Z-Test Z-Test 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 x=0.05. Inpt: Data Statsu #70 Ho:70 o:3.75 x:68.75 n:40 H:HoHo Ho Calculate Draw z=-2.10818511 p=0.03501498 x=68.75 n=40
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Q: The test statistic in a right-tailed test is z=2.15. Determine the P-value and decide whether, at…
A: Solution: From the given information, the test statistic is z=2.15 and the test is right-tailed.
Q: Use technology to find the P-value for the hypothesis test described below. The claim is that for a…
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A: Given Data : n=500 , p^=0.25 , p0=0.2 and α = 0.025 The null hypothesis is H0 : p ≤0.2 Alternative…
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A: Denote μ as the population mean.
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A: a)This is right tailed test.
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Q: Use technology to find the P-value for the hypothesis test described below. The claim is that for a…
A: Claim: The claim is that for a smartphones carriers data speeds at airports, the mean is μ=13.00…
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A: From the given data: P-value=0.00918 level of significance=0.01
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A: Given information: Samples 254, 328 μ=2899
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Q: Z-Test Inpt: Data Statsu = 70 Ho:70 0:5.25 x:68.75 In:40 u* Ho Ho Z-Test Use the calculator displays…
A: If p- value is greater than alpha then the null hypothesis is accepted. If p- value is less than…
Q: Use the calculator displays to the right to make a decision to reject or fail to reject the null…
A: Given, Level of significance = α =0.05 P value = 0.07751053
Q: 0.3.16-T A data set lists earthquake depths. The summary statistics are n = 400, x = 5.82 km, s =…
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Q: Use technology to find the P-value for the hypothesis test described below. The claim is that for a…
A: The degrees of freedom is, 11–1 = 10.The value of test statistic is –1.298.
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A: The null and alternative hypotheses are, H0: µ = 80 vs. H1: µ ≠ 80.
Q: Use technology to find the P-value for the hypothesis test described below. The claim is that for a…
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Q: Compute test statistic (Independent t-score) Pooled variance S2p:…
A: 1. Pooled variance S2p: Sp2=SS1+SS2df1+df2=932+70613+13 (df=n-1)=63 Standard error:…
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Q: Use technology to find the P-value for the hypothesis test described below. The claim is that for a…
A: It is given that the sample size n is 29. Therefore, the degrees of freedom is (n-1)=29-1=28.
Q: Use the calculator displays to the right to make a decision to reject or fail to reject the null…
A: Decision rule: If p-value ≤ α, then reject the null hypothesis. Otherwise, fail to reject the null…
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A: The p-value is obtained by using Excel.
Q: Use technology to find the P-value for the hypothesis test described below. The claim is that for a…
A: Given n=20 Mean=15 t=-2.405
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A: Claim : The two populations of cereal have equal means. The hypotheses being tested is , Ho : µ1…
Q: Compute test statistic (Independent t-score) Pooled variance S2p:…
A: 1. The formula for pooled variance is given by : S2p=(n1-1).s21+(n2-1).s22n1+n2-2=SS1+SS2n1+n2-2…
Q: of men who own cats is smaller than the proportion on nce level. The null and alternative hypothesis…
A: Given: Level of significance, α=0.005 Set the hypothesis: H0 : pM=pFHa : pM<pF The test is: right…
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A: Given: Sample data table is shown below,
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Q: Use technology to find the P-value for the hypothesis test described below. The claim is that for a…
A: We have to find pvalue.
