Find the z-score boundaries that separate the middle 70% from the 30% in the tails in a normal distribution. Lower z-score: Upper z-score:
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Find the z-score boundaries that separate the middle 70% from the 30% in the tails in a
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Upper z-score:
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- Does 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 0.01 significance level, test the claim that the mean difference has increased after hypnotism.(Be sure to subtract in the same direction). Before (cm) After (cm) Difference (cm) 7 3 6.2 4.2 11 6 7.3 6.3 10.1 10.1 7.5 6.5 10.9 6.9 8.6 3.6 What are the correct hypotheses? (Select the correct symbols and values.)H0: Select an answer s₁² x̄₂ μ₁ μ p x̄₁ μ(d) σ₁² ? = ≠ ≥ > < ≤ Select an answer μ₁ σ₁² μ₂ μ 0 s₁² x̄₁ p x̄₂ H1: Select an answer s₂² μ μ(d) x̄₂ σ₂² x̄₁ μ₁ μ₂ p ? ≠ = < ≥ > ≤ Select an answer s₁² p 0 μ₂ x̄₂ x̄₁ σ₁² μ₁ μ Original Claim = Select an answer H₀ H₁ df = Based on the hypotheses, find the following: Test Statistic = (Round to three decimal places.)Critical value(s) = (Round to three decimal places.)p-value = (Round…The height of women ages 20-29 is normally distributed, with a mean of 64.6 inches. Assumeo=2.7 inches. Are you more likely to randomly select 1 woman with a height less than 67.1 inches or are you more likely to select a sample of 8 women with a mean height less than 67.1 inches? Explain. E 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 67.1 inches? (Round to four decimal places as needed.)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 30 20 5 25 59 10 25 16 2 24 19 4 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. (a) What is the level of significance?State the null and alternate hypotheses. Ho: ?1 = ?2 = ?3; H1: Exactly two means are equal.Ho: ?1 = ?2 = ?3; H1: At least two means are equal. Ho: ?1 = ?2 = ?3; H1: All three means are different.Ho: ?1 = ?2 = ?3; H1: Not all the means are equal. (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.)…
- Calculate the z-score for the largest value and interpret it in terms of standard deviations. Do the same for the smallest value N: 50 Mean:28.766 SE Mean: 0.476 St.Dev: 3.369 Min: 21.300 Q1: 26.375 Median: 29.400 Q3: 31.150 Max: 35.100A survey was conducted two years ago asking college students their top motivations for using a credit card. To determine whether this distribution has changed, you randomly select 425 college students and ask each one what the top motivation is for using a credit card. Can you conclude that there has been a change in the claimed or expected distribution? Use a = 0.05. Complete parts (a) through (d). New Survey Frequency, f 112 Response Old Survey % Rewards 28% 23% 22% 7% 97 Low rates Cash back Discounts Other 109 48 20% 59 a. State Hn and H, and identify the claim. What is the null hypothesis, H,? O A. The distribution of motivations is 28% rewards 23% low rate, 22% cash back, 7% discounts, and 20% other. O B. The distribution of motivations differs from the expected distribution. O C. The distribution of motivations is 112 rewards, 97 low rate, 109 cash back, 48discounts, and 59 other. What is the alternate hypothesis, Ha? O A. The distribution of motivations is the same as the…The test scores of 16 students are listed below: 44, 46, 51, 57, 60, 63, 65, 70, 75, 76, 85, 87, 90, 94, 95, 97. (c) Find the third quartile of the above data.
- 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 30 20 5 25 59 10 25 16 2 24 19 4 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. (a) What is the level of significance?State the null and alternate hypotheses. Ho: ?1 = ?2 = ?3; H1: Exactly two means are equal.Ho: ?1 = ?2 = ?3; H1: At least two means are equal. Ho: ?1 = ?2 = ?3; H1: All three means are different.Ho: ?1 = ?2 = ?3; H1: Not all the means are equal. (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.)…Age discrimination: the following table presents the number of employees, by age group, who were promoted, or not promoted, in a sample drawn from a certain industry during the past year.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 Femaining
- In a statistics class, 10 scores were randomly selected with the following results (mean = 71.5): 74, 73, 77, 77, 71, 68, 65, 77, 67, 66 What is the standard deviation?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…
