Test at a 5% level of significance whether the PowerPoint presentation makes a difference in the average understanding of the statistical concept. Assume that samples are drawn from a normal population.
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- The height of women ages 20-29 is normally distributed, with a mean of 63.6 inches. Assume o = 2.5 inches. Are you more lrely to randomly select 1 woman with a height less than 64.1 inches or are you more likely to select a sample of 20 women with a mean height less than 64.1 inches? Explain. E Click the icon to view page 1 of the standard normal table. E Click the icon to view page 2 of the standard normal table. What is the probability of randomly selecting 1 woman with a height less than 64.1 inches? (Round to four decimal places as needed.) Enter your answer in the answer box and then click Check Answer. Check Answer Clear All parts 2 FemainingDoes Hypnotism Relieve Pain? The table shows the pain levels of patients before and after hypnotism. Pain level is measured on a cm scale. Assume that the two samples are randomly selected. At the 0.05 significance level, test the claim that the mean difference has increased after hypnotism.(Be sure to subtract in the same direction).Forest rangers believe that 30% of trees in a large national forest have a leaf fungus. A survery team will inspect a sample of 200 trees. a. Show the steps needed to justify and use a normal model to describe the percentage of fungus-infected trees that may appear in this sample. b. What proportion of infected trees in this sample would lead you to believe the true proportion of ifected trees is higher than the rangers believe?
- The height of women ages 20-29 is normally distributed, with a mean of 63.8 inches. Assume o = 2.8 inches. Are you more likely to randomly select 1 woman with a height less than 65.2 inches or are you more likely to select a sample of 23 women with a mean height less than 65.2 inches? Explain. Click the icon to view page 1 of the standard normal table. Click the icon to view page 2 of the standard normal table. What is the probability of randomly selecting 1 woman with a height less than 65.2 inches? (Round to four decimal places as needed.)The height of women ages 20-29 is normally distributed, with a mean of 64.4 inches. Assume o = 2.4 inches. Are you more likely to randomly select 1 woman with a height less than 65.3 inches or are you more likely to select a sample of 26 women with a mean height less than 65.3 inches? Explain. Click the icon to view page 1 of the standard normal table. Click the icon to view page 2 of the standard normal table. What is the probability of randomly selecting 1 woman with a height less than 65.3 inches? (Round to four decimal places as needed.) 4 What is the probability of selecting a sample of 26 women with a mean height less than 65.3 inches? (Round to four decimal places as needed.) ... Are you more likely to randomly select 1 woman with a height less than 65.3 inches or are you more likely to select a sample of 26 women with a mean height less than 65.3 inches? Choose the correct answer below. O A. It is more likely to select a sample of 26 women with a mean height less than 65.3…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…
- Fill in the missing entries in the ANOVA table.e) What proportion of the total variability of the stopping distance isexplained by the regression model?A marketing research consultant interested in determining the proportion of customers who use online dating services. A random sample of 2001 people were asked and the count revealed that 15% used online dating services. Data 10 Samples Generate 100 Samples Generate 1000 Samples Reset Plot portion- ht Tail samples - do00 mean- o, 150 std. error=0,008O 0.025 0.950 0.025 0.130 o135 o50 0.16 0.140 0.145 0.155 0.160 0.170 0.175 0.15 0.134 0.166 Compute a 95% confidence interval for the true proportion of people who use online dating. 95% Confidence interval What is the numerical value of the margin of error? If the researcher is using a confidence level of 95% what percent of confidence intervals would not capture the true parameter?The height of women ages 20-29 is normally distributed, with a mean of 64.9 inches. Assume o = 2.5 inches. Are you more likely to randomly select 1 woman with a height less than 66.7 inches or are you more likely to select a sample of 17 women with a mean height less than 66.7 inches? Explain. Click the icon to view page 1 of the standard normal table. Click the icon to view page 2 of the standard normal table. What is the probability of randomly selecting 1 woman with a height less than 66.7 inches? (Round to four decimal places as needed.) What is the probability of selecting a sample of 17 women with a mean height less than 66.7 inches? (Round to four decimal places as needed.) Are you more likely to randomly select 1 woman with a height less than 66.7 inches or are you more likely to select a sample of 17 women with a mean height less than 66.7 inches? Choose the correct answer below. O A. It is more likely to select a sample of 17 women with a mean height less than 66.7 inches…
- The height of women ages 20-29 is normally distributed, with a mean of 64.9 inches. Assume a = 2.5 inches. Are you more likely to randomly select 1 woman with a height less than 65.8 inches or are you more likely to select a sample of 29 women with a mean height less than 65.8 inches? Explain. E Click the icon to view page 1 of the standard normal table. E Click the icon to view page 2 of the standard normal table, What is the probability of randomly selecting 1 woman with a height less than 65.8 inches? O (Round to four decimal places as needed.)Which best illustrates the distinction between statistical significance and practical importance? Multiple Choice "This year, 360 of 440 statistics students at Oxnard Technical College rented their textbooks, compared with 110 of 330 students last year. This is a significant increase." "Our new manufacturing technique has increased the life of the 80 GB USB AsimoDrive external hard disk significantly, from 240,000 hours to 250,000 hours." "In 50,000 births, the new vaccine reduced the incidence of infant mortality in Morrovia significantly from 14.2 deaths per 1000 births to 10.3 deaths per 1000 births." "The new Sky Penetrator IV business jet’s cruising range has increased significantly from 3,975 miles to 4,000 miles."Where are the deer? Random samples of square-kilometer plots were taken in different ecological locations of a national park. The deer counts per square kilometer were recorded and are shown in the following table. Mountain Brush Sagebrush Grassland Pinon Juniper 33 18 10 27 57 4 22 18 2 29 26 13 Shall we reject or accept the claim that there is no difference in the mean number of deer per square kilometer in these different ecological locations? Use a 5% level of significance. (b) Find SSTOT, SSBET, and SSW and check that SSTOT = SSBET + SSW. (Use 3 decimal places.) SSTOT = SSBET = SSW = Find d.f.BET, d.f.W, MSBET, and MSW. (Use 2 decimal places for MSBET, and MSW.) dfBET = dfW = MSBET = MSW = Find the value of the sample F statistic. (Use 3 decimal places.)What are the degrees of freedom? (numerator) (denominator) Make a summary table for your ANOVA test. Source ofVariation Sum ofSquares Degrees ofFreedom MS…