stat exam 7 SG
Use the given statement to represent a claim. Write its complement and state which is Upper H 0 and which is Ha. mu greater than or equals 300
mu ___ 300 < H 0: mugreater than or equals300 Ha: mu <300
Can a critical value for the chi squared-test be negative? Explain.
No, in a chi squared-distribution, all chi squared-values are greater than or equal to 0.
Find the critical value(s) and rejection region(s) for a two-tailed chi-square test with a sample size n equals 20 and level of significance alphaequals0.05.
Critical Chi-Square Values - MathCracker.com = 8.907,32.852 blue onsides not middle
The P-value for a hypothesis test is shown. Use the P-value to decide whether to reject Upper H 0 when the level of significance is (a) alphaequals0.01, (b) alphaequals0.05, and (c) alphaequals0.10. Pequals0.0284
Fail to reject Upper H 0 because the P-value, 0.0284, is greater than alphaequals0.01. Reject Upper H 0 because the P-value, 0.0284, is less than alphaequals0.05. Reject Upper H 0 because the P-value, 0.0284, is less than alphaequals0.10.
Determine whether the statement is true or false. If it is false, rewrite it as a true statement. A large P-value in a test will favor a rejection of the null hypothesis.
False. A small P-value in a test will favor a rejection of the null hypothesis.
Find the critical value(s) for a left-tailed z-test with alphaequals0.05. Include a graph with your answer.
Z Critical Value Calculator LEFT so use negative and opposite graph -1.64 blue on left tail
Use a t-test to test the claim about the population mean mu at the given level of significance alpha using the given sample statistics. Assume the population is normally distributed. Claim: muequals52 comma 700; alphaequals0.01 Sample statistics: x overbarequals53 comma 743, sequals2700, nequals16
0: = 52700 a: =/= 52700 ------------------------- Standardized Test Statistic Calculator | Hypothesis Testing Calculator z Test standardized test statistic =1.55 ----------------------------- Critical T-values - MathCracker.com (= so 2-tailed) -2.947, 2.947 ------------- Fail to reject Upper H 0. There is not enough evidence to reject the claim.
Match the alternative hypothesis shown below with its graph to the right. Then state the null hypothesis and sketch its graph. Upper H Subscript a: sigmanot equals7
6<-- open7 -->9 = closed just 7
For the statement below, write the claim as a mathematical statement. Use proportions to state the null and alternative hypotheses, and identify which represents the claim. According to a recent survey, 67% of college students did not use student loans to pay for college.
= = =/= null =
Test the claim about the population variance sigma squared at the level of significance alpha. Assume the population is normally distributed. Claim: sigma squarednot equals32.7; alphaequals0.01 Sample statistics: s squaredequals40.1, nequals91
= 32.7 =/= 32.7 ----------------- Critical Chi-Square Values - MathCracker.com 59.196, 128.299 --------------------- chi^2=((n-1)s^2)/SD^2 =110.367 ------------------ DRAW IT OUT Fail to reject Upper H 0. There is not enough evidence at the 1% level of significance to support the claim.
Find the critical value(s) and rejection region(s) for a left-tailed chi-square test with a sample size n equals 21 and level of significance alphaequals0.10.
Critical Chi-Square Values - MathCracker.com =12.443 left tailed
Find the critical value(s) and rejection region(s) for the indicated t-test, level of significance alpha, and sample size n. Right-tailed test, alphaequals0.05, nequals14
Critical T-values - MathCracker.com 1.771 t>1/771
Determine whether the statement is true or false. If it is false, rewrite it as a true statement. In a hypothesis test, you assume the alternative hypothesis is true.
False. In a hypothesis test, you assume the null hypothesis is true.
Explain how to decide when a normal distribution can be used to approximate a binomial distribution.
If npgreater than or equals5 and nqgreater than or equals5, the normal distribution can be used.
Find the P-value for a left-tailed hypothesis test with a test statistic of zequalsnegative 1.59. Decide whether to reject Upper H 0 if the level of significance is alphaequals0.10.
Quick P Value from Z Score Calculator .0643 Since Pless than or equalsalpha, reject Upper H 0.
Determine whether the statement is true or false. If it is false, rewrite it as a true statement. If you decide to reject the null hypothesis, then you can support the alternative hypothesis.
This statement is true.
Find the critical value(s) for a left-tailed z-test with alphaequals0.01. Include a graph with your answer.
