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
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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.
One-sample t-test are used in the parametric test which analyzes the means of populations. The t-test for independent groups are statistics that relates difference between treatment means to the amount of variability expected between any two samples of data within the same population (Hansen & Myers, 2012). Critical values are used in significant testing provide a range of t distribution that is used in whether a null hypothesis is rejected. Based on the data below as the level of significance is at .05, thus the critical values would fall under ±1.860 and the t value for this is 1.871 would suggest for the null to be rejected as it is greater than the critical value (Privitera, 2015, p. 267). Based on the population mean of 70 there was a mean difference of
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.
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).
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.
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 degrees of freedom (df) of an estimate is the number or function of sample size of information on which the estimate is based and are free to vary relating to the sample size (Jackson, 2012; Trochim & Donnelly, 2008).
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.
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.