The p-value for the one-sided test Ho: µ = 3 ppm vs. H:u> 3 ppm was 0.23. Do you reject the null hypothesis at a 5% significance level? Explain your answer.
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- A researcher wants to know if there is a significant difference in the amount of time professional track cyclists and road cyclists spend per week training on their bike. For this test the level of significance is 0.05, the degrees of freedom are 7, and the critical value is 2.37. Number of hours for track cyclists (Mean = 26.00; SD = 2.71): 22, 28, 27, and 27 Number of hours for road cyclists (Mean = 30.20; SD = 1.92): 31, 28, 33, 29, and 30 For this question, on your hand calculation document clearly state: a) the null hypothesis, b) the alternative hypothesis, c) the alpha level you are using, d) the critical value, e) your standard error, f) the t-value, g) and your decision about the null hypothesis. This is what I will be marking. You may also enter your responses here, but it is not required. Priority will be given to the hand calculation document.The Nero Match Company sells matchboxes that are supposed to have an average of 40 matches per box, with σ = 7. A random sample of 92 matchboxes shows the average number of matches per box to be 42.0. Using a 1% level of significance, can you say that the average number of matches per box is more than 40? What are we testing in this problem? single proportion single mean (a) What is the level of significance? State the null and alternate hypotheses. H0: μ = 40; H1: μ ≠ 40 H0: p = 40; H1: p ≠ 40 H0: p = 40; H1: p > 40 H0: μ = 40; H1: μ > 40 H0: p = 40; H1: p < 40 H0: μ = 40; H1: μ < 40 (b) What sampling distribution will you use? What assumptions are you making? The Student's t, since we assume that x has a normal distribution with unknown σ. The standard normal, since we assume that x has a normal distribution with known σ. The standard normal, since we assume that x has a normal distribution with unknown σ. The Student's t, since we assume that x has a normal…Please answer all sub-parts Test the claim that the proportion of people who own cats is larger than 10% at the 0.10 significance level.The null and alternative hypothesis would be: H0:μ=0.1H0:μ=0.1Ha:μ≠0.1Ha:μ≠0.1 H0:p=0.1H0:p=0.1Ha:p≠0.1Ha:p≠0.1 H0:μ≤0.1H0:μ≤0.1Ha:μ>0.1Ha:μ>0.1 H0:μ≥0.1H0:μ≥0.1Ha:μ<0.1Ha:μ<0.1 H0:p≥0.1H0:p≥0.1Ha:p<0.1Ha:p<0.1 H0:p≤0.1H0:p≤0.1Ha:p>0.1Ha:p>0.1 The test is: two-tailed right-tailed left-tailed Based on a sample of 500 people, 12% owned catsThe test statistic is: (Round to 2 decimals)The p-value is: (Round to 2 decimals)Based on this we: Reject the null hypothesis Do not reject the null hypothesis
- A bit confused on this problem.You are conducting a study to see if the proportion of women over 40 who regularly have mammograms is significantly less than 0.87. You use a significance level of α=0.05. H0:p=0.87 H1:p<0.87You obtain a sample of size n=388n=388 in which there are 325 successes.What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value =A sample of n=10,000 (x, y) pairs resulted in r= .022. Test H0: p=0 versus Ha: p=? 0 at significance level .05. Is the result statistically significant? Comment on the practical significance of your analysis.
- Test the claim that the proportion of men who own cats is significantly different than 60% at the 0.02 significance level.The null and alternative hypothesis would be: H0:μ=0.6H0:μ=0.6H1:μ<0.6H1:μ<0.6 H0:p=0.6H0:p=0.6H1:p>0.6H1:p>0.6 H0:p=0.6H0:p=0.6H1:p<0.6H1:p<0.6 H0:μ=0.6H0:μ=0.6H1:μ≠0.6H1:μ≠0.6 H0:μ=0.6H0:μ=0.6H1:μ>0.6H1:μ>0.6 H0:p=0.6H0:p=0.6H1:p≠0.6H1:p≠0.6 The test is: two-tailed right-tailed left-tailed Based on a sample of 60 people, 66% owned catsThe test statistic is: (to 2 decimals)The positive critical value is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesisYou wish to test the claim that u = 45 at a level of significance of a = 0.05 and are given sample statistics n = 6, x = 60, and s= 15. Compute the value of the standardized test statistic, t, round your answer to three decimal places.You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly different from 0.28. You use a significance level of α=0.01 H0:p=0.28 H1:p≠0.28You obtain a sample of size n=613 in which there are 163 successes.What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value =
- For the following hypothesis, state the IV and DV with their respective measurement scales: "The gender of the interviewer affects the interviewee's anxiety levels (low/med/high)." Independent Variable = IV scale of measurement = Dependent Variable = DV scale of measurement =Test the claim that the proportion of people who own cats is significantly different than 80% at the 0.02 significance level. The null and alternative hypothesis would be: Ho:µ 0.8 Ho:p= 0.8 Ho:µ = 0.8 Ho:p > 0.8 H1:µ > 0.8 H1:p > 0.8 H1:µ < 0.8 H1:p 0.8 H1:µ + 0.8 H1:p < 0.8 The test is: two-tailed right-tailed left-tailed Based on a sample of 100 people, 84% owned cats (to 2 decimals) The p-value is: Based on this we: Fail to reject the null hypothesis Reject the null hypothesisYou wish to test the following claim (Ha) at a significance level of a = 0.02. H.:P1 = P2 Ha:P1 < P2 You obtain a random sample of size 476 from the first population, with 344 successes. You obtain a random sample of size 398 from the second population, with 331 successes. What is the test statistic for this sample? (Report answer accurate to 2 decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to 3 decimal places.) p-value = The p-value is... less than a greater than a This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... Because our p-value is less than alpha, we fail to reject the Ho. There is not enough evidence to support the claim that the first population proportion is less than the second population proportion. Because our p-value is greater than alpha, we reject the Ho. There is enough evidence to support the claim that the first population…