Z Critical Value Calculator -2.33 LEFT so use negative and left tailed graph
Use the given statement to represent a claim. Write its complement and state which is Upper H 0 and which is Upper H Subscript a. p less than 0.27
p greater than or = to .27 . Upper H 0: p greater than or equals 0.27 Upper H Subscript a: p less than 0.27
What are the two decisions that you can make from performing a hypothesis test?
reject the null hypothesis fail to reject the null hypothesis
Find the P-value for the indicated hypothesis test with the given standardized test statistic, z. Decide whether to reject Upper H 0 for the given level of significance alpha. Right-tailed test with test statistic zequals1.32 and alphaequals0.06
Quick P Value from Z Score Calculator .0934 Fail to reject Upper H 0 because the P-value is greater than the level of significance alpha.
State whether the standardized test statistic t indicates that you should reject the null hypothesis. Explain. (a) tequalsnegative 1.935 (b) tequals1.811 (c) tequals1.876 (d) tequalsnegative 1.892
Reject Upper H 0, because t less than minus 1.865. Fail to reject Upper H 0, because negative 1.865 less than t less than 1.865. Reject Upper H 0, because t greater than 1.865. Reject Upper H 0, because t less than minus 1.865.
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 most 41 days.
Since the claim contains a statement of equality, it represents 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 most 41 days. There is insufficient evidence to reject the claim that the mean incubation period for the eggs of a species of bird is at most 41 days. There is not enough evidence to support the claim that more than 74% of households in a county struggle to afford basic necessities.
A null and alternative hypothesis are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Upper H 0: p less than or equals 0.2 Upper H Subscript a: p greater than 0.2
left?
Test the claim about the population mean, mu, at the given level of significance using the given sample statistics. Claim: muequals40; alphaequals0.03; sigmaequals3.37. Sample statistics: x overbarequals38.9, nequals59
0: = 40 a: =/= 40 Standardized Test Statistic Calculator | Hypothesis Testing Calculator z Test -2.51 Calculator of Critical Z-Values - MathCracker.com +/- 2.17 Quick P Value from Z Score Calculator z=-2.51 so p=.006<.03 so Reject Upper H 0. At the 3% significance level, there is enough evidence to reject the claim.
Test the claim about the population mean, mu, at the given level of significance using the given sample statistics. Claim: munot equals6000; alphaequals0.07; sigmaequals383. Sample statistics: x overbarequals5700, nequals36
0: =6000 a: =/= 6000 ------------------ Standardized Test Statistic Calculator | Hypothesis Testing Calculator z Test -4.7 ----------------- Calculator of Critical Z-Values - MathCracker.com +/- 1.81 -------------------------- Quick P Value from Z Score Calculator -4.7 = p=.000<.07 so Reject Upper H 0. At the 7% significance level, there is enough evidence to support the claim.
A null and alternative hypothesis are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Upper H 0: p equals 0.2 Upper H Subscript a: p not equals 0.2
2 tailed
An education researcher claims that 59% of college students work year-round. In a random sample of 300 college students, 177 say they work year-round. At alphaequals0.05, is there enough evidence to reject the researcher's claim? Complete parts (a) through (e) below.
59% of college students work year-round. ----------------------------- 0: = .59 a: =/= .59 ----------------------------- Critical T-values - MathCracker.com TWO TAILED NEG,POS -1.97,1.97 -------------------------- The rejection regions are zless than z< -1.971 and z>1.97 --------------------------- MedCalc's Test for one proportion calculator OP: 177/300 = .59 = 59% SS: 300 NHV: 59% = 0.00 ------------------------ DRAW IT OUT Fail to reject the null hypothesis. There is not enough evidence to reject the researcher's claim.
Use a chi squared-test to test the claim sigma squared equals 0.52 at the alpha equals 0.10 significance level using sample statistics s squared equals 0.509 and n equals 15. Assume the population is normally distributed.
=.52 =/= .52 ------------------ chi^2=((n-1)s^2)/SD^2 =((15-1).509)/(.52) =13.70 ---------------- Critical Chi-Square Values - MathCracker.com TWO TAILED BC = 6.57, 23.69 ----------------- DRAW IT OUT Fail to reject Upper H 0. There is not enough evidence at the 10% level of significance to reject the claim.
For the statement below, write the claim as a mathematical statement. State the null and alternative hypotheses and identify which represents the claim. The standard deviation of the base price of a certain type of all-terrain vehicle is at least $347.
