwise, fail to reject the null hypothesis. Fail to reject H0. At the 10% significance level, there is insufficient evidence to reject the car company's claim that the mean gas mileage for the luxury sedan is at least 25 miles per gallon.

Test the claim below about the mean of the differences for a population of paired data at the level of significance α. Assume the samples are random and​ dependent, and the populations are normally distributed. ​Claim: μd=​0; α=0.10. Sample​ statistics: d=2.9​, sd=8.27​, n=8

STAT T-Test InvT 2 Tailed (a) Identify the hypotheses ​Ho: μd = 0 (claim) Ha​: μd ≠ 0 (b) The test statistic STAT TESTS 2: T-Test µ: 0 x: 2.9 Sx: 8.27 n: 8 µ: = t = .99 P = .35 (c) Critical values PRGM InvT AREA LEFT: .05 (.1/2 bc 2 tailed) DF: 7 To = -1.89,1.89 (d) Inference Since the test statistic is not in the rejection​ region, fail to reject the null hypothesis. There is not statistically significant evidence to reject the claim. 2 Tailed (a) Identify the hypotheses ​Ho: μd = 0 (claim) Ha​: μd ≠ 0 (b) The test statistic STAT TESTS 2: T-Test µ: 0 x: 2.9 Sx: 8.27 n: 8 µ: = t = .99 P = .35 (c) Critical values PRGM InvT AREA LEFT: .05 (.1/2 bc 2 tailed) DF: 7 To = -1.89,1.89 (d) Inference Since the test statistic is not in the rejection​ region, fail to reject the null hypothesis. There is not statistically significant evidence to reject the claim.

7.3 Use a​ t-test to test the claim about the population mean μ at the given level of significance α using the given sample statistics. Assume the population is normally distributed. ​Claim: μ≥7900​ α=0.05 Sample​ statistics: x=7600​ s=470​ n=24

STAT T-Test Left tailed test (a) null and alternative hypotheses H0​: μ ≥ 7900 Ha​: μ < 7900 (b) VAlue of standardized test statistic STAT TESTS 2: T-Test Stats μ: 7900​ x: 7600 Sx: 470 n: 24 ​μ: >μ Calculate ENTER = -3.13 (c) P value 1-.9976 = .002 (d) Reject of fail to reject Reject H0. At the 5% level of significance, there is enough evidence to reject the claim.

7.3 Use technology and a​ t-test to test the claim about the population mean μ at the given level of significance α using the given sample statistics. Assume the population is normally distributed. ​Claim: μ>80​ α=0.10 Sample​ statistics: x=83.2​ s=3.5​ n=26

STAT T-Test Right tailed test (a) Hypotheses H0​: μ ≤ 80 HA​: μ > 80 (b) value of standardized test statistics STAT TESTS 2: T-Test Stats μ: 80​ x: 83.2 Sx: 3.5 n: 26 ​μ: <μ Calculate ENTER = 4.66 (c) P-value 1-.99999 = 0 (d) reject or fail to reject null hypothesis Reject H0. There is enough evidence to support the claim.

7.3 A consumer group claims that the mean minimum time it takes for a sedan to travel a quarter mile is greater than 14.9 seconds. A random sample of 21 sedans has a mean minimum time to travel a quarter mile of 15.5 seconds and a standard deviation of 2.11 seconds. At α=0.01 is there enough evidence to support the consumer​ group's claim? Complete parts​ (a) through​ (d) below. Assume the population is normally distributed. α= 0.01 μ > 14.9 x= 15.5 s= 2.11 n= 21

STAT T-Test Right tailed test (a) Identify the claim H0: μ ≤ 14.9 Ha: μ > 14.9 (b) standardized test statistic STAT TESTS 2: T-Test Stats μ: 14.9​ x: 15.5 Sx: 2.11 n: 21 ​μ: < μ Calculate ENTER t = 1.3 p: 1-.8963 p = .104 (c) reject or fail to reject Fail to reject H0 because the​P-value is greater than α. (d) Interpret There is not enough evidence at the 1​% level of significance to support the claim that the mean minimum time it takes for a sedan to travel a quarter mile is greater than 14.9 seconds.

