Use the z-table, find the critical region/s for each: a) a = 0.015, two-tailed test b) a = 0.025, left-tailed test c) a = 0.0125, right-tailed
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Use the z-table, find the critical region/s for each:
a) a = 0.015, two-tailed test
b) a = 0.025, left-tailed test
c) a = 0.0125, right-tailed test
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- The average American gets a haircut every 41 days. Is the average smaller for college students? The data below shows the results of a survey of 13 college students asking them how many days elapse between haircuts. Assume that the distribution of the population is normal. 44, 36, 30, 41, 41, 34, 41, 30, 44, 31, 36, 38, 43 What can be concluded at the the αα = 0.05 level of significance level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? p μ Select an answer > ≠ = < H1:H1: ? μ p Select an answer ≠ = > < The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 3 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer fail to reject accept reject the null hypothesis. Thus, the final conclusion is that ... The data suggest…The average American gets a haircut every 43 days. Is the average larger for college students? The data below shows the results of a survey of 11 college students asking them how many days elapse between haircuts. Assume that the distribution of the population is normal. 38, 37, 38, 45, 53, 55, 35, 37, 35, 41, 38 What can be concluded at the the α = 0.05 level of significance level of significance? For this study, we should use? Select an answer: t-test for a population mean? z-test for a population mean? The null and alternative hypotheses would be: H0:H0? p or μ Select an answer > = ≠ ≤ < ≥ H1:H1? μ or p Select an answer ≠ ≥ > ≤ = < 2. The test statistic ? t z = (please show your answer to 3 decimal places.) The p-value = _______ (Please show your answer to 3 decimal places.) The p-value is? > or ≤ α 3. Based on this, we should?? Select an answer reject ? accept ? or fail to reject ?…Before the furniture store began its ad campaign, it averaged 186 customers per day. The manager is investigating if the average is larger since the ad came out. The data for the 16 randomly selected days since the ad campaign began is shown below: 170, 172, 203, 178, 200, 219, 181, 172, 218, 181, 184, 188, 192, 204, 205, 210 Assuming that the distribution is normal, what can be concluded at the αα = 0.05 level of significance? For this study, we should use : z-test for a population proportion or t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer ≠ < = > H1:H1: ? μ p Select an answer ≠ = < > The test statistic ? t or z = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should: accept, reject or fail to reject the null hypothesis. Thus, the final conclusion is that ... The data…
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- The distribution of Mathematics subject scores on the ACT in 2016 was approximately Normal with mean μ = 20.6 and standard deviation o = 6.15. (a) Students with a Math subject ACT score between 18 and 21 can enroll in MAT110E at Missouri Western. What percent of students taking the ACT may enroll in MAT110E? (b) Students must have at least a 22 Math subject ACT score to enroll in College Algebra. What percent of students taking the ACT qualified to enroll in College Algebra? (c) What percent of students taking the ACT had a Math subject score below 15? (d) How high must a student score on the Math subject portion of the ACT to be in the top 10%?Before the furniture store began its ad campaign, it averaged 158 customers per day. The manager is investigating if the average is smaller since the ad came out. The data for the 11 randomly selected days since the ad campaign began is shown below: 147, 142, 149, 133, 146, 159, 132, 137, 148, 172, 150 Assuming that the distribution is normal, what can be concluded at the αα = 0.10 level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer > ≠ < = H1:H1: ? p μ Select an answer < ≠ > = The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? ≤ > αα Based on this, we should Select an answer accept fail to reject reject the null hypothesis. Thus, the final conclusion is that ... The data suggest…Facebook: A study showed that two years ago, the mean time spent per visit to Facebook was 20.8 minutes, Assume the standard devlation is o= 8.0 minutes. Suppose that a simple random sample of 107 visits was selected this year and has a sample mean of x= 18.8 minutes. A social scientist is interested to know whether the mean time of Facebook visits has decreased. Use the a=0.10 level of significance and the critical value method with the table. (a) State the appropriate null and alternate hypotheses. (b) Compute the value of the test statistic. (c) State a conclusion. Use the a = 0.10 level of significance.
- Determine the cri tical z-score corresponding to a = 0.2 in a left-tail test. crititical z-score = [three decimal accuracy]Solve using R commands and manually Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 104 andstandard deviation 5 (“Mathematical Model of Chloride Concentration in Human Blood,” J. of Med.Engr. and Tech., 2006: 25–30”).(a) What is the probability that chloride concentration equals, less than, and at most 104?(b) What is the probability that someone’s blood chloride concentration level is between 99 and 114?(c) What is the probability that someone’s blood chloride concentration level is above 120?Before the furniture store began its ad campaign, it averaged 242 customers per day. The manager is investigating if the average is larger since the ad came out. The data for the 12 randomly selected days since the ad campaign began is shown below: 246, 261, 265, 253, 238, 267, 225, 272, 254, 259, 228, 255 Assuming that the distribution is normal, what can be concluded at the αα = 0.05 level of significance? For this study, we should use The null and alternative hypotheses would be: H0:H0: H1:H1: The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is αα Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest that the population mean number of customers since the ad campaign began is not significantly more than 242 at αα = 0.05, so there is insufficient evidence to conclude that the…