> or = >/= < The null hypothesis Upper H 0: sigmagreater than or equals347 is the claim.
For the statement below, write the claim as a mathematical statement. State the null and alternative hypotheses and identify which represents the claim. An amusement park claims that the mean daily attendence at the park isnbsp at least 19 comma 000 people.
>/= >/= < The null hypothesis Upper H 0: mugreater than or equals19 comma 000 is the claim.
Use a chi squared-test to test the claim sigma less than 44 at the alpha equals 0.01 significance level using sample statistics s equals 39.7 and n equals 12. Assume the population is normally distributed.
>/= 44 < 44 ----------------- chi^2=((n-1)s^2)/SD^2 =8.955 ---------------- Critical Chi-Square Values - MathCracker.com LEFT =3.053 --------------- DRAW IT OUT Fail to reject Upper H 0. There is not enough evidence at the 1% level of significance to support the claim.
Describe type I and type II errors for a hypothesis test of the indicated claim. A culinary school publicizes that at least 10% of applicants become full time students.
A type I error will occur when the actual proportion of applicants who become full time students is at least 0.10, but you reject Upper H 0: pgreater than or equals0.10. A type II error will occur when the actual proportion of applicants who become full time students is less than 0.10, but you fail to reject Upper H 0: pgreater than or equals0.10.
Describe type I and type II errors for a hypothesis test of the indicated claim. A public works department publicizes that at least 40% of applicants become groundskeepers.
A type I error will occur when the actual proportion of applicants who become groundskeepers is at least 0.40, but you reject Upper H 0: pgreater than or equals0.40. A type II error will occur when the actual proportion of applicants who become groundskeepers is less than 0.40, but you fail to reject Upper H 0: pgreater than or equals0.40.
An education researcher claims that at most 6% of working college students are employed as teachers or teaching assistants. In a random sample of 600 working college students, 7% are employed as teachers or teaching assistants. At alphaequals0.10, is there enough evidence to reject the researcher's claim? Complete parts (a) through (e) below.
At most 6% of working college students are employed as teachers or teaching assistants. ------------------------------ 0: </= .06 a: > .06 ------------------------------- Critical T-values - MathCracker.com RIGHT SIDE POSITIVE 1.28 ---------------------------- The rejection region is zgreater than 1.28. ------------------------- MedCalc's Test for one proportion calculator 1.03 ----------- z=1.03 <1.28=rejection region DRAW IT OUT Fail to reject the null hypothesis. There is not enough evidence to reject the researcher's claim.
Find the critical value(s) and rejection region(s) for the type of z-test with level of significance alpha. Include a graph with your answer. Two-tailed test, alphaequals0.005
Calculator of Critical Z-Values - MathCracker.com TWO TAILED TEST so -2.81, 2.81 The rejection regions are zless than -2.81 negative 2.81 and zgreater than 2.81. middle not blue
Find the critical value(s) and rejection region(s) for the indicated t-test, level of significance alpha, and sample size n. Left-tailed test, alphaequals0.005, nequals22
Critical T-values - MathCracker.com -2.831 t<-2.831
Find the critical value(s) and rejection region(s) for the indicated t-test, level of significance alpha, and sample size n. Two-tailed test, alphaequals0.02, nequals16
Critical T-values - MathCracker.com -2.602,2.602 t<-2.602 and t> 2.602
State whether each standardized test statistic chi squared allows you to reject the null hypothesis. Explain. 0 40 chi squaredchi Subscript Upper L Superscript 2 Baseline equals 7.261chi Subscript Upper R Superscript 2 Baseline equals 24.996 A coordinate system has a horizontal axis labeled chi squared from 0 to 40 in increments of 5 and a vertical axis. A curve is over a horizontal axis. Vertical line segments extend from the horizontal axis to the curve at chi Subscript Upper L Superscript 2=7.261 and chi Subscript Upper R Superscript 2=24.996. The area under the curve between chi Subscript Upper L Superscript 2=7.261 and chi Subscript Upper R Superscript 2=24.996 is shaded. (a) chi squaredequals24.995 (b) chi squaredequals7.260 (c) chi squaredequals7.262 (d) chi squaredequals24.997
Do not reject the null hypothesis because the standardized test statistic is not in the rejection region. Reject the null hypothesis because the standardized test statistic is in the rejection region. Do not reject the null hypothesis because the standardized test statistic is not in the rejection region. Reject the null hypothesis because the standardized test statistic is in the rejection region. BLUE IS REJECTION
State whether each standardized test statistic chi squared allows you to reject the null hypothesis. Explain. 0 45 chi squaredchi Subscript 0 Superscript 2 Baseline equals 23.5420 A coordinate system has a horizontal axis labeled chi squared from 0 to 45 in increments of 5 and a vertical axis. A curve is over a horizontal axis. Vertical line segment extends from the horizontal axis to the curve at chi Subscript 0 Superscript 2=23.5420. The area under the curve to the right of chi Subscript 0 Superscript 2=23.5420 is shaded. (a) chi squaredequals13.425 (b) chi squaredequals0 (c) chi squaredequals1.574 (d) chi squaredequals23.504
Do not reject the null hypothesis because the standardized test statistic is not in the rejection region. Do not reject the null hypothesis because the standardized test statistic is not in the rejection region. Do not reject the null hypothesis because the standardized test statistic is not in the rejection region. Do not reject the null hypothesis because the standardized test statistic is not in the rejection region.