7.3 You receive a brochure from a large university. The brochure indicates that the mean class size for​ full-time faculty is fewer than 33 students. You want to test this claim. You randomly select 18 classes taught by​ full-time faculty and determine the class size of each. The results are shown in the table below. At α=0.10​, can you support the​ university's claim? Complete parts​ (a) through​ (d) below. Assume the population is normally distributed. 37 31 31 31 32 40 26 22 28 31 33 35 32 27 27 29 29 24

STAT T-Test Right tailed test (a) Identify the claim H0​: μ ≥ 33 Ha​: μ < 33 (b) standardized test statistic STAT TESTS 2: T-Test Data μ: 33​ List: L1 (or whichever list you put data in) ​μ: > μ Calculate ENTER p: 1-.9907 p = .009 (c) reject or fail to reject Reject H0 because the​P-value is less than α. (d) Interpret At the 10​% level of​ significance, there is sufficient evidence to support the claim that the mean class size for​ full-time faculty is fewer than 33 students.

QR Test the claim about the population​ mean, μ​, at the given level of significance using the given sample statistics. (a) identify hypothesis (b) standardized test stat z (c) critical value(s) (d) reject / fail to reject ​Claim: μ=40​; α=0.09​; σ=3.12. Sample​ statistics: x=39.3​, n=68

STAT TEST 1: Z ZCRIT 2 tailed (a) Identify Hypothesis Ho: μ=40 (CLAIM) Ha: µ ≠ 40 (b) Standardized test statistic since σ is given, use z test STAT TEST 1: Z z = -1.85 p = .064 (c) Critical values PRGM ZCRIT AREA LEFT: .91 (1-.09) zₒ = ±1.70 (d) Reject / Fail to reject REJECT Ho. At the 9% sig level, there IS enough evidence to REJECT the claim.

7.2 Use technology to help you test the claim about the population​ mean, μ​, at the given level of​ significance, α​, using the given sample statistics. Assume the population is normally distributed. ​Claim: μ>1220​ α=0.05​ σ=207.29 Sample​ statistics: x=1236.74​ n=250

STAT TESTS normalcdf if sigma is known, use z-test otherwise, use t test STAT TESTS 1: Z-Test Stats INPUT INFO select the opposite μ>1220​ RESULTS μ<1220 z=1.28 p=.8992 x=1236.74 n=250 ​ H0 is always the alternative hypostheses (a) identify the null and alternative hypotheses H0​: μ≤1220 Ha​:μ>1220 (CLAIM) (b) Calculate the standardized test statistic = 1.28 (c) Determine the P-value 2nd vars 2: normalcdf (1.28,9999) = .100 Reject H0 if the​ P-value is less than or equal to α. ​Otherwise, fail to reject H0. If the claim is the null hypothesis and H0 is​ rejected, then there is enough evidence to reject the claim. If H0 is not​ rejected, then there is not enough evidence to reject the claim. If the claim is the alternative hypothesis and H0 is​ rejected, then there is enough evidence to support the claim. If H0 is not​ rejected, then there is not enough evidence to support the claim. (d) Determine the outcome and conclusion of the test Fail to reject H0. At the 5% significance level, there is not enough evidence to reject the claim.

7.2 Use the calculator displays to the right to make a decision to reject or fail to reject the null hypothesis at a significance level of α=0.10.

Since the​ P-value is less than α​, reject the null hypothesis.

Explain how to perform a​ two-sample z-test for the difference between two population proportions.

State the hypotheses and identify the claim. Specify the level of significance. Find the critical​ value(s) and rejection​ region(s). Find p and q. Find the standardized test statistic. Make a decision and interpret it in the context of the claim.

Test the claim below about the mean of the differences for a population of paired data at the level of significance α. Assume the samples are random and​ dependent, and the populations are normally distributed. ​Claim: μd<​0; α=0.10. Sample​ statistics: d=1.3​, sd=3.8​, n=11

T-Test InvT Left tailed (a) Identify the hypotheses H0​: μd ≥ 0 Ha​: μd < 0 (claim) (b) Test statistic STAT TESTS 2: T-Test µ: 0 x: 1.3 Sx: 3.8 n: 11 µ: > t = 1.13 P = .141 (c) Critical values PRGM InvT AREA LEFT: .1 DF: 10 To = -1.37 (d) Inference Since the test statistic is not in the rejection​ region, fail to reject the null hypothesis. There is not statistically significant evidence to support the claim.

