Test the claim that the mean GPA of night students is smaller than 2.6 at the 0.005 significance level. The null and alternative hypothesis would be: Ho: µ > 2.6 Ho:p = 0.65 Ho:< 2.6 Ho:p < 0.65 Ho:u = 2.6 Ho:p> 0.65 H1:µ < 2.6 H1:p# 0.65 H:p > 2.6 H1:p > 0.65 H1:µ + 2.6 H1:p< 0.65 The test is: right-tailed left-tailed two-tailed Based on a sample of 80 people, the sample mean GPA was 2.59 with a standard deviation of 0.08 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis

The lowest level of significance to reject the null hypothesis of no linear association between blood pressure and age is: OA: 0.003 OB: 0.05 OC: 0.0002 OD: 0.0001 OE: 0.04

A claim is made that the proportion of children who play sports is less than it used to be. Assume that you plan to use a significance level of αα =  0.05  to test the claim that p<0.5p<0.5. Find the critical value for this hypothesis test, then use x=396x=396 and n=1320n=1320 from the sample data to decide if the statistics support the claim. -1.96; yes, they do support the claim -1.96; no, they do not support the claim -1.645; yes, they do support the claim -1.645; no, they do not support the claim

You wish to test the following claim (Ha) at a significance level of α=0.01.       Ho:μ1=μ2       Ha:μ1>μ2 You obtain a sample of size n1=13 with a mean of M1=65.1 and a standard deviation of SD1=15.8 from the first population. You obtain a sample of size n2=14 with a mean of M2=49.4 and a standard deviation of SD2=8.3 from the second population.  Use the Theory-based inference applet to conduct the test of significance. (a)  What is the t-score for this sample?       t-score =   (Round to 2 decimal places.)  (b)  What is the p-value for this sample?      p-value =   (Round to 4 decimal places.) (c)  The p-value is... less than (or equal to) α   greater than α     (d)  This p-value leads to a decision to... strong evidence in support of the alternative accept the null null is plausible   (e)  As such, the final conclusion is that... We conclude that μ1>μ2 . We conclude that μ1=μ2 . We have statistically significant evidence to support the claim that μ1>μ2 ..…

You wish to test the following claim (HaHa) at a significance level of α=0.002α=0.002. For the context of this problem, μd=μ2−μ1μd=μ2-μ1 where the first data set represents a pre-test and the second data set represents a post-test.      Ho:μd=0Ho:μd=0      Ha:μd<0Ha:μd<0You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain pre-test and post-test samples for n=12n=12 subjects. The average difference (post - pre) is ¯d=−44.4d¯=-44.4 with a standard deviation of the differences of sd=39.4sd=39.4.What is the critical value for this test? (Report answer accurate to three decimal places.)critical value = What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = The test statistic is... in the critical region not in the critical region This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final…

A test of H₁: μ=7 versus H₁: u<7 is performed using a significance level of a=0.01. The P-value is 0.20. Part: 0/3 Part 1 of 3 (a) Is H rejected? Since P ≤ Part: 1 / 3 Part 2 of 3 (b) If the true value of u is 1, is the result a Type I error, a Type II error, or a correct decision? a, we do not reject Ho at the a= 0.01 level. The result is a Type I error Part: 2/3 Part 3 of 3 X 3 (c) If the true value of u is 7, is the result a Type I error, a Type II error, or a correct decision? The result is a [Choose one) ▾ Español

Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .05 significance level.The null and alternative hypothesis would be:   H0:μM=μFH1:μM<μF   H0:pM=pFH1:pM<pF   H0:pM=pFH1:pM>pF   H0:pM=pFH1:pM≠pF   H0:μM=μFH1:μM≠μF   H0:μM=μFH1:μM>μF The test is: right-tailed left-tailed   two-tailed Based on a sample of 20 men, 40% owned catsBased on a sample of 60 women, 55% owned catsThe test statistic is:  (to 2 decimals)The p-value is:  (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis

You wish to test the following claim (HaHa) at a significance level of α=0.002α=0.002. For the context of this problem, μd=μ2−μ1μd=μ2-μ1 where the first data set represents a pre-test and the second data set represents a post-test.      Ho:μd=0Ho:μd=0      Ha:μd>0Ha:μd>0You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain pre-test and post-test samples for n=40n=40 subjects. The average difference (post - pre) is ¯d=3.2d¯=3.2 with a standard deviation of the differences of sd=9.6sd=9.6.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 = The p-value is... less than (or equal to) αα greater than αα 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... There is…

Let's examine the mean of the numbers 1, 2, 3, 4, 5, 6, 7, and 8 by drawing samples from these values, calculating the mean of each sample, and then considering the sampling distribution of the mean. To do this, suppose you perform an experiment in which you roll an eight-sided die two times (or equivalently, roll two eight-sided dice one time) and calculate the mean of your sample. Remember that your population is the numbers 1, 2, 3, 4, 5, 6, 7, and 8. The true mean (p) of the numbers 1, 2, 3, 4, 5, 6, 7, and 8 is The number of possible different samples (each of size n = 2) is the number of possibilities on the first roll (8) times the number of possibilities on the second roll (also 8), or 8(8) = 64. If you collected all of these possible samples, the mean of your sampling distribution of means (HM) would equal , and the standard deviation of your sampling distribution of means (that is, the standard error or GM) would be The following chart shows the sampling distribution of the…

Test the claim that the proportion of men who own cats is smaller than 90% at the .05 significance level.The null and alternative hypothesis would be: H0:μ=0.9H0:μ=0.9H1:μ>0.9H1:μ>0.9 H0:p=0.9H0:p=0.9H1:p>0.9H1:p>0.9 H0:p=0.9H0:p=0.9H1:p≠0.9H1:p≠0.9 H0:μ=0.9H0:μ=0.9H1:μ≠0.9H1:μ≠0.9 H0:μ=0.9H0:μ=0.9H1:μ<0.9H1:μ<0.9 H0:p=0.9H0:p=0.9H1:p<0.9H1:p<0.9 The test is: right-tailed two-tailed left-tailed Based on a sample of 65 people, 87% owned catsThe test statistic is:  (to 2 decimals)The critical value is:  (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis I dont understand this.