State whether each standardized test statistic chi squared allows you to reject the null hypothesis. Explain. 0 40 chi squaredchi Subscript 0 Superscript 2 Baseline equals 21.0640 A coordinate system has a horizontal axis labeled chi squared from 0 to 40 in increments of 5 and a vertical axis. A curve is over a horizontal axis. Vertical line segment extends from the horizontal axis to the curve at chi Subscript 0 Superscript 2=21.0640. The area under the curve to the right of chi Subscript 0 Superscript 2=21.0640 is shaded. (a) chi squaredequals11.804 (b) chi squaredequals0 (c) chi squaredequals1.729 (d) chi squaredequals21.153
Do not reject the null hypothesis because the standardized test statistic is not in the rejection region. Do not reject the null hypothesis because the standardized test statistic is not in the rejection region. Do not reject the null hypothesis because the standardized test statistic is not in the rejection region. Reject the null hypothesis because the standardized test statistic is in the rejection region.
State whether the standardized test statistic z indicates that you should reject the null hypothesis. (a) zequals1.948 (b) zequals2.027 (c) zequalsnegative 1.751 (d) zequalsnegative 2.114
Fail to reject Upper H 0 because negative 1.960 less than z less than 1.960. Reject Upper H 0 because z greater than 1.960. Fail to reject Upper H 0 because negative 1.960 less than z less than 1.960. Reject Upper H 0 because z less than minus 1.960.
State whether the standardized test statistic z indicates that you should reject the null hypothesis. (a) zequals2.565 (b) zequals2.598 (c) zequalsnegative 2.352 (d) zequalsnegative 2.789 -
Fail to reject Upper H 0 because negative 2.575 less than z less than 2.575. Reject Upper H 0 because z greater than 2.575. Fail to reject Upper H 0 because negative 2.575 less than z less than 2.575. Reject Upper H 0 because z less than minus 2.575.
The P-value for a hypothesis test is shown. Use the P-value to decide whether to reject Upper H 0 when the level of significance is (a) alphaequals0.01, (b) alphaequals0.05, and (c) alphaequals0.10. Pequals0.0603
Fail to reject Upper H 0 because the P-value, 0.0603, is greater than alphaequals0.01. Fail to reject Upper H 0 because the P-value, 0.0603, is greater than alphaequals0.05. Reject Upper H 0 because the P-value, 0.0603, is less than alphaequals0.10.
State whether the standardized test statistic z indicates that you should reject the null hypothesis. (a) zequals2.241 (b) zequalsnegative 2.424 (c) zequalsnegative 2.474 (d) zequalsnegative 2.189
Fail to reject Upper H 0 because z greater than minus 2.330. Reject Upper H 0 because z less than minus 2.330. Reject Upper H 0 because z less than minus 2.330. Fail to reject Upper H 0 because z greater than minus 2.330.
State whether the standardized test statistic t indicates that you should reject the null hypothesis. Explain. (a) tequals2.264 (b) tequals0 (c) tequalsnegative 2.179 (d) tequalsnegative 2.249
Fail to reject Upper H 0, because tgreater thannegative 2.227. Fail to reject Upper H 0, because tgreater thannegative 2.227. Fail to reject Upper H 0, because tgreater thannegative 2.227 Reject Upper H 0, because tless thannegative 2.227..
