A die is rolled 30 times and the results are recorded in the table.Number Rolled 123456Frequency694425A goodness-of-fit chi-square test is to be used to test the null hypothesis that the die is fair. At a significance level of a = 0.01, what is the value of chi-square and the appropriate conclusion?Ox²=5.6; fail to reject the null hypothesis=0x =5.6; reject the null hypothesis= 11.2; fail to reject the null hypothesisOx11.2; reject the null hypothesisO 15.33; fail to reject the null hypothesis=

For a fair die, each possible outcome should occur with equal frequency. If we are tossing a fair…

Answered: A die is rolled 30 times and the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A sample mean, sample size, and population standard deviation are provided below. Use the one-mean z-test to perform the required hypothesis test at the 1% significance level. x=34, n=20, σ=7, H0: μ=37, Ha: μ≠37 The test statistic is z=___________ (Round to two decimal places as needed.) Identify the critical value(s). Select the correct choice below and fill in the answer box within your choice. (Round to two decimal places as needed.) A. The critical values are ±zα/2=±_________ B. The critical value is zα=____________ C. The critical value is −zα=____________ ________________ (Do not reject, Reject) the null hypothesis. The data ____________ ( provide, do not provide) sufficient evidence to conclude that the mean is ______________ (not equal to 37, less than 37, equal to 37, greater than 37)
Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is µ< 98.6°F. The sample size is n = 7 and the test statistic is t= P-value = (Round to three decimal places as needed.) hs
15% of all Americans suffer from sleep apnea. A researcher suspects that a lower percentage of those who live in the inner city have sleep apnea.Of the 352 people from the inner city surveyed, 39 of them suffered from sleep apnea.Test the calim at the level of significance of αα = 0.01? What type of test should be run? t-test of a mean z-test of a proportion The alternative hypothesis indicates a two-tailed test one-tailed test
Nonparametric: A small study is conducted to compare birth weight in babies born to mothers who do not smoke to those that smoke more than ½ pack/day. Is there a significant difference between birth weights due to maternal smoking status? Use the Mann-Whitney U test with a 5% level of significance. What is U1 (non-smokers)? Birth weights of infants in two groups of mothers. (Observed data) Nonsmokers Heavy Smokers (≥1/2 pack/day) 8.6 7.0 8.5 5.2 6.3 6.1 9.3 6.7 8.0 6.6 6.8 7.4 4.9 6.3
Hypotheses for a chi-square goodness-of-fit test are given, along with the sample data and the chi-square statistic.Hypotheses:H0:pA=0.3,pB=0.3,Ha: Some pi is not as givenSample Data: A B C Observed(Expected) 28(38) 50(38) 48(50) Chi-square statistic=6.5 (a) What are the degrees of freedom for the test? (b) What is the p-value?Round your answer to five decimal places. (c) What is the conclusion, using a 5% significance level? Reject H0: There is evidence that the assumed proportions are wrong. Reject H0: There is evidence to support the assumed proportions. Do not reject H0: The assumed proportions appear to be wrong. Do not reject H0: The assumed proportions seem correct.
(see attachment) Use technology to conduct a one-factor ANOVA to determine if the group means are equal using α=0.05α=0.05. Group means (report to 2 decimal places): Group 1: Internship: Group 2: Co-op: Group 3: Work Study: ANOVA summary statistics:F-ratio = (report accurate to 3 decimal places)p=p= (report accurate to 4 decimal places)
Test the hypothesis that 0₁ #02 at the a= 0.01 level of significance for the given sample data. Assume that the populations are normally distributed. Use the P-value approach. Ho: 01 H₁: 01 Jhumu Search S Use technology to find the P-value for this test. The P-value is 0.072 (Round to three decimal places as needed.) Make a statement regarding the null hypothesis and draw a conclusion for this test. Since the P-value is the = 0.01 level of significance, the null hypothesis. There sufficient evidence to conclude that the standard deviation of Population 1 is significantly different from the standard deviation of Population 2 at the x=0.01 level of significance. LOL Population 1 Population 2 19 19 3.1 4.8 DELL M
The test statistic in a left-tailed test is z=−0.59. Determine the P-value and decide whether, at the 10% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The P-value is ___________ (Round to four decimal places as needed.) This P-value ________________ (provides, does not provide) sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is_____________(greater than, less than) the significance level.
In a one-tail hypothesis test where you reject H0 only in the lower tail, it was found that the p-value is 0.1469 if ZSTAT=−1.05. What is your statistical decision if you test the null hypothesis at the 0.05 level of significance? Question content area bottom Part 1 Choose the correct answer below. A. Reject the null hypothesis because the p-value is less than the level of significance. B. Reject the null hypothesis because the p-value is greater than the level of significance. C. Fail to reject the null hypothesis because the p-value is greater than the level of significance. D. Fail to reject the null hypothesis because the p-value is less than the level of significance.
what is the chi-square test for goodness of fit for the study (use the 0.01 level)State the null and research hypotheses?Determine the cutoff sample score at which the null hypothesis should be rejected.?what is the cutoff score ? Solve for A-F based on the chart attached?
A survey by the National Institutes of Health asked a random sample of young adults (aged 19 to 25 years), "Where do you live now? That is, where do you stay most often?" Given is Minitab output from a chi-square test. (omitting a few who refused to answer and one who reported being homeless). Chi-Squared Test: Female, Male Expected counts are printed below observed Which cell contributes most to the chi-square test counts Chi-Square contributions are printed below statistic? In what way does this cell differ from what we expected counts would expect if there is no association between gender and "where do you live now?" O The largest component comes from females who currently live in their parents' homes. This cell had the largest expected count. O The largest component comes from males who currently live in their parents' homes. There were more males living in their parents' homes than would have been expected. O The smallest component comes from males who currently live in "group"…

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