QR A used car dealer says that the mean price of a​ three-year-old sports utility vehicle is ​$23000. You suspect this claim is incorrect and find that a random sample of 21 similar vehicles has a mean price of ​$23837 and a standard deviation of ​$1971. Is there enough evidence to reject the claim at α=0.01​? ​Complete parts​ (a) through​ (e) below. Assume the population is normally distributed. (a) Identify hypothesis (b) Critical values (c) Standardized test stat T (d) Reject / Fail to reject Claim: µ =23000 x̅ = 23837 s = 1971 n = 21 α = .01

TCRIT STAT TEST 2: t-test 2 tailed (a) Identify hypothesis µ = 2300 (CLAIM) µ ≠ 2300 (b) Critical values PRGM TCRIT C: .99 N: 21 zₒ = ± 2.845 (c) Standardized test stat T STAT TEST 2: T t = 1.95 p = .066 (d) Reject / Fail to reject Fail to reject Ho because the test stat is not in rejection regions.

What conditions are necessary in order to use a t​-test to test the differences between two population​ means?

The following conditions are necessary to use a t​-test for independent samples. 1. The population standard deviations are unknown. 2. The samples are randomly selected 2. The samples are independent. 3. The populations are normally distributed or each sample size is at least 30.

7.2 Match each​ P-value with the graph that displays its area without performing any calculations. Explain your reasoning. P=0.0162 and P=0.1867.

The greater the shaded area, the larger the p value. (a) displays the area for P=0.0162 (b) displays the area for P=0.1867 because the​ P-value is equal to the shaded area.

7.2 A random sample of 88 eighth grade​ students' scores on a national mathematics assessment test has a mean score of 265. This test result prompts a state school administrator to declare that the mean score for the​ state's eighth graders on this exam is more than 260. Assume that the population standard deviation is 33. At α=0.05​, is there enough evidence to support the​ administrator's claim? Complete parts​ (a) through​ (e). α=0.05​ x=265 σ=33 μ > 260 n=88

Z-Test Left Tailed Test (a) Write the claim H0: μ≤260 Ha​: μ>260 claim (b) find the standard test statistic z STAT TESTS 1: Z-test μ: 260 σ: 33 x̅: 265 n: 88 μ: < z=1.42 (c) find the P-value (above steps) p=.078​ (d) Reject or fail to reject Fail to reject H0 (e) Interpret There is not enough evidence to support the claim

7.4 An education researcher claims that 63​% of college students work​ year-round. In a random sample of 500 college​ students, 315 say they work​ year-round. At α=0.05​, is there enough evidence to reject the​ researcher's claim? Complete parts​ (a) through​ (e) below. An education researcher claims that 63​% H0​: p = .63 (claim) Ha​: p ≠ .63 2 tailed z0 = -1.96 , 1.96 z = 0

invNorm PRGM PROPZCLM 2 tailed test (a) claim scenario 63​% of college students work​ year round. hypotheses H0​: p = .63 (claim) Ha​: p ≠ .63 (b) z critical .05 / 2 = .025 2nd vars invNorm 3: (.025) = -1.96,1.96 Rejection regions The rejection regions are z < −1.96 and z > 1.96 (c) standardized test stat z phat = 315 / 500 = .63 PRGM PROPZCLM p: .63, phat: .63 , n:500 = 0 (d) Reject or fail to reject H0 claim is true Ha is false Fail to reject the null hypothesis. There is not enough evidence to reject the​researcher's claim.

7.4 For the following​ information, determine whether a normal sampling distribution can be​ used, where p is the population​ proportion, α is the level of​ significance, p is the sample​ proportion, and n is the sample size. If it can be​ used, test the claim. ​Claim: p>0.28​ α=0.06 Sample​ statistics: p=0.36​ n=175 q=.72

invNorm PRGM PROPZCLM Right tailed test (a) Hypotheses H0: p ≤ .28 Ha​: p > .28 (b) Critical values 1-.06=.94 2nd vars invNorm 3: (.94) = 1.55 (c) Rejection Regions The rejection region is z > 1.55 (d) Standardized test stat z PRGM PROPZCLM = 2.36 (e) Reject or fail to reject Reject the null hypothesis. There is enough evidence to support the claim.

