Scores on the SAT Mathematics test (SAT-M) are believed to be Normally distributed with mean µ. The scores of a random sample of three students who recently took the exam are 550, 620, and 480. A 95% confidence interval for u based on these data is (148.89, 951.11) a
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- Traffic Highway planners investigated the relationshipbetween traffic Density (number of automobiles per mile)and the average Speed of the traffic on a moderately largecity thoroughfare. The data were collected at the samelocation at 10 different times over a span of 3 months.They found a mean traffic Density of 68.6 cars per mile(cpm) with standard deviation of 27.07 cpm. Overall, the cars’ average Speed was 26.38 mph, with standard devia-tion of 9.68 mph. These researchers found the regression line for these data to be Speed = 50.55 - 0.352 Density.a) What is the value of the correlation coefficientbetween Speed and Density?b) What percent of the variation in average Speed isexplained by traffic Density? c) Predict the average Speed of traffic on the thorough-fare when the traffic Density is 50 cpm. d) What is the value of the residual for a traffic Densityof 56 cpm with an observed Speed of 32.5 mph?e) The data set initially included the point Density =125 cpm, Speed = 55 mph. This…According to a research study, college students spent 24.2 hours doing homework per week last year, on average. A random sample of 16 college students was surveyed and the mean amount of time per week each college student spent on homework was 23.7. This data has a sample standard deviation of 2.2. (Assume that the scores are normally distributed.) Researchers conduct a one-mean hypothesis at the 1% significance level, to test if the mean amount of time college students spend on homework per week is less than the mean amount of time last year. Which answer choice shows the correct null and alternative hypotheses for this test? Select the correct answer below: H0:μ=24.2; Ha:μ>24.2, which is a right-tailed test. H0:μ=24.2; Ha:μ<24.2, which is a left-tailed test. H0:μ=23.7; Ha:μ>23.7, which is a right-tailed test. H0:μ=23.7; Ha:μ<23.7, which is a left-tailed test.Á study was conducted that examined the relationship between alcohol consumption and risk of heart attack. In all, 417 men, age 40 or older, were tracked over ten years and the number of that experienced a heart attack was recorded. The participants were grouped by those who totally abstained from alcohol and those that drank moderately. For the results given in the table below, use the chi-squared test to find the p value for NHST at the .05 level. (In SPSS use the Yates continuity correction value.) Heart attack Alcohol Crosstabulation Count Alcohol Abstainer Moderate Total Heart attack No 197 192 389 Yes 9. 19 28 Total 206 211 417 Op= .090 Op= 121 !! Op=.067 Op = .048
- A sample of 76 female workers and another sample of 48 male workers from a state produced mean weekly earnings of $743.50 for the females and $777.63 for the males. Suppose that the population standard deviations of the weekly earnings are $80.05 for the females and $88.68 for the males. The null hypothesis is that the mean weekly earnings are the same for females and males, while the alternative hypothesis is that the mean weekly earnings for females is less than the mean weekly earnings for males. Directions: • Label your answers with the correct statistical symbols. • If you use the Ti, identify which function and values you used to calculate. If you solve by hand, show all steps 2.5 The significance level for the test is 1%. What is/are the critical value(s)? 2.6. What is the value of the test statistic, rounded to three decimal places? 2.7. What is the p-value for this test, rounded to four decimal places? 2. 8. Using the p-value approach, do you reject or fail to reject the null…The coach of a very popular men’s basketball team claims that the average distance the fans travel to the campus to watch a game is 35 miles. The team members feel otherwise. A sample of 16 fans who travel to games was randomly selected and yielded a mean of M= 36 miles and s= 5 miles. Test the coach’s claim at the 5% (.05) level of significance. one-tailed or two-tailed test: State the hypotheses: df= tα or t value for the critical region = sM = t (test statistic)= Decision:An educator has developed a program to improve math scores on the Texas STAAR test. The average STAAR Mathematics test for the third graders in the district was a mean= 1472. A sample of n= 30 third graders took the new math training prgram before taking the STAAR Math test. The average score for the students was M= 1497. with a standard deviation of s= 145 (note, this is standard deviation, not variance). Was there a significant change in test scores ( use stard devistion of 0.05, two tail test). Calculate the effect size using Cohen's d.
- The mean potassium content of a popular sports drink is listed as 132 mg in a 32-oz bottle. Analysis of 24 bottles indicates a sample mean of 131.2 mg. (a) State the hypotheses for a two-tailed test of the claimed potassium content.. a. Ho: μ= 132 mg vs. H₁: b. He: ≤132 mg vs. H₁: c. He: ≥132 mg vs. H₁: Oa Ob Oc (b) Assuming a known standard deviation of 2.1 mg, calculate the z test statistic to test the manufacturer's claim. (Round your answer to 2 decimal places. A negative value should be indicated by a minus sign.) Test statistic 132 mg >132 mg <132 mg (c) At the 2 percent level of significance (a = .02) does the sample contradict the manufacturer's claim? Decision Rule: Reject HoChoose 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"Researchers examined the bill color (in hue degree) of male and female zebra finches. Here are the summary statistics from the study. Sex N Mean Standard deviation Males 59 2.91 1.46 Females 60 7.42 2.48 How different are male and female zebra finches in bill color? The margin of error for a 95% confidence interval for the difference in mean bill color μF – μM (in hue degree) is
- A Two-Sample Hypothesis TestThe mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? (Assume a 5% level of significance.)Useful tools:Normal Distribution Calculatort-Distribution Calculator 1. Which of the following null and alternative hypotheses match this scenario? A---H0:Δμ=0H0:Δμ=0Ha:Δμ≠0Ha:Δμ≠0 B---H0:Δμ≠0H0:Δμ≠0Ha:Δμ=0Ha:Δμ=0 C---H0:Δμ=0H0:Δμ=0Ha:Δμ>0 2. Which type of test should be applied? A---The alternative hypothesis indicates a right-tailed test. B---The alternative hypothesis indicates a left-tailed test. C---The alternative hypothesis indicates a two-tailed test. 3. Which type of distribution should be…The design of controls and instruments affects how easily people can use them. A student project investigated this effect by asking 25 right handed students to turn a knob with their right hands that moved an indicator by screw action. There were two identical instruments, one with a right-hand thread and the other with a left-hand thread. The table given has the data from the student project.1. Calculate and interpret a 90% confidence interval for the mean time advantage of right hand over left hand threads in the setting above.A sports writer wished to see if a football filled with helium travels farther, on average, than a football filled with air. To test this, the writer used 18 adult male volunteers. These volunteers were randomly divided into two groups of nine men each. Group 1 kicked a football that was filled with helium to the recommended pressure. Group 2 kicked a football that was filled with air to the recommended pressure. The mean yardage for Group 1 was ?¯1=300 yards with a standard deviation of ?1=8 yards. The mean yardage for Group 2 was ?¯2=296 yards with a standard deviation of ?2=6 yards. Assume the two groups of kicks are independent. Let ?1 and ?2 represent the mean yardage we would observe for the entire population represented by the volunteers if all members of this population kicked, respectively, a helium‑filled football and an air‑filled football. Assuming two‑sample ? procedures are safe to use and using Option 1 for the degrees of freedom, a 90% confidence interval for ?1−?2 is:…