Stats something
Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.
Note Card: Two-Tailed: qt(1 - alpha / 2, dof) One-Tailed: qt(1 - alpha, dof)
A standardized test is given to a seventh grade class and a fifth grade class. The superintendent believes that the variance in performance for the two grades is different. The sample variance of a sample of 6 test scores from the seventh grade class is 29.59. The sample variance of a sample of 5 test scores from the fifth grade class is 18.79. Construct the 90% confidence interval for the ratio of the population variances. Round your answers to four decimal places.
# Given data s1_squared <- 29.59 # Sample variance for seventh grade s2_squared <- 18.79 # Sample variance for fifth grade df1 <- 5 # Degrees of freedom for seventh grade df2 <- 4 # Degrees of freedom for fifth grade alpha <- 0.10 # Significance level (for 90% confidence) # Calculate the ratio of the variances variance_ratio <- s1_squared / s2_squared # Critical F-values F_alpha2 <- qf(1 - alpha / 2, df1, df2) # F at alpha/2 F_1_alpha2 <- qf(alpha / 2,df1, df2) # F at 1-alpha/2 # Confidence interval lower_bound <- variance_ratio / F_alpha2 upper_bound <- variance_ratio / F_1_alpha2 # Output the confidence interval lower_bound upper_bound
Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
# Calculate degrees of freedom (Welch-Satterthwaite formula) df <- ((s1^2 / n1 + s2^2 / n2)^2) / (((s1^2 / n1)^2 / (n1 - 1)) + ((s2^2 / n2)^2 / (n2 - 1))) # Critical value for a 90% confidence interval alpha <- 0.1 critical_value <- qt(1 - alpha / 2, df)
Compute the value of the test statistic. Round your answer to three decimal places. (Chapter 12)
# Data for the two populations population1 <- c(62, 60, 71, 67, 63, 64, 80, 65) population2 <- c(64, 54, 79, 62, 72, 72, 73, 75) # Calculate the paired differences differences <- population1 - population2 # Calculate the mean and standard deviation of the differences mean_differences <- mean(differences) sd_differences <- sd(differences) # Number of pairs n <- length(differences) # Compute the test statistic t_statistic <- mean_differences / (sd_differences / sqrt(n)) # Print the test statistic rounded to three decimal places round(t_statistic, 3)
Find the value of the standard deviation of the paired differences. Round your answer to one decimal place. (Chapter 12)
# Data for the two populations population1 <- c(62, 60, 71, 67, 63, 64, 80, 65) population2 <- c(64, 54, 79, 62, 72, 72, 73, 75) # Calculate the paired differences differences <- population1 - population2 # Calculate the standard deviation of the differences sd_differences <- sd(differences) # Print the result sd_differences
Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to the nearest integer.
# Given data s1 <- 28 n1 <- 11 s2 <- 18 n2 <- 9 # Calculate the standard error SE <- sqrt((s1^2 / n1) + (s2^2 / n2))
A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 65 feet, with a population standard deviation of 7.9. The mean braking distance for SUVs equipped with tires made with compound 2 is 68 feet, with a population standard deviation of 5.1. Suppose that a sample of 57 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1 be the true mean braking distance corresponding to compound 1 and μ2 be the true mean braking distance corresponding to compound 2. Use the 0.1 level of significance. Step 2 of 4 : Compute the value of the test statistic. Round your answer to two de
# Given data x1 <- 65 sigma1 <- 7.9 n1 <- 57 x2 <- 68 sigma2 <- 5.1 n2 <- 57 # Calculate the difference in means mean_diff <- x1 - x2 # Calculate the standard error SE <- sqrt((sigma1^2 / n1) + (sigma2^2 / n2)) # Calculate the test statistic z_statistic <- mean_diff / SE
Construct the 90% confidence interval. Round your answers to the nearest integer.
# Given data x1 <- 80 n1 <- 11 s1 <- 28 x2 <- 99 n2 <- 9 s2 <- 18 # Calculate the degrees of freedom df <- ((s1^2 / n1 + s2^2 / n2)^2) / (((s1^2 / n1)^2 / (n1 - 1)) + ((s2^2 / n2)^2 / (n2 - 1))) # Calculate the critical value alpha <- 0.1 t_critical <- qt(1 - alpha / 2, df) # Calculate the standard error SE <- sqrt((s1^2 / n1) + (s2^2 / n2)) # Calculate the confidence interval mean_diff <- x1 - x2 lower_bound <- mean_diff - t_critical * SE upper_bound <- mean_diff + t_critical * SE
Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to two decimal places.
# Significance level alpha <- 0.1 # Critical value for a one-tailed test critical_value <- qnorm(alpha)
