Assuming the populations of measurements to be ap- proximately normally distributed, test the hypothesis that oA = OB against the alternative that o A # OB. Use a P-value.
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Rate of Change
The relation between two quantities which displays how much greater one quantity is than another is called ratio.
Slope
The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:
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- Question 20Question 2: Consider testing the hypotheses Ho :p= 0.4 versus Ha :p>0.4. Four possible sample statistics, along with four possible p-values, are given. Match the statistics to their p-values. Ap30.42 Вр30.38 C p = 0.51 Dp= 0.46 a. p-value = 0.72 b. p-value = 0.293 С. p-value = 0.138 correct d. p-value = 0.019 %3DChoose the correct answer and show a hand written solution
- Question 4 く Test the claim that the mean GPA of night students is larger than the mean GPA of day students at the .01 significance level. The null and alternative hypothesis would be: Ho:PN = PD Ho:HN H1:PN > PD H1:µN > HD H1:PN < PD H1:PN # PD H1:µN < HD H1:µN # µD HD Ho:PN = PD Ho:PN = pD Ho:HN = HD Ho:UN = D The test is: two-tailed left-tailed right-tailed The sample consisted of 70 night students, with a sample mean GPA of 2.22 and a standard deviation of 0.03, and 50 day students, with a sample mean GPA of 2.21 and a standard deviation of 0.02. The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the nul hypothesisQUESTION 20 An environmental researcher wants to know whether the mean amount of sulfur dioxide in the air in UAE cities is less than 1.22 parts per billion (ppb). She tested the hypotheses Ho: 21.22 versus Ha: < 1.22 using a significance level a = 0.05. The p-value of the test is 0.045. If the true value of µ is 1.23, the conclusion results in Correct decision. Type I error. Both Type I and Type II errors. Type II error.12 PM Tue Dec 14 ... ohm.lumenlearning.coma Question 7 > Test the claim that the mean GPA of night students is larger than 3.2 at the .05 significance level. The null and alternative hypothesis would be: Ho:p = 0.8 Ho:H = 3.2 Ho:u = 3.2 Ho:p = 0.8 Ho:p= 0.8 Ho:µ = 3.2 H1:p 3.2 H1:p# 0.8 H1:p> 0.8 H1:µ + 3.2 The test is: two-tailed right-tailed left-tailed Based on a sample of 45 people, the sample mean GPA was 3.21 with a standard deviation of 0.02 The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis Question Help: D Post to forum Submit Question
- Question 3It is believed that the waiting time for a service has a population mean value of µ =13minutes. One wants to test whether the population mean is 13 minutes. The waiting time of a sample of 60 services is collected. The significance level alpha is set at 5%. The Excel data analysis results are presented in table 2.1) Give appropriate null and alternative hypotheses.Null hypothesis H 0 .Alternative hypothesis H 1 2) Find the p value for this test.3) What decision should be made about the hypotheses? Give evidence for your choice.Table 2: One sample t-test results VariableMean 14.72Variance 34.38Observations 60.00Hypothesized Mean 0.00 DifferenceDf 59.00t Stat 2.27P(T<=t) one-tail 0.02t Critical one-tail 1.67P(T<=t) two-tail 0.04t Critical two-tail 2.00Index of Biotic Integrity. Refer to the Journal of Agricultural, Biological, and Environmental Sciences (June 2005) analysis of the Index of Biotic Integrity (IBI). The IBI measures the biological condition or health of an aquatic region. Summary data on the IBI for sites located in two Ohio river basins, Muskingum and Hocking, are given in the next table. Conduct a test of hypothesis (at α = .10) to test that the mean IBI values of Hocking river are higher then IBI levels of Muskingum river. River Basin Sample Size Mean Standard Deviation Muskingum 53 .035 1.046 Hocking 51 .340 .960 State the null and alternate hypothesis significance level used Test statistic p-value Decision ConclusionQuestion 2: Consider testing the hypotheses Ho :p = 0.4 versus Ha : p> 0.4. Four possible sample statistics, along with four possible p-values, are given. Match the statistics to their p-values. Ap = 0.42 В р%30.38 C p = 0.51 D p = 0.46 a. p-value = 0.72 b. p-value = 0.293 p-value = 0.138 d. p-value = 0.019 %3D C.
- Suppose the average blood sugar level in 35- to 44-year-olds is 4.86 (mmol/L). Do sedentary people have a different blood sugar level than that of the general population? To answer that question, a hypothesis test is planned to collect data from a group of 200 sedentary people in this age group. State the null and alternative hypothesis for this study. A. H0: u (mean) ≠ 4.86 mmol/L vs Ha: u (mean) = 4.86 mmol/L B. H0: u (mean) = 4.86 mmol/L vs Ha: xbar ≠ 4.86 mmol/L C. H0: u (mean) = 4.86 mmol/L vs Ha: u (mean) > 4.86 mmol/L D. H0: u(mean) = 4.86 mmol/L vs Ha: u (mean) ≠ 4.86 mmol/LQUESTION 2 To test the effect of a new medication on pulse rate, the researcher randomly selected the 60 subjects into two groups of 30. New medicine and placebo were given to Group 1 and Group 2, respectively. Below is the Minitab output. Two-Sample T-Test and CI Group Мean StDev SE Mean 1 30 65.20 7.80 1.4 2 30 70.30 8.40 1.5 Difference = mu (1) - mu (2) Estimate for difference: 95% CI for difference: T-Test of difference = 0 (vs not =): T-Value = -2.44 -5.10 (-9.29, -0.91) P-Value=0.018 DF=57 ii) Using to.025,57 = 2.003 shows that the 95% confidence interval for the difference in mean pulse rate between new medication and placebo is (-9.29, -0.91). Hence, interpret its meaning.Denny studied factors affecting cardiovascular reactivity to a postural change in a sample of healthy men and women. Change in systolic blood pressure (SBP) from supine to standing position was used as the measure of cardiac reactivity in a sample of 842 adults. Supine measurement of blood pressure was taken after participants had lain on an examination table for 15 minutes. Blood pressure was determined every 30 seconds using a Dinamap device for 2 minutes, and the measurements were averaged. Participants then stood up immediately, and additional measurements were made for another 2 minutes. Below is a table describing the characteristics of Denny’s sample: Which variables in the table, if any, were measured on a nominal scale? Which variables in the table, if any, were measured on an ordinal scale? Which variables in the table, if any, were measured on an interval scale? Which variables in the table, if any, were measured on a ratio scale? Which measures of central tendency, if any,…