Find the critical value(s) and rejection region(s) for a right-tailed chi-square test with a sample size n equals 14 and level of significance alphaequals0.10.
Free Critical Chi-Square Value Calculator - Free Statistics Calculators 19.812 right tailed blue
Match each P-value with the graph that displays its area without performing any calculations. Explain your reasoning. Pequals0.0212 and Pequals0.2148.
Graph (a) displays the area for Pequals0.0212 and graph (b) displays the area for Pequals0.2148 because the P-value is equal to the shaded area.
Match each P-value with the graph that displays its area without performing any calculations. Explain your reasoning. Pequals0.0801 and Pequals0.2891.
Graph (b) displays the area for Pequals0.0801 and graph (a) displays the area for Pequals0.2891 because the P-value is equal to the shaded area.
Match each P-value with the graph that displays its area without performing any calculations. Explain your reasoning. Pequals0.1711 and Pequals0.0096.
Graph (b) displays the area for Pequals0.1711 and graph (a) displays the area for Pequals0.0096 because the P-value is equal to the shaded area.
Explain how to find the critical values for a t-distribution. Give the first step. Choose the correct answer below.
Identify the level of significance alpha and the degrees of freedom, d.f.equalsnminus1. If the hypothesis test is left-tailed, use the "One tail comma alpha " column with a negative sign. If the hypothesis test is right-tailed, use the "One tail comma alpha " column with a positive sign. If the hypothesis test is two-tailed, use the "Two tails comma alpha " column with a negative and a positive sign.
A medical researcher says that less than 74% of adults in a certain country think that healthy children should be required to be vaccinated. In a random sample of 200 adults in that country, 70% think that healthy children should be required to be vaccinated. At alphaequals0.01, is there enough evidence to support the researcher's claim? Complete parts (a) through (e) below.
Less than 75 75% of adults in the country think that healthy children should be required to be vaccinated. ---------------------- 0: p>/= .75 a: p<.75 ------------- Critical T-values - MathCracker.com -1.65 ------------------ going left so z< The rejection region is z<-1.65 ---------------------- MedCalc's Test for one proportion calculator MAKE NEG BC LEFT -2.26 --------------------------- DRAW IT OUT -2.26<-1.65 so Reject the null hypothesis. There is enough evidence to support the researcher's claim.
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 report claims that more than 74% of households in a specific county struggle to afford basic necessities.
Since the claim does not contain a statement of equality, it represents the alternative hypothesis. There is enough evidence to support the claim that more than 74% of households in a county struggle to afford basic necessities.
State Upper H 0 and Upper H Subscript a in words and in symbols. Then determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Explain your reasoning. Sketch a normal sampling distribution and shade the area for the P-value. A report claims that lung cancer accounts fornothing 39% of all cancer diagnoses.
The null hypothesis expressed in words is the proportion of cancer diagnoses attributable to lung cancer is 39.39 . 39. The null hypothesis is expressed symbolically as Upper H 0: p equals 39 0.39. The alternative hypothesis expressed in words is the proportion of cancer diagnoses attributable to lung cancer is not .39 . 39. The alternative hypothesis is expressed symbolically as Upper H Subscript a: p not equals .39 . 39. The hypothesis test is two dash tailed because the alternative hypothesis contains not equals . blue on both sides tan in middle
A baseball team claims that the mean length of its games is at least 1.3 hours. State Upper H 0 and Upper H Subscript a in words and in symbols. Then determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed. Explain your reasoning.
The null hypothesis expressed in words is, "the mean length of a baseball team's games isnbsp at least 1.3 hours." The null hypothesis is expressed symbolically as, "Upper H 0: mugreater than or equals1.3." The alternative hypothesis expressed in words is, "the mean length of a baseball team's games is less than 1.3 hours." The alternative hypothesis is expressed symbolically as, "Upper H Subscript a: muless than1.3." The hypothesis test is left dash tailed because the alternative hypothesis contains less than .
A baseball team claims that the mean length of its games is less than 1.9 hours. State Upper H 0 and Upper H Subscript a in words and in symbols. Then determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed. Explain your reasoning.
The null hypothesis expressed in words is, "the mean length of a baseball team's games isnbsp at least 1.9 hours." The null hypothesis is expressed symbolically as, "Upper H 0: mugreater than or equals1.9." The alternative hypothesis expressed in words is, "the mean length of a baseball team's games is less than 1.9 hours." The alternative hypothesis is expressed symbolically as, "Upper H Subscript a: muless than1.9." The hypothesis test is left dash tailed because the alternative hypothesis contains less than .
