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- Question 2 There is a claim that drink driving has been very common in high ways. The Highway Police randomly caught 300 drivers for breath alcohol test and found that 14% of drivers have breath alcohol over the legal limit. At 4% level of significance, use the p-value approach to test the hypothesis that the percentage of drivers who are drink driving is more than 10%.Question 2 A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130 degrees. A sample of 9 systems when tested yields a sample average activation temperature of 131.08 degrees. If the distribution of activation times is normal with standard deviation 1.5 degrees, test at the 1% level of significance to see if the data shows evidence that is different from the manufacturers claim. (d) Calculate the p-value for this test. (e) State the conclusion of this test. Give a reason for your answer.Question 2 A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130 degrees. A sample of 9 systems when tested yields a sample average activation temperature of 131.08 degrees. If the distribution of activation times is normal with standard deviation 1.5 degrees, test at the 1% level of significance to see if the data shows evidence that is different from the manufacturers claim. (a) State the null and alternative hypotheses for the test. (b) What is the test statistic? (c) Calculate the value of the test statistic for this test. (d) Calculate the p-value for this test. (e) State the conclusion of this test. Give a reason for your answer.
- Question 1 An educator claims that the average salary of substitute teachers in school districts in Allegheny County, PA is less than $60 per day. A random sample of eight school districts is selected, and the daily average salary (in dollars) is $58.88 with a sample standard deviation of $5.08. Set alpha = 0.10 for this test. Is there enough evidence to support the educator’s claim? Question 2 A tobacco company claims that its best-selling cigarettes contain less than 40mg of nicotine. This claim is tested at the 1% significance level by using the results of 15 randomly selected cigarettes. The mean is 42.6 mg and the standard deviation is 3.7mg. Evidence suggests that nicotine is normally distributed. Is there enough evidence from this study to support the claim of the tobacco company? Question 3 A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute in his city is less and wants to…You obtain a t comp of .975 in a two sample independent t test with alpha at .05. Is it significant?Question 14 State the conclusion of the test based on a p-value of 0.2681, if we use a 5% significance level. Reject H0 Do not reject H0
- Question 4 Compute the z score for a score of 90 on a test that has a mean of 92 and a standard deviation of 2.QUESTION 5 PART A- Carry out a formal t-test for the association between extroversion and neuroticism in personalities. H0: The population correlation coefficient is 0. HA: The population correlation coefficient is not 0. What is the critical value of the t-statistic required to reject the null hypothesis at a confidence level of 0.05? (Report the absolute value. The t-distribution is symmetric, thus it is easier to compare the absolute value of the t-statistic with the positive values in statistical table C). QUESTION 5 PART B- The data in chocolate_nobels is not bivariate normal. Plot a scatter plot of the data. Based on the scatter plot, how does the data deviate from bivariate normality? There are outliers The standard deviation of both variables seems to increase as the variables increase. The relationship between the two variables is not linear. The standard deviation of the Nobel prize winners seems to increase as chocolate…QUESTION 18 A manufacturer of car batteries claims that his product will last at least 4 years on average. A sample of 50 is taken and the mean and standard deviation are found. The test statistic is calculated to be -1.656. Using a 5% significance level, the conclusion would be: there is sufficient evidence for the manufacturer’s claim to be considered correct there is insufficient evidence for the manufacturer’s claim to be considered correct there is insufficient evidence for the manufacturer’s claim to be considered incorrect there is sufficient evidence for the manufacturer’s claim to be considered incorrect
- question 3 Sixty gym members were randomly selected and their weights were recorded and inputted into MINITAB. The results are summarized in Exhibit 1 below. One-Sample Z: WeightTest of mu = 195 vs < 195The assumed standard deviation = 22.11 Variable N Mean StDev SE MeanWeight 60 193.13 22.11 * i. Calculate the value of *? ii. State the null and alternative hypotheses for the test. iii. What is the value of the test statistic for this test.? iv. Calculate the p-value for this test. v. State the conclusion of this test. Give a reason for your answer.QUESTION 4 PART A- What is the standard error of the correlation coefficient for the association of extroversion and neuroticism in personalities? QUESTION 4 PART B- Carry out a formal t-test for the association between extroversion and neuroticism in personalities.xls. H0: The population correlation coefficient is 0. HA: The population correlation coefficient is not 0. What is the value of the t-statistic corresponding to this test? (Report the absolute value. The t-distribution is symmetric, thus it is easier to compare the absolute value of the t-statistic with the positive values in statistical table C). DATA IS ATTACHED.QUESTION 16 A researcher wants to compare the mean concentration of two medications considered biologically equivalent, i.e., two medications that are able to produce the same therapeutic effect at the same level of concentration in the blood. The group of individuals on medication one (n = 32) had a mean blood concentration of 21.7 micrograms per milliliter with a standard deviation of 8.7 micrograms per milliliter. The group of individuals on medication two (n= 32) had a mean blood concentration of 19.4 micrograms per milliliter with a standard deviation of 5.2 micrograms per milliliter. Construct and interpret a 95% confidence interval demonstrating the difference in means for the individuals on medication one when compared to the group of individuals on medication two. The researchers are 95% confident that the true mean difference in medication concentration levels between individuals on medication one and individuals on medication two is between 4.867 micrograms…