To evaluate the effect of a treatment, one sample of n=8 is obtained from a population mean: u=40 and the treatment is administered to the individuals in the sample After the treatment, the sample mean is found to be M=35. 2. If the sample variance is S2=73, are the dat sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05?
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To evaluate the effect of a treatment, one sample of n=8 is obtained from a population
2. If the sample variance is S2=73, are the dat sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05?
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- The amount of chlorine in 100 cm3 water taken from a water tank was measured, the average was 2.15 mg was found. The expert claims that the amount of chlorine in the water is greater than that measured. For this reason, 100 households were randomly selected, with an average chlorine content of 2.55mg and a variance of 0.49 mg. According to this data, find the account value z. 13 - O A) 8,15 O B) 5,71 O C) -5,71 O D) -8, 15 O E) 0.54The blood pressure and weight of Jellystone Park bears are correlated with and .6 is the correlation coefficient of blood pressure with weight. The average blood pressure is 150 with variance 100, for Jellystone Park bears. For a randomly selected bear if we predict its blood pressure is 150 before weighing it, what is our expected squared error?Two pain-relieving drugs were compared for effectiveness on the basis of length of time elapsing between administration of the drug and cessation of pain. Thirteen patients received drug 1, and 13 received drug 2. The sample variances were = 64 and = 16. Test the null hypothesis that the two populations variances are equal. Let α = .05.
- A normal population has mean u (unknown) and variance 9. A sample of size 9 observations has been taken and its variance is found to be 5.4. Test the null hypothesis H,: o = 9 against H: o > 9 at 5% level of significance.Two different formulations of an oxygenated motor fuel are being tested to study their road octane numbers. The variance of road octane number for formulation 1 is o? = 2.2 and for formulation 2 it is o = 1.8. Two random samples of size n = 42 and %3D n2 = 45 are tested, and the mean road octane numbers observed %3D are X = 89.6 and X, = 92.5. Assume the samples are random and independent from a normal population. Construct an 80% two-sided confidence interval on the difference in mean road octane number. < 1 - H2<The blood pressure and weight of Jellystone Park bears are correlated and .5 is the correlation coefficient of blood pressure with weight. The average blood pressure is 150 with variance 36, for Jellystone Park bears. For a randomly selected bear with average weight if we predict its blood pressure is 150 after weighing it, using SLR, what is our expected squared error?
- A survey is conducted to determine if there is a The conditions for inference are met. The chi-square difference in the proportion of students, parents, and test statistic is 0.19, df = 2, and the P-value is greater than any reasonable significance level. What teachers who volunteer at least once a month. To investigate, a random sample of 45 students, 25 parents, and 12 teachