The test statistic of z = -2.39 is obtained when testing the claim that p 0.596. This is a two-tailed test. Using a 0.10 significance level, complete parts (a) and (b).Click here to view the standard normal distribution table for negative z scores. Click here to view the standard normal distribution table for positive z scores.a. Find the critical value(s).Select the correct choice below and fill in the answer box(es) within your choice.(Round to two decimal places as needed.)OA. There are two critical values; the lower critical value isand the upper critical value isOB. There is one critical value; the critical value is Standard Normal (z) Distribution Table, Negative z ScoresStandard Normal (z) Distribution Table, Positive z ScoresNEGATIVE z ScoresStandard Normal (2) Distribution802.03000517POSITIVE z ScoresStandard Normal (z) Distribution6255NOTE: For values ofve 3.49, use 0.9999 for the area

Step 1Here given,α=0.10It is a two-tailed test.For two tailed test two critical value need to be…

Answered: The test statistic of z = -2.39 is…

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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Complete parts (a) and (b) below. Click here to view Page 1 of the table of areas under the standard normal curve. Click here to view Page 2 of the table of areas under the standard normal curve. a. The test statistic in a right-tailed test is z= 1.71. Determine the P-value and decide whether, at the 5% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The P-value is (Round to three decimal places as needed.) This P-value V sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is V the significance level. b. The test statistic in a right-tailed test is z= - 0.82. Determine the P-value and decide whether, at the 5% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The P-value is (Round to three decimal places as needed.) This P-value sufficient evidence to reject the null hypothesis in…
Assume 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…
The test statistic of z=0.91 is obtained when testing the claim that p>0.29. This is a right-tailed test. Using a 0.10 significance level, complete parts (a) and (b). Click here to view the standard normal distribution table for negative z scores. Click here to view the standard normal distribution table for positive z scores. a. Find the critical value(s). Select the correct choice below and fill in the answer box(es) within your choice. (Round to two decimal places as needed.) OA. There are two critical values; the lower critical value is B. There is one critical value; the critical value is and the upper critical value is
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What are the values for SS and variance for the following sample of n = 3 scores? Sample: 8, 5, 2
kshe The height of women ages 20-29 is normally distributed, with a mean of 64.6 inches. Assume o = 2.9 inches. Are you more likely to randomly select 1 woman with a e score les (fo height less than 66.4 inches or are you more likely to select a sample of 30 women with a mean height less than 66.4 inches? Explain. E . Click the icon to view page 1 of the standard normal table. signme ee score A Click the icon to view page 2 of the standard normal table. radeboc What is the probability of randomly selecting 1 woman with a height less than 66.4 inches? e score mail O (Round to four decimal places as needed.) What is the probability of selecting a sample of 30 women with a mean height less than 66.4 inches? Text (Round to four decimal places as needed.) Accessibl Resource Are you more likely to randomly select 1 woman with a height less than 66.4 inches or are you more likely to select a sample of 30 women with a mean height less than 66.4 inches? Choose the correct answer below. Tools…
A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is less than 0.5. Assume that sample data consists of 78 girls in 169 births, so the sample statistic of 613 results in a z score that is 1 standard deviation below 0. Complete parts (a) through (h) below.Click here to view page 1 of the Normal table. LOADING... Click here to view page 2 of the Normal table. LOADING...Question content area bottomPart 1a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.A.H0: p≠0.5H1: p<0.5B.H0: p=0.5H1: p≠0.5C.H0: p=0.5H1: p<0.5D.H0: p=0.5H1: p>0.5Part 2b. What is the value of α?α=enter your response here (Type an integer or a decimal.)Part 3c. What is the sampling distribution of the sample statistic? Normal distribution Student (t) distribution χ2Part 4d. Is the test two-tailed, left-tailed, or right-tailed? Right-tailed…
A 0.01 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is different from 0.5. Assume that sample data consists of 55 girls in 100 births, so the sample statistic of 1120 results in a z score that is 1 standard deviation above 0. Complete parts (a) through (h) below.Click here to view page 1 of the Normal table. LOADING... Click here to view page 2 of the Normal table. LOADING...Question content area bottomPart 1a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.A.H0: p=0.5H1: p≠0.5B.H0: p=0.5H1: p>0.5C.H0: p=0.5H1: p<0.5D.H0: p≠0.5H1: p=0.5Part 2b. What is the value of α?α=enter your response here (Type an integer or a decimal.)Part 3c. What is the sampling distribution of the sample statistic? χ2 Normal distribution Student (t) distribution
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
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The test statistic of z = - 2.06 is obtained when testing the claim that p = 2/ 3. a. Using a significance level of a = 0.01, find the critical value(s). b. Should we reject Ho or should we fail to reject Ho? Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. The critical value(s) is/are z = (Round to two decimal places as needed. Use a comma to separate answers as needed.)
The test statistic of z = - 1.88 is obtained when testing the claim that p = 1/2. a. Using a significance level of a = 0.10, find the critical value(s). b. Should we reject Ho or should we fail to reject Ho? Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. The critical value(s) is/are z= (Round to two decimal places as needed. Use a comma to separate answers as needed.) b. Choose the correct conclusion below. O A. Reject Ho. There is sufficient evidence to warrant rejection of the claim that p=1/2. B. Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that p = 1/2. O C. Fail to reject Ho. There is not sufficient evidence to warrant rejection of the claim that p =1/2. D. Reject Ho- There is not sufficient evidence to warrant rejection of the claim that p = 1/2.

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