The figure below shows a randomization distribution for testing H0 : µ = 50 vs Ha : µ < 50. Use the distribution to decide if the p-value for the observed sample mean x = 34 is closer to 0.02 or 0.35. Answer?
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The figure below shows a randomization distribution for testing H0 : µ = 50 vs Ha : µ < 50.
Use the distribution to decide if the p-value for the observed sample
0.02 or 0.35.
Answer?
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- Statistics students believe that the mean score on a first statistics test is 65. The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores: Grades 74.4 96 83.2 65 63.9 64.3 74.4 69 68.4 88 Test grades are believed to be normally distributed.Use a significance level of 5%. State the alternative hypothesis: HA:HA: μ>65μ>65 μ<65μ<65 μ≠65μ≠65 State the mean of the sample: (Round to two decimal places.) State the standard error of the sample means: (Round to four decimal places.) State the test statistic: t=t= (Round to four decimal places.) State the p-value: (Round to four decimal places.) Decision: Reject the null hypothesis. Do not reject the null hypothesis.Q2: Find the value of 'p’ if the mean of the distribution is 14.45, and hence find median, mode, variance and standard deviation: Class 2.525-7.524 7.525-12.524 | 12.525-17.524 17.525-22.524 22.525-27.524 Frequency 9 15 34 PStatistics students believe that the mean soore on a first statistics test is 65. The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores: Grades 96 69 83.2 69 65 85.5 74.4 65 85.5 66.5 Test grades are believed to be normally distributed. Use a significance level of 5%. A. State the alternative hypothesis: H: Ομ 65 B. State the mean of the sample: (Round to two decimal places.) C. State the standard error of the sample means: (Round to four decimal places.) D. State the test statistic: t = (Round to four decimal places.) E. State the D-value; (Round to four decimal places.) F. Decision: O Do not reject the null hypothesis. O Reject the null hypothesis. Submit Question
- Use technology to help you test the claim about the population mean, u, at the given level of significance, a, using the given sample statistics. Assume the population is nomally distributed. Claim: u> 1230; a = 0.03; o = 209.07. Sample statistics: x= 1251.28, n 250 O A. H,: µ2 1251.28 Hg: µ 1230 Hg: us 1230 O C. Ho us 1251.28 Ha: u> 1251.28 D. Ho: us 1230 Ha: p> 1230 Ο Ε Η : μ2 1230 OF. Ho: u>1251.28 Ha: µ<1230 H3: µS 1251.28 Calculate the standardized test statistic. The standardized test statistic is 1.61. (Round to two decimal places as needed.) Determine the P-value. P= 0.0537 (Round to three decimal places as needed.) Determine the outcome and conclusion of the test. Fail to reject Ho: At the 3% significance level, there is enough evidence to support the claim.A psychologist believes that youths identified as lesbian, gay, or bisexual already know their sexual orientation before they turned 18. To verify this, she took a random sample of 654 young adults that identified as lesbian, gay, or bisexual and asked them the age when they first realized their sexual orientation. It is known that the mean age when a young adult first realized they might be lesbian, gay, or bisexual is normally distributed. R COMMANDER OUTPUT sd se (mean) mean n 14.72199 14.9814 0.57568 654 95 percent confidence interval: 13.572 15.872Assume the samples are random and independent, the populations are nomally distributed, and the population variances are equal. The table available below shows the prices (in dollars) for a sample of automobile batteries. The prices are classified according to battery type. At a = 0.10, is there enough evidence conclude that at least one mean battery price is different from the others? Complete parts (a) through (e) below. E Click the icon to view the battery cost data. (a) Let u1. P2. H3 represent the mean prices for the group size 35, 65, and 24/24F respectively. Identify the claim and state Ho and H. H Cost of batteries by type The claim is the V hypothesis. Group size 35 Group size 65 Group size 24/24F 101 111 121 124 D 146 173 182 278 124 140 141 89 (b) Find the critical value, Fo, and identify the rejection region. 90 79 84 The rejection region is F Fo, where Fo = (Round to two decimal places as needed.) (c) Find the test statistic F. Print Done F= (Round to two decimal places as…
- Help is neededesc Introduction A chain of restaurants has historically had a mean wait time of 9 minutes for its customers. Recently, the restaurant added several very popular dishes back to their menu. Due to this, the manager suspects the wait time, μ, has increased. He takes a random sample of 44 customers. The mean wait time for the sample is 9.6 minutes. Assume the population standard deviation for the wait times is known to be 3.3 minutes. Can the manager conclude that the mean wait time is now greater than 9 minutes? Perform a hypothesis test, using the 0.05 level of significance. (a) State the null hypothesis Ho and the alternative hypothesis H₁. Ho: O H₁:0 (b) Perform a Z-test and find the p-value. Here is some information to help you with your Z-test. The value of the test statistic is given by Standard Normal Distribution Step 1: Select one-tailed or two-tailed. O One-tailed O Two-tailed Explanation μ Check □<口 X X OSO 020 0=0 x-H √n • The p-value is the area under the curve to the right…Solve this question please
- Statistics students believe that the mean score on a first statistics test is 65. The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores: Grades 73.5 88 63.9 85.5 62.7 65 83.2 61.9 88 62.7 Test grades are believed to be normally distributed.Use a significance level of 5%. State the alternative hypothesis: HA:HA: μ>65μ>65 μ<65μ<65 μ≠65μ≠65 State the mean of the sample: State the standard error of the sample means: State the test statistic: t=t= State the p-value: Decision: Fail to reject the null hypothesis. Reject the null hypothesis.A sample of 100 clients of an exercise facility was selected. Let X = the number of days per week that a randomly selected client uses the exercise facility. Frequency 2 1 15 30 3 27 4 10 8 6. 8 Find the 80th percentile.Using this data set What is the standard error of the relevant distribution? Ignoring that the number of men in the 2012 Survey was about 17,675, and the number of women was about 18,634, assume the samples were 100 men and 100 women in each Survey My worked solution is: P1: men proportion change = 15.5/100 = 0.155 N1 = 100 P2: women proportion change = 57.9/100 = 0.579 N2 = 100 {[0.155 * (1 - 0.155) /] + [0.579 * (1 - 0.579) / 100]} = 0.0037 Standard Error = 0.0037 is this correct?