Chapter 7 HW, review quiz with TI-83 INSTRUCTIONS combined, Chapter 7 Quiz Review
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
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
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.
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=0.05 H0: μ≤1220 Ha:μ>1220 p=.100
Fail to reject H0 not enough evidence to reject 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 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
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 thesociety'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.
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 theP-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.
7.2 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
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
7.4 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 theresearcher'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.
7.2 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.
7.4 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 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.
7.4 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 thecenter'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 theresearcher'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 theresearcher'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
7.2 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
7.4 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 than30, 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 thet-test, meaning the criticalvalue(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. 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
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
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
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
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 (≠).
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 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.
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.
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 theP-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 theP-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.
