A simple random sample of size n = 40 is drawn from a population. The sample mean is found to be 108.6, and the sample standard deviation is found to be 19.6. Is the population mean greater than 100 at the a = 0.01 level of significance? Determine the null and alternative hypotheses. Ho H₁: 4 COD

The average fees paid by students at state universities across the country were $2236 per student last year. The average for a random sample of 500 students this year is $2350. Given the following hypotheses, Ho: 4 = 2236 (The average fees are the same as last year.) Ha: H = 2236 (The average fees are not the same as last year.) Compute the p-value for the sample result. Assume the population standard deviation is $724. If the significance level is 5%, can the null hypothesis be rejected? A. none of these B. The p-value here is .04. Since .04 is less than the significance level, that is, since .04 < .05, we can reject HO C. The p-value here is .02. Since 02 is less than the significance level, that is, since 02 < 05, we can't reject HO. D. The p-value here is .0004. Since .0004 is less than the significance level, that is, since .0004 < .05, we can reject HO.

A sample of 33 observations is selected from a normal population. The sample mean is 53, and the population standard deviation is 6. Conduct the following test of hypothesis using the 0.05 significance level. Hei H = 57 Hi:H # 57 a. Is this a one- or two-tailed test? O One-tailed test O Two-tailed test es b. What is the decision rule? Reject Hg if -1.960 1.960

The null and alternate hypotheses are:   H0 : μ1 = μ2H1 : μ1 ≠ μ2   A random sample of 10 observations from one population revealed a sample mean of 22 and a sample standard deviation of 3.7. A random sample of 7 observations from another population revealed a sample mean of 26 and a sample standard deviation of 5.0.   At the 0.10 significance level, is there a difference between the population means?   State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.)           Compute the pooled estimate of the population variance. (Round your answer to 3 decimal places.)           Compute the test statistic. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)           State your decision about the null hypothesis.   multiple choice 1 Reject H0. Do not reject H0.     The  p-value is   multiple choice 2 between 0.1 and 0.05 less than 0.001 between…

he average cost of tuition and room and board for a small private liberal arts college is reported to be $8,500 per term, but a financial administrator believes that the average cost is higher. A study conducted using 350 small liberal arts colleges showed that the average cost per term is $8,745. The population standard deviation is $1,200. Let α = 0.05. What are the null and alternative hypotheses for this study?     H0: µ ≤ $9,000; H1: µ > $9,000     H0: µ ≥ $9,000; H1: µ < $9,000     H0: µ ≤ $8,500; H1: µ > $8,500     H0: µ ≥ $8,500; H1: µ < $8,500

10:13 AM Thu Dec 10 * 90% +: 0 ---- 3) Jack is a golfer, and his actual mean drive distance last year was 260 yards. He has been practicing a new technique all year. He now wants to test to see if he has made an improvement. The distances are given below. Perform a hypothesis test at the 10% significance level to see if his mean drive distance is now greater than last year's average of 260. 270 280 249 260 264 250 258 259 280 270 262 272 290 255 261 259 Assume that the true population of distances is normally distributed with standard deviation 9.8. a) Calculate the sample mean. (To save you some time, the sum is 4239) b) Set up the null and alternative hypotheses c) State the significance level, a, in probability form. d) Compute the test statistic. PLEASE SHOW THE STEPS. No credit will be given if work is not show. 8 >

I. A study reports that teenagers spend 27 hours a week online (www.telegraph.co.uk). A researcher wanted to check if this claim is true. A random sample of 200 teenagers taken showed that they spend an average of 25.7 hours per week online with a standard deviation of 8.2 hours. Perform a hypothesis test with the alternate hypothesis that teenagers spend less than 27 hours per week online. Use c = 0.05. Perform all of the steps outline in your notes and write a complete conclusion in words. Assume the distribution of hours spent online is normally distributed.

Do men take more time than women to get out of bed in the morning? The 46 men observed averaged 6.5 minutes to get out of bed after the alarm rang. Their standard deviation was 1.1. The 45 women observed averaged 6 minutes and their standard deviation was 2.9 minutes. What can be concluded at the αα = 0.01 level of significance? For this study, we should use        The null and alternative hypotheses would be:        H0:H0: u1 or p1, <,>,=,   u2, p2            (please enter a decimal)     H1:H1:  u1 or p1, <,>,=,   u2, p2               (Please enter a decimal) The test statistic z or t  is 1.083. Enter it here     (please show your answer to 3 decimal places.) The p-value is 0.1417. Enter it here      (Please show your answer to 4 decimal places.) The p-value is   <,>  αα Based on this, we should fail to reject, accept,or reject      the null hypothesis. Thus, the final conclusion is that ... The results are statistically significant at αα = 0.01, so there is…

Claim: The mean systolic blood pressure of all healthy adults is less than than 119 mm Hg. Sample data: For 283 healthy adults, the mean systolic blood pressure level is 118.53 mm Hg and the standard deviation is 15.06 mm Hg. The null and alternative hypotheses are Ho μ=119 and H₁: μ< 119. Find the value of the test statistic. The value of the test statistic is (Round to two decimal places as needed.)

A simple random sample of size n = 40 is drawn from a population. The sample mean is found to be 103.7, and the sample standard deviation is found to be 20.4. Is the population mean greater than 100 at the x = 0.10 level of significance? Determine the null and alternative hypotheses. Ho: H₁: Compute the test statistic. = (Round to two decimal places as needed.) Determine the P-value. The P-value is . (Round to three decimal places as needed.) What is the result of the hypothesis test? the null hypothesis because the P-value is the level of significance. At the α = 0.10 level of significance, the population mean