7.2 homework

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

7.2 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

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 invnorm .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 ​(≠​).

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

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

InvNorm 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

7.2​ 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.

7.2 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 (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.

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 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

7.2 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 2nd Vars normalcdf (-99999,-1.96) p = .025 Since P≤α​, rejectH0.

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 2nd Vars normalcdf (1.90,9999) p = .0287 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. Two-tailed test with test statistic z=−2.29 and α=0.05

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

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

7.2 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 H0: μ≤260 Ha​: μ>260 claim z=1.42 p=.078

​The original claim that the mean score for the​ state's eighth graders on the exam is more than 260 is the alternative​ hypothesis, Ha. If we reject the null​ hypothesis, there is enough evidence to support the claim. If we fail to reject the null​ hypothesis, there is not enough evidence to support the claim. Thus, at the 15​% significance​ level, there is not enough evidence to support the​ administrator's claim that the mean score for the​ state's eighth graders on the exam is more than 290. Fail to reject H0 There is not enough evidence to support the claim