1- During period of high electricity demand, especially during the hot summer months, the power output from a gas turbine engine can drop dramatically………….. Conduct a test to determine if the mean heat rate of gas turbines augmented with high-pressure inlet fogging exceeds 10,000 kJ/kWh. Use α = .05. (data for this question (GASTURBINE.sav) is in the CD accompanying the text book. This is also provided in the D2L content area in Module 3, Assessment section. Note that here we are interested in the heat rate only.) Do not forget to state the hypothesis. The hypothesis can be expressed as: H0: μ ≤10000 kJ/kWh Ha: μ >10000 kJ/kWh Where µ is the heat rate. It is one sample t- test. One-Sample Statistics | | N | Mean | …show more content…
State the hypothesis, conduct the appropriate test using SPSS and interpret the results. The hypothesis can be expressed as: H0: μ ≥5 Ha: μ< 5 μ: is the mean rating for machine performing. Random sample: 10 mixing colors. One-Sample Statistics | | N | Mean | Std. Deviation | Std. Error Mean | Gallons per minute | 10 | 4.6600 | .58157 | .18391 | One-Sample Test | | Test Value = 5 | | t | df | Sig. (2-tailed) | Mean Difference | 95% Confidence Interval of the Difference | | | | | | Lower | Upper | Gallons per minute | -1.849 | 9 | .098 | -.34000 | -.7560 | .0760 | P value= 0.098/2 = 0.049 Since P value is 0.049 which less than o.o5, the null hypothesis is rejected. The sample supports the alternative hypothesis that the mixing rate of the machine is less than 5 gallons per minute. 4-The data in the table, obtained from Business Week’s (June 22, 2006) technology section, represents typical salaries of technology professionals in 13 metropolitan areas for 2003 and 2005… (a) Set up the null and alternative hypothesis for the test. (f) Make the appropriate conclusion. More specifically, I want you to examine whether analyzing the data with SPSS should reject the null hypothesis. Use α = .05 (Yes, the α level has been changed from .10 to .05). The hypothesis can be expressed as: 1: The average
From the above output, we can see that the p-value is 0.000186, which is smaller than 0.05 (if we select a 0.05 significance level).
A. A three year study done by Boston College found that “tests profoundly shape what
15 In testing the hypotheses: H0 β1 ’ 0: vs. H1: β 1 ≠ 0 , the following statistics are available: n = 10, b0 = 1.8, b1 = 2.45, and Sb1= 1.20. The value of the test statistic is:
Because the p-value of .035 is less than the significance level of .05, I will reject the null hypothesis at 5% level.
Testing allows the p-value that represents the probability showing that results are unlikely to occur by chance. A p-value of 5% or lower is statistically significant. The p value helps in minimizing Type I or Type II errors in the dataset that can often occur when the p value is more than the significance level. The p value can help in stopping the positive and negative correlation between the dataset to reject the null hypothesis and to determine if there is statistical significance in the hypothesis. Understanding the p value is very important in helping researchers to determine the significance of the effect of their experiment and variables for other researchers
Select one (1) project from your working or educational environment that you would use the hypothesis test technique. Next, propose the hypothesis structure (e.g., the null hypothesis, data collection process, confidence interval, test statistics, reject or not reject the decision, etc.) for the business process of the selected project. Provide a rationale for your response.
Write a conclusion paragraph analyzing your results. In your conclusion, be sure to address the following questions:
Consider the ski resort satisfaction data we used in class (“satisfaction.sav” in Week 7 folder). This is data based on a survey of 500 users of a ski resort. In class we did part of the analysis. In this question you will be asked to redo some of the analysis and complete the rest. For part a), b) and c), write down your null and alternative hypotheses and then discuss the hypothesis testing results based on SPSS output. Please show your SPSS output for part a), b), and c).
Please enter the numbers here so that I know that you know how to read the SPSS output. Please paste your SPSS output right here where it belongs.
To test the null hypothesis, if the P-Value of the test is less than 0.05 I will reject the null hypothesis.
Conclusion : Fails to reject the null hypothesis. The sample does not provide enough evidence to support the claim that mean is significantly different from 12 .
I rejected the null hypothesis and found out the p-value is smaller than the significance level. The p-value is greater than the significance level 0.05.
With a P-value of 0.00, we have a strong level of significance. No additional information is needed to ensure that the data given is accurate.
d) Among all employees whose starting salary is below the median ($37,750), find a 95% confidence interval for the proportion who stay with D&Y for at least 3 years.