Question 2 It would be a one-tailed test. Hypothesis for an one-tail test: Ho:μHo≥μHa Compute Zobt: The Z critical Value = ±1.645 Ho: should be rejected, the researcher should conclude that the new toothpaste prefix better at helping prevent cavities. 95% confident that the population mean is within the range: Question 4 This is a type two error Question 6 Yes, this is a two-tailed Ho:μo≠μ1,μo<μ1 95% Confidence interval: 56.80479 < µ < 61.19521 2.160 No, The researcher should fail to reject Question 8 X^(2 )=2.55 df=1 X^(2 )=3.84 X^2 (1,N=120)=2.55,p>0.05 Reject the Null Hypothesis and concluded the national rate of exercise is greater. Jackson even-numbered Chapter exercises (pp. 273-275) Question 2 He should use an independent t-Test Ho: μo music=μa no music | Ho: μo music ≤ μa no music 6-7.6 √(2.88/9+4.56/9=-1.76 ) Yes, he should reject; he should conclude that studies without music produce higher test grades. yes; it is significant Test Statistic, t: -2.6515, Critical t: ±2.146376, P-Value: 0.0191, Degrees of freedom: 14, 95% Confidence interval: -2.895174 < µ1-µ2 < -0.3048258 Degrees of freedom Degrees of freedom are a collection of sample data comprised of a number of sample values that varies according to dynamic restrictions that may be present. They are the number of data points (N) that can change without affecting the mean of the data set. Jackson p. 210 without affecting the mean. Additionally, they are
Note: the program we used on this worksheet said the results were not significant, but then in statistical notion it had “p < 0.05”. That is confusing because I believe it should be “p >
Exercise level-as mentioned before- needs to be reported in a different way. You have to mention the percentage of participants belonged to each exercise category and compare the proportions between the two groups. The p value related to the chi-square test needs to be reported too.
Since the P-value (0.386) is greater than the significance level (0.05), we fail to reject the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis.
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.
The degree of freedoms was 9 and the significance level was 0.05. For these conditions, the chi-square value must be above 16.92. The test statistic provided no convincing evidence that the
A researcher predicts that listening to music while solving math problems will make a particular area of the brain more active. To test this, a research participant has her brain scanned while listening to music and solving math problems, and the brain area of interest has a percentage signal change of 58. From many previous studies with the same math problems procedure (but not listening to music), it is known that the signal change in this brain area is normally distributed with a mean of 35 and a standard deviation of 10. (a) Using the .01 level, what should the researchers conclude? Solve this problem explicitly using all five steps of hypothesis testing, and illustrate your answers with a sketch showing the comparison distribution, the cutoff (or cutoffs), and the score of the sample on this distribution. (b) Then explain your answer to someone who has never had a course in statistics (but who is familiar with mean, standard deviation, and Z scores).
Under tax law, the Accounting Standards Codification (ASC) states four possible sources of taxable income. These four sources may realize a tax benefit for deductible temporary differences and carryforwards. The sources as stated directly in ASC 740-10-30-18 are:
So, we should reject the null hypothesis H0. At a 0.05 level of significance level, we conclude that there is a significant difference between the average height for females and the average height for the males.
Truman Capote Truman Capote was an astonishing, well-known writer. He wrote many great novels and short stories in the fifty-nine years of his life. Many of his works were made into movies, with many great actors starring in them. Even though his life was very reckless, he still managed to show how creative he was in his works.
It tells that the t-statistic with 97 degrees of freedom was 2.14, and the corresponding p-value was less than .05, specifically around 0.035. Therefore, it is appropriate to conclude the research study was statistically significant.
1. Using the data in the file named Ch. 11 Data Set 2, test the research hypothesis at the .05 level of significance that boys raise their hands in class more often than girls. Do this practice problem by hand using a calculator. What is your conclusion regarding the research hypothesis? Remember to first decide whether this is a one- or two-tailed test.
Since 3.27 the t statistic is in the rejection area to the right of =1.701, the level of
This was apparent with the case of Colgate ' Palmolive introducing a new toothpaste that fights not only cavities but bad breath, yellowing teeth, sensitive gums and also gingivitis.
Attention getter statement: Toothpaste: We use toothpaste to brush our teeth everyday (hopefully). We here in America love to have white teeth and from the time we are very young, we are told by our parents and our dentists that we need to brush twice daily with fluoride in order to prevent cavities. But what if I told you that toothpaste was poisonous?
The n-HAp present in the toothpaste precipitated as closely packed grins on the enamel surface during the simulated one year brushing as revealed by the AFM in the current study (Fig. 2 A, B) and confirmed by the SEM in a previous study 16. Therefore, the n-HAp function is to protect the teeth with the formation of a new layer of synthetic enamel around the tooth rather than altering its surface roughness under conventional conditions 3, 16. It worth mentioning, that the obtained Ra (nm) mean values of enamel were slightly higher than those obtained from previous studies 13, 22-25. This could be related to the topical application of the toothpaste instead of brushing and the method of specimen preparation. Additionally, the higher SD value noticed among the control group could be related to the human variation among the teeth used.