7.4 Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the population proportion p at the given level of significance α using the given sample statistics. ​Claim: p≠0.29 α=0.01 Sample​ statistics: p=0.25​ n=120

invNorm PRGM PROPZCLM two tailed test (a) Can the normal sampling distribution be​ used? ​Yes, because both np and nq are greater than or equal to 5. (b) State the null and alternative hypotheses. H0​: p = 0.29 Ha​: p ≠ 0.29 (c) Determine critical values Since 2 tailed, must divide α=0.01 by 2 2nd vars 3: invNorm (.005) = -2.58, 2.58 (d) Standardized test statistic z PRGM PROPZCLM = -.97 (e) Results Fail to reject H0. The data do not provide sufficient evidence to support the claim. since z does not fall in the rejection area, fail to reject. Since the claim is Ha, there is not enough evidence to support it. basically failing to reject the alternative hypothesis

7.2 Find the critical​ value(s) and rejection​ region(s) for the type of​ z-test with level of significance α. Include a graph with your answer. Two​-tailed test, α=0.04

invnorm since 2-tailed, divide 0.04 / 2 = .02 2nd vars 3: invnorm (.02) = -2.05 z< -2.05 and z > 2.05

7.4 To manually find the P value given z z = 1.45

look 1.45 up on z chart = .9265 1-.9265 = .0735 for left & .0735 for right .0594 + .0594 = .147 P = .147

7.2 Find the​ P-value for the indicated hypothesis test with the given standardized test​ statistic, z. Decide whether to reject H0 for the given level of significance α. Two-tailed test with test statistic z=−2.29 and α=0.05

normalcdf 2 tailed 2nd vars normalcdf (-9999,-2.29) = .011 * 2 = .022 p = .022 Reject H0

7.2 Find the​ P-value for the indicated hypothesis test with the given standardized test​ statistic, z. Decide whether to reject H0 for the given level of significance α. Right-tailed test with test statistic z=1.90 and α=.06

normalcdf Right tailed 2nd Vars normalcdf (1.90,9999) p = .0287 Reject H0.

7.4 A medical researcher says that less than 79​% of adults in a certain country think that healthy children should be required to be vaccinated. In a random sample of 300 adults in that​ country, 77​% think that healthy children should be required to be vaccinated. At α=0.10​, is there enough evidence to support the​ researcher's claim? Complete parts​ (a) through​ (e) below.

nvNorm PRGM PROPZCLM Left tailed test (a) claim scenario Less than 79​% of adults in the country think that healthy children should be required to be vaccinated. hypotheses ​H0: p ≥ .79 Ha​: p < .79 (b) z critical 2nd vars invNorm 3: (.10) = -1.28 Rejection regions The rejection region is z < −1.28. (c) standardized test stat z PRGM PROPZCLM p: 79, phat: .77, n:300 = -.85 (d) Reject or fail to reject Fail to reject the null hypothesis. There is not enough evidence to support the ​researcher's claim.

7.4 An education researcher claims that at most 8​% of working college students are employed as teachers or teaching assistants. In a random sample of 400 working college​ students, 9​% are employed as teachers or teaching assistants. At α=0.10​, is there enough evidence to reject the​ researcher's claim? Complete parts​ (a) through​ (e) below.

nvNorm PRGM PROPZCLM Right tailed test (a) claim scenario At most 88​% of working college students are employed as teachers or teaching assistants. hypotheses ​H0: p ≤ .08 Ha​: p > .08 (b) z critical 1-.10=.90 2nd vars invNorm 3: (.90) = 1.28 Rejection regions The rejection region is z > 1.28. (c) standardized test stat z PRGM PROPZCLM p: .08, phat: .09, n:400 = .74 (d) Reject or fail to reject Fail to reject the null hypothesis. There is not enough evidence to support the ​researcher's claim.

7.2 The lengths of time​ (in years) it took a random sample of 32 former smokers to quit smoking permanently are listed. Assume the population standard deviation is 5.7 years. At α=0.08​, is there enough evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 15 ​years? Complete parts​ (a) through​ (e). 19.3 8.3 18.9 22.2 20.3 13.8 12.3 10.2 19.2 9.2 17.6 21.6 21.2 21.4 19.9 13.8 15.4 21.8 13.5 14.3 15.5 17.9 20.2 18.1 22.2 8.2 20.5 13.3 12.7 12.8 17.5 7.1 α=0.08​ x= σ=5.7 μ=15 n=32 H0: μ=15 (claim) Ha​: μ≠15 z=1.25 p=.106 * 2 = .211 fail to reject not enough evidence to reject the claim

stat tests 1: z test data

7.4 Fail or reject A research center claims that at least 26​% H0​: p ≥ .26 (claim) Ha​: p < .26 left tailed z0 = -1.28 z = -2.41

z is in the rejection zone H0 claim is false Ha is true Reject the null hypothesis. There is enough evidence to reject the​center's claim.