A baseball team claims that the mean length of its games is 2.9 hours. State Upper H 0 and Upper H Subscript a in words and in symbols. Then determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed. Explain your reasoning.
The null hypothesis expressed in words is, "the mean length of a baseball team's games isnothing 2.9 hours." The null hypothesis is expressed symbolically as, "Upper H 0: muequals2.9." The alternative hypothesis expressed in words is, "the mean length of a baseball team's games is not equal to 2.9 hours." The alternative hypothesis is expressed symbolically as, "Upper H Subscript a: munot equals2.9." The hypothesis test is two dash tailed because the alternative hypothesis contains not equals .
A security expert claims that exactly 10% of all homeowners have a home security alarm. State Upper H 0 and Upper H Subscript a in words and in symbols. Then determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed. Explain your reasoning.
The null hypothesis expressed in words is, "the proportion of all homeowners who own a home security alarm isnothing 0.10." The null hypothesis is expressed symbolically as, "Upper H 0: pequals0.10." The alternative hypothesis expressed in words is, "the proportion of all homeowners who own a home security alarm is not equal to 0.10." The alternative hypothesis is expressed symbolically as, "Upper H Subscript a: pnot equals0.10." The hypothesis test is two dash tailed because the alternative hypothesis contains not equals .
A golf analyst claims that the standard deviation of the 18-hole scores for a golfer is 3.1 strokes. State Upper H 0 and Upper H Subscript a in words and in symbols. Then determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed. Explain your reasoning.
The null hypothesis expressed in words is, "the standard deviation of the 18-hole scores for a golfer isnothing 3.1 strokes." The null hypothesis is expressed symbolically as, "Upper H 0: sigmaequals3.1." The alternative hypothesis expressed in words is, "the standard deviation of the 18-hole scores for a golfer is not equal to 3.1 strokes." The alternative hypothesis is expressed symbolically as, "Upper H Subscript a: sigmanot equals3.1." The hypothesis test is two dash tailed because the alternative hypothesis contains not equals .
Determine whether the statement shown below is true or false. If it is false, rewrite is as a true statement. If you want to support a claim, write it as your null hypothesis.
The statement is false. If you want to support a claim, write it as your alternative hypothesis.
Use a 0.025 significance level to test the claim that peanut candies have weights that vary more than plain candies. The standard deviation for the weights of plain candies is 0.431. A sample of 71 peanut candies has weights with a standard deviation of 0.38. Assume the population is normally distributed.
Upper H 0 : sigma less than or equals 0.431; Upper H Subscript a Baseline : sigma greater than 0.431 (Claim) --------------------------- Critical Chi-Square Values - MathCracker.com = 95.023 ----------------------------- right tail colored -------------------- chi^2=((n-1)s^2)/SD^2 S= .38 sigma= .431 =((71-1).38^2)/(.431^2) =(70*.1444)/( .185761) = 10.108/.185760 =54.414 ------------------ DRAW IT OUT fail to reject ------------------ Since the null hypothesis is not rejected, the weights do not appear to vary more for peanut candies.
Find the critical value(s) and rejection region(s) for the type of z-test with level of significance alpha. Include a graph with your answer. Right-tailed test, alphaequals0.02
Z Critical Value Calculator RIGHT so keep positive and right tailed graph The rejection region is zgreater than 2.05.
For the statement below, write the claim as a mathematical statement. State the null and alternative hypotheses and identify which represents the claim. A laptop manufacturer claims that the mean life of the battery for a certain model of laptop is more than 9 hours.
mu > 9 Upper H 0: muless than or equals9 Upper H Subscript a: mugreater than9 The alternative hypothesis Upper H Subscript a: mugreater than9 is the claim.
A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 743 hours. A random sample of 24 light bulbs has a mean life of 720 hours. Assume the population is normally distributed and the population standard deviation is 56 hours. At alphaequals0.02, do you have enough evidence to reject the manufacturer's claim? Complete parts (a) through (e).
mu>/=743 n=24 xbar=720 SD= 56 alpha=.02 ------------------ 0: mu >/= 743 a: mu < 743 --------------------- Z Critical Value Calculator LEFT so do negative and opposite graph -2.05 graph on left tail ------------------------- Standardized Test Statistic Calculator | Hypothesis Testing Calculator z Test = -2.01 -------------------- p=.0222 > .02 so Fail to reject Upper H 0. There is not sufficient evidence to reject the claim that mean bulb life is at least 743 hours.