Q: Use technology to find the P-value for the hypothesis test described below. The claim is that for a…
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Q: Z-Test Inpt: Data Stats u 60 Ho 60 a4.75 58 25 n 45 Use the calculator displays to the right to make…
A: Based on the p-value, the rejection rule is "If the p-value is less than the level of significance,…
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- Use technology to find the P-value for the hypothesis test described below. The claim is that for a smartphone carrier's data speeds at airports, the mean is u = 15.00 Mbps. The sample size is n = 20 and the test statistic is t= 2.079. P-value = (Round to three decimal places as needed.)Interpret computer output: A sample of college students was asked whether they had a job outside of school. The following MINITAB output presents the results of a hypothesis test regarding the proportion of college students who have a job outside of school. Test of p=0.3 vs p>0.3 95% Lower X N Sample p Bound Z-Value P-Value 411 1198 0.343072 0.320511 3.25 0.001 Part: 0 / 3 Part 1 of 3 (a) What are the null and alternate hypotheses? Ho: OO D=0 H:Do men have a higher body temperature than women? Test the indicated claim about the means of two populations. Assume that the two samples are randomly selected, independent, the population standard deviations are not know and not considered equal. The table shows results from a study of body temperatures of men and women. At the 0.05 significance level, test the claim that men have a higher body temperature than women. Men Women n1 = 13 n2 = 16 xˉx̄1 = 97 °F xˉx̄2 = 95.17 °F s1 = 0.26 °F s2 = 0.57 °F What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)H0: Select an answer x̄₂ μ₁ p p₂ μ₂ μ μ(men) p̂₁ x̄₁ σ₁² p₁ s₁² ? > < ≥ ≠ ≤ = Select an answer p̂₁ p₁ μ μ₁ p x̄₁ μ₂ s₁² x̄₂ μ(women) σ₁² p₂ H1: Select an answer σ₂² μ p̂₂ p₁ p p₂ x̄₁ μ(men) x̄₂ μ₂ μ₁ s₂² ? = ≤ > < ≥ ≠ Select an answer s₁² x̄₁ μ₂ μ p̂₁ x̄₂ σ₁² μ(women) p₁ p p₂ μ₁ Original Claim = Select an answer H₀ H₁ df = Based on the hypotheses, find…
- A researcher is testing a hypothesis of a single mean. The critical t value for a = .05 and a one-tailed test is 2.0639. The observed t value from sample data is 1.742. The decision made by the researcher based on this information is to the null hypothesis. reject not reject redefine change the alternate hypothesis into restateA researcher studying air quality in a metropolitan city is interested in testing the claim that less than 20% of people smoke cigarettes to test this claim the researcher collects the following data on a sample of 700 adults and 110 of them smoke cigarettes the following is the data from the study. Sample size= 700 adults Alternative hypothesis= ha:pThe test statistic in a right-tailed test is z=−1.83. Determine the P-value and decide whether, at the 5% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The P-value is _____________ (Round to three decimal places as needed.) This P-value _____________________ (does not provide or provides) sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is_____________ (greater than or less than) the significance level.
- Mark performed a two-sample z-test for proportions to test the hypothesis that there was no difference in the proportion who support increasing student fees between male and female students at a particular university. Mark obtained a z-statistic of 0. Based on this information, which of the following is always true. The sample proportions for male and females will be the same as the hypothesized proportions for male and female students at this university. The sample proportions for both male and female students at this university will be 0. All male and female students in the sample will feel the same way about supporting the increase in student fees. The sample proportions will be the same for both male and female students at this university. Test whether p1>p2 The sample data are x1=116,n1=258,x2=135,n2=311. Conduct a test at the α=0.10 level of significance. Determine the correct null and alternative hypothesis below. Determine the test statistic. Find the P-value: What is…Big fish: A sample of 250 flounder of a certain species have sample mean weight 50.5 grams. Scientists want to perform a hypothesis test to determine how strong the evidence is that the mean weight is less than 49 grams. State the appropriate null and alternate hypotheses.Use the AustStates data. Perform a repeated measures ANOVA using the 9 years as the paired variables and the states as the observations. Based on the Greenhouse-Geisser p-value of p=0.130, sphericity can be assumed for these data. The result of the analysis is F=9.563 on degrees of freedom, for a p-value of You the null hypothesis. 8 and 56, p<0.001, reject 8 and 56, p<0.001, fail to reject 1.0 and 7.3, p=0.016, reject 1.0 and 7.3, p=0.016, fail to reject
- A claim is given. Select the corresponding null hypothesis and using a significance level of α = 0.05 and the given p-value, determine if you reject the null hypothesis or fail to reject the null hypothesis. Claim: Americans spend on average less than 8 minutes in the shower. Data: In a sample of Americans (n = 1,803), the average time spent in the shower is 7.8 minutes with a p-value of 0.0546. A. H0 :μ=8and we reject the null. B. H0 :μ=8and we fail reject the null. C. H0 :μ>8and we reject the null. D. H0 :μ>8and we fail reject the null. E. H0 :μ<8and we reject the null. F. H0 :μ<8and we fail reject the null.Test the claim about the population variance o² at the level of significance α. Assume the population is normally distributed. Claim: o² ≥8.5; α = 0.10 2 Sample statistics: s² = 7.78, n = 25 Write the null and alternative hypotheses. Ho: 0² ≥ 8.5 Ha: 0² < 8.5 (Type integers or decimals. Do not round.) Calculate the standardized test statistic. x² = 21.97 (Round to two decimal places as needed.) Determine the P-value. P-value = (Round to three decimal places as needed.)Test the daim that the proportion of men who own cats is significan tly different than the proportion of women who own cats at the 0.02 significance level. The null and alternative hypothesis would be: Ho:PM = PF Ho:HM PF Ho:PM µp H1:µM PF H1:µM + HF The test is left-tailed two-tailed right-tailed Based on a sample of 80 men, 40% owned cats Based on a sample of 20 women, 50% owned cats positive Critical Value = [three decimal accuracy] Test Statistic = [three decimal accuracy] Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis