Which of the following formulas gives a 99% confidence interval for the mean GPA of the professor's students?
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A: sample size(n)=49 Mean(x)=7.2 standard deviation(s)=2.1 confidence level=99%
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A: xbar = 48 mu = 32 s = 6 n =20
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A: Given For males, μ=4.296σ=0.907 For females μ=3.764σ=0.866
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A: a)To testH0:μ1=μ2H1:μ1<μ2 It is given that n1=39x1=19.8s1=2.1n2=46x2=18.4s2=2.8
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- One year Terry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 2.98. Also, Karla had the lowest ERA of any female pitcher at the school with an ERA of 3.37. For the males, the mean ERA was 4.456 and the standard deviation was 0.557. For the females, the mean ERA was 4.166 and the standard deviation was 0.635. Find their respective z-scores. Which player had the better year relative to their peers, Terry or Karla? (Note: In general, the lower the ERA, the better the pitcher.) Terry had an ERA with a z-score of Karla had an ERA with a z-score of (Round to two decimal places as needed.) Which player had a better year in comparison with their peers? OA. Karla had a better year because of a lower z-score. B. Terry had a better year because of a lower z-score. OC. Terry had a better year because of a higher z-score. OD. Karla had a better year because of a higher z-score.A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 44 participants in the treatment group lowered their cholesterol levels by a mean of 20.7 points with a standard deviation of 5.1 points. The 34 participants in the control group lowered their cholesterol levels by a mean of 18.6 points with a standard deviation of 1.3 points. Assume that the population variances are not equal and test the company's claim at the 0.02 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the mean square between samples?
- A random sample of 30 words from Jane Austen's Pride and Prejudice had a mean length of 4.08 letters with a standard deviation of 2.40. A random sample of 30 words from Henry James's What Maisie Knew had a mean length of 3.85 letters with a standard deviation of 2.26. Which of the following is a correct expression for the 99% confidence interval for the difference in mean word length for these two novels? 2.40, 2.26 2.40, 2.26 (a) (4.08– 3.85)±2.756| (b) (4.08 – 3.85) ±2.045 30 30 30 30 2.402 2.26? |2.40? 2.26? (c) (4.08 – 3.85)± 2.045| 30 (d) (4.08 – 3.85)±2.045 30 30 30 |2.40² + 30 2.262 (e) (4.08 – 3.85)±2.756| 30Words were displayed on a computer screen with one of two background colors: red or blue. The subjects were then asked to recall the words that were displayed. There were 35 subjects who were shown words on red backgrounds; they scored a mean score of 15.89 words on the memory test, with a standard deviation of 5.90 words. The 36 subjects shown words on a blue background had a mean score of 12.31 words on the memory test, with a standard deviation of 5.48 words. Constructa 95% confidence interval to estimate the difference between the two population means. a) List the degrees of freedom for this experiment. Find the critical number that is relevant to the results using a 95% confidence level. b) Find the margin of error E that corresponds to a 95% confidence level. RoundEto 5 decimal places. c) Find the 95% confidence interval estimate of the difference of the population means, (μ1−μ2). Follow rounding rules and use appropriate notation. d) Based on the confidence interval estimate, is…One year Perry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 2.81. Also, Jane had the lowest ERA of any female pitcher at the school with an ERA of 3.34. For the males, the mean ERA was 4.953 and the standard deviation was 0.724. For the females, the mean ERA was 5.079 and the standard deviation was 0.804. Find their respective z-scores. Which player had the better year relative to their peers, Perry or Jane? (Note: In general, the lower the ERA, the better the pitcher.)
- A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 44 participants in the treatment group lowered their cholesterol levels by a mean of 20.7 points with a standard deviation of 5.1 points. The 34 participants in the control group lowered their cholesterol levels by a mean of 18.6 points with a standard deviation of 1.3 points. Assume that the population variances are not equal and test the company's claim at the 0.02 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. TH = It: °HIn one study (survey), participants gave their answers on a scale of 1-5 (1=strongly disagree with 5=strongly agree). The average of the responses of 120 men participating in the study was found to be 4.06 and the standard deviation was 0.6, and the average of the responses of 162 women was found to be 3.46 and the standard deviation was 0.4. which of the following is the test statistic calculated when conducting a test of whether the actual average of men's responses at the meaning level of 0.01 is higher than that of women?Answer options group A)4,82 B)2,33 C)7,10 D)9.50Construct a 90% confidence interval for the mean when x̅=50 and there are 31 observations. Suppose σ=7. What is the confidence interval?
- You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the mean square within samples?You would like to estimate the variance and standard deviation for credit scores based on FICO. These scores are thought to be normally distributed. You randomly select 36 credit scores to get a standard deviation of 56.5. Using a 95% confidence interval, estimate the variance and standard deviation of credit scores.A researcher thinks that people under forty have vocabularies different from those over sixty years of age. The researcher administers a vocabulary test to a group of 50 younger subjects and a group of 50 older subjects. Higher scores reflect better performance. The mean score for younger subjects was 9.0, and the standard deviation of younger subject's scores was 5.0. The mean score for older subjects was 20.0, and the standard deviation of older subject's scores was 8.0. Does this experiment provide evidence for the researcher's theory? As part of your answer: a. Please provide, in words, the null and alternative hypotheses (NO symbols). Note that you can choose whether to perform a one- or two-tailed test Answer: Null hypothesis: Alernative hypothesis: b. Provide a brief (one or two sentence(s) rationale for your decision to use a one or two tailed test Answer c. Calculate the Standard Error (SE) of the mean difference (demonstrate calculation steps to get an entire point) and…