A random sample of 86 eighth grade students' scores on a national mathematics assessment test has a mean score of 288. 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 280. Assume that the population standard deviation is 33. At alphaequals0.09, is there enough evidence to support the administrator's claim? Complete parts (a) through (e).
n=86 xbar=288 mu>280 SD= 33 alpha = .09 ----------------- H0: mu</=280 Ha: mu>280(claim) -------------- z=2.25 Standardized Test Statistic Calculator | Hypothesis Testing Calculator z Test ----------------- p=.012 Quick P Value from Z Score Calculator ------------------------- .012<.09 so REJECT ----------------------- At the 9% significance level, there is enough evidence to support the administrator's claim that the mean score for the state's eighth graders on the exam is more than 280.
What are the two types of hypotheses used in a hypothesis test? How are they related?
null and alternative They are complements.
Match the alternative hypothesis shown below with its graph to the right. Then state the null hypothesis and sketch its graph. Upper H Subscript a: mugreater than8
open circle on 8 ---> 10 Upper H 0: mu less than or equals 8 7<-- closed circle 8
A company that makes cola drinks states that the mean caffeine content per 12-ounce bottle of cola is 50 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 49.4 milligrams. Assume the population is normally distributed and the population standard deviation is 7.1 milligrams. At alphaequals0.05, can you reject the company's claim? Complete parts (a) through (e).
same websites as above = 50 =/= 50 --------------- = so two tailed +/- 1.96 ------------------------- blue on outsides --------------------- z= -.46 ------------------ Since z is not in the rejection region, fail to reject the null hypothesis. ------------------------ At the 5% 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 zs 50 milligrams.
A company claims that the mean monthly residential electricity consumption in a certain region is more than 860 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 66 residential customers has a mean monthly consumption of 880 kWh. Assume the population standard deviation is 129 kWh. At alphaequals0.01, can you support the claim? Complete parts (a) through (e).
same websites as above except for critical use Z Critical Value Calculator amd make POSITIVE AND RIGHTbecause MORE THAN </= 860 >860 (claim) ----------------- 2.33 --------------- The rejection region is z>2.33. ------------------ z=1.26 -------------- so p= .1038 >.01 so Fail to reject Upper H 0 because the standardized test statistic is not in the rejection region. ------------------------- At the 1% significance level, there is not enough evidence to support the claim that the mean monthly residential electricity consumption in a certain region is greater than 860 kWh.
A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 738 hours. A random sample of 23 light bulbs has a mean life of 718 hours. Assume the population is normally distributed and the population standard deviation is 64 hours. At alphaequals0.02, do you have enough evidence to reject the manufacturer's claim? Complete parts (a) through (e).
same websites as above except for critical use Z Critical Value Calculator amd make negative and left because AT LEAST >/= 738 (claim) <738 ----------------- z=-1.5 -------------- so p= .0668 >.02 so Fail to reject Upper H 0. There is not sufficient evidence to reject the claim that mean bulb life is at least 738 hours.
Use the given statement to represent a claim. Write its complement and state which is Upper H 0 and which is Ha. sigma equals 12
sigma =/= 12 H0: sigmaequals12 Ha: sigmanot equals12
Use technology to help you test the claim about the population mean, mu, at the given level of significance, alpha, using the given sample statistics. Assume the population is normally distributed. Claim: mugreater than1270; alphaequals0.04; sigmaequals209.04. Sample statistics: x overbarequals1296.46, nequals250
use calcs and exp above </= > 2.00 p=.023 which is less than alpha so Reject Upper H 0. At the 4% significance level, there is enough evidence to support the claim.
A nutritionist claims that the mean tuna consumption by a person is 3.8 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.4 pounds per year. Assume the population standard deviation is 1.09 pounds. At alphaequals0.06, can you reject the claim?
xbar= 3.4 n=90 mu=3.8 sd= 1.09 alpha=.06 -------------- 0: = a: =/= --------------- z=-3.48 Standardized Test Statistic Calculator | Hypothesis Testing Calculator z Test -------------- p=.000 Quick P Value from Z Score Calculator ------------------ .000<.09 so Reject Upper H 0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.8 pounds.