7.4​ Claim: p>0.36​ H0​: p ≤ .36 Ha​: p >.36 (claim) right tailed z0 = 1.55 z = 3

z is in the rejection zone H0 is false Ha claim is true Reject the null hypothesis. There is enough evidence to support the claim.

7.4 A medical researcher says that less than 79​% H0​: p ≥ .79 Ha​: p < .79 (claim) left tailed z0 = -1.28 z = -.85

z is not in the rejection range H0 is true Ha claim is false Fail to reject the null hypothesis. There is not enough evidence to support the​researcher's claim.

7.4 Fail or reject An education researcher claims that at most 8​% ​H0: p ≤ .08 (claim) Ha​: p > .08 right tailed z0 = 1.28 z = .74

z is not in the rejection zone H0 claim is true Ha is false Fail to reject the null hypothesis. There is not enough evidence to reject the​ researcher's claim.

7.4 An education researcher claims that at most 3​% H0​: p ≤ .03 (claim) Ha​: p > .03 right tailed z0 = 2.33 z = 1.31

z is not in the rejection zone H0 claim is true Ha is false Fail to reject the null hypothesis. There is not enough evidence to reject the​researcher's claim.

QR State whether the standardized test statistic t indicates that you should reject the null hypothesis. Explain. tₒ = -2.188 ​(a) t=2.209 ​(b) t=0 ​(c) t=−2.159 ​(d) t=-2.253

​(a) t=2.209 Fail to reject because 2.209 > -2.188 ​(b) t=0 Fail to reject because 0 > -2.188 ​(c) t=−2.159 Fail to reject because -2.159 > -2.188 ​(d) t=-2.253 Reject because -2.253 < -2.188

7.4 A research center claims that 30​% of adults in a certain country would travel into space on a commercial flight if they could afford it. In a random sample of 1100 adults in that​ country, 32​% say that they would travel into space on a commercial flight if they could afford it. At α=0.10​, is there enough evidence to reject the research​ center's claim? Complete parts​ (a) through​ (d) below. (a) hypotheses (b) find Z & P (c) reject or fail (d) interpret H0​: p = .30 (claim) Ha​: p ≠ .30 2 tailed z = 1.45 P = .147 α=0.10

​invNorm PRGM PROPZCLM 2 tailed test (a) claim scenario 30​%of adults in the country would travel into space on a commercial flight if they could afford it. hypotheses H0​: p = .3 Ha​: p ≠ .3 (b) Z = PRGM PROPZCLM p: .30, phat: .32 , n:1100 = 1.45 P = get z score from table 1.45 = .9265 1-.9265 = .0735 left side & .0735 right side = .147 (d) Reject or fail to reject H0 claim is true Ha is false Fail to reject the null hypothesis. There is not enough evidence to reject the​ researcher's claim.

7.2​ Claim: μ>1150​; α=0.03​; σ=203.25. x=1167.16​, n=300 H0​: μ≤1150 Ha​: μ>1150 z=1.46 p=.072

Fail to reject H0 There is not enough evidence to support the claim

7.2 α=0.02​ H0​: μ≤1220 Ha​:μ>1220 p=.118

Fail to reject H0 not enough evidence to support the claim

7.2 Find the​ P-value for a​ left-tailed hypothesis test with a test statistic of z=−1.96. Decide whether to reject H0 if the level of significance is α=0.10.

normalcdf left tailed 2nd Vars normalcdf (-99999,-1.96) p = .025 Since P≤α​, rejectH0.

7.2 State whether the standardized test statistic z indicates that you should reject the null hypothesis. ​(a) z=1.587 ​(b) z=1.715 ​(c) z=−1.503 ​(d) z=1.833

(a) z=1.587 Fail to reject H0 because −1.645<z<1.645. (b) z=1.715 Reject H0 because z>1.645 (c) z=1.503 Fail to reject H0 because -1.645<z<1.645 (d) z=-1.833 Reject H0 because z<-1.645