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We considered the change in the number of days exceeding 90°F from 1948 and 2018 at 197 randomly sampled locations from the NOAA database in Exercise 7.19. The mean and standard deviation of the reported differences are 2.9 days and 17.2 days. Calculate a 90% confidence interval for the average difference between number of days exceeding 90°F between 1948 and 2018. We’ve already checked the conditions for you.
We checked conditions in an earlier exercise, so we'll jump right to the calculations. First we compute the standard error and find z*: SE = 17.2 / sqrt(197) = 1.23; Z* = 1.65. Next, we can compute the confidence interval: 2.9 ± 1.65 x 1:23 = (0.87; 4.93)
We considered the change in the number of days exceeding 90°F from 1948 and 2018 at 197 randomly sampled locations from the NOAA database in Exercise 7.19. The mean and standard deviation of the reported differences are 2.9 days and 17.2 days. Interpret the 90% interval in context.
We are 90% confident that there was an increase of 0.87 to 4.93 in the average number of days that hit 90°F in 2018 relative to 1948 for NOAA stations.
We considered the change in the number of days exceeding 90°F from 1948 and 2018 at 197 randomly sampled locations from the NOAA database in Exercise 7.19. The mean and standard deviation of the reported differences are 2.9 days and 17.2 days. Does the 90% confidence interval provide convincing evidence that there were more days exceeding 90°F in 2018 than in 1948 at NOAA stations? Explain.
Yes, since the interval lies entirely above 0.
We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey in Exercise 7.20. The mean and standard deviation of the differences are x(read−write) = −0.545 and 8.887 points. Calculate a 95% confidence interval for the average difference between the reading and writing scores of all students.
A 95% confidence interval can be calculated using xdiff as follows: -0.545 ± 1:98 x 8.887 / sqrt(200) = (-1.79, 0.70)
We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey in Exercise 7.20. The mean and standard deviation of the differences are x(read−write) = −0.545 and 8.887 points. Interpret the 95% confidence interval in context.
We are 95% confident that on the reading test students score, on average, 1.79 points lower to 0.70 points higher than they do on the writing test.
We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey in Exercise 7.20. The mean and standard deviation of the differences are x(read−write) = −0.545 and 8.887 points. Does the 95% confidence interval provide convincing evidence that there is a real difference in the average scores? Explain.
No, since 0 is included in the interval.
Casein is a common weight gain supplement for humans. Does it have an effect on chickens? Using data provided in Exercise 7.27, test the hypothesis that the average weight of chickens that were fed casein is different than the average weight of chickens that were fed soybean. If your hypothesis test yields a statistically significant result, discuss whether or not the higher average weight of chickens can be attributed to the casein diet. Assume that conditions for inference are satisfied.
The hypotheses are H0: µc = µs and HA: µc ≠ µs. We are told to assume that conditions for inference are satisfied. T = ((323.58 – 246.43) – 0) / sqrt((64.4322 / 12) + (54.1322 / 14)) = 3.48; df = 11; p-value = P(|T| > 3.27) < 0.01. Since p-value < 0:05, reject H0. The data provide strong evidence that the average weight of chickens that were fed casein is different than the average weight of chickens that were fed soybean. Since this is a randomized experiment the type of diet is the only difference between the various groups of chicken, and hence the observed differences are can be attributed to differences in diet.
Subjects from Central Prison in Raleigh, NC, volunteered for an experiment involving an “isolation” experience. The goal of the experiment was to find a treatment that reduces subjects’ psychopathic deviant T scores. This score measures a person’s need for control or their rebellion against control, and it is part of a commonly used mental health test called the Minnesota Multiphasic Personality Inventory (MMPI) test. The experiment had three treatment groups: (1) Four hours of sensory restriction plus a 15 minute “therapeutic” tape advising that professional help is available. (2) Four hours of sensory restriction plus a 15 minute “emotionally neutral” tape on training hunting dogs. (3) Four hours of sensory restriction but no taped message.
Let µdiff = µpre - µpost. Then, for each treatment the hypotheses are: H0: µdiff = 0: Treatment has no effect. HA: µdiff > 0: Treatment has an effect on P.D.T. scores, either positive or negative. The conditions that need to be satisfied are: 1. Independence: The subjects are randomly assigned to treatments, so independence within and between groups is satisfied. 2. Normality: All three sample sizes are smaller than 30, so we look for clear outliers. There is a borderline outlier in the first treatment group. Since it is borderline, we will proceed, but we should report this caveat with any results. The calculation of the test statistics and p-values are shown below. Treatment 1: T = (6.21 – 0) / (12.3 / sqrt(14) = 1.89; df = 13; p-value = 2 x P(T > 1.89) = 0.081. Treatment 2: T = (2.86 – 0) / (7.94 / sqrt(14) = 1.35; df = 13; p-value = 2 x P(T > 1.35) = 0.200. Treatment 3: T = (-3.21 – 0) / (8.57 / sqrt(14) = -1.40; df = 13; p-value = 2 x P(T > -1.40) = 0.185. We do not reject the null hypothesis for any of these groups. As earlier noted, there is some uncertainty about if the method applied is reasonable for the first group.
Determine if the following statement is true or false, and explain your reasoning for statements you identify as false: When comparing means of two samples where n1 = 20 and n2 = 40, we can use the normal model for the difference in means since n2 ≥ 30.
False, in order to be able to use a Z test both sample sizes need to be above 30.
Determine if the following statement is true or false, and explain your reasoning for statements you identify as false: As the degrees of freedom increases, the t-distribution approaches normality.
True.
Determine if the following statement is true or false, and explain your reasoning for statements you identify as false: We use a pooled standard error for calculating the standard error of the difference between means when sample sizes of groups are equal to each other.
False, we use the pooled standard deviation when the variability in groups is constant.
A large farm wants to try out a new type of fertilizer to evaluate whether it will improve the farm’s corn production. The land is broken into plots that produce an average of 1,215 pounds of corn with a standard deviation of 94 pounds per plot. The owner is interested in detecting any average difference of at least 40 pounds per plot. How many plots of land would be needed for the experiment if the desired power level is 90%? Use α = 0.05. Assume each plot of land gets treated with either the current fertilizer or the new fertilizer.
Difference we care about: 40. Single tail of 90%: 1.28 x SE. At the 5% significance level the rejection region bounds span ±1.96 x SE. 40 = 1.28 x SE + 1.96 x SE = 3.24 x SE; SE = 12.35 = sqrt(942 / n + 942 / n). n = 115.86. We will need 116 plots of land for each fertilizer.
A medical research group is recruiting people to complete short surveys about their medical history. For example, one survey asks for information on a person’s family history in regards to cancer. Another survey asks about what topics were discussed during the person’s last visit to a hospital. So far, as people sign up, they complete an average of just 4 surveys, and the standard deviation of the number of surveys is about 2.2. The research group wants to try a new interface that they think will encourage new enrollees to complete more surveys, where they will randomize each enrollee to either get the new interface or the current interface. How many new enrollees do they need for each interface to detect an effect size of 0.5 surveys per enrollee, if the desired power level is 80%? Use α = 0.05.
Difference we care about: 0.5. Single tail of 90%: 1.28 x SE. At the 5% significance level the rejection region bounds span ±1.96 x SE. 0.5 = 0.84 x SE + 1.96 x SE = 32.8 x SE; SE = 0.1786 = sqrt(2.22 / n + 2.22 / n). n = 303.47. We will need 304 plots of land for each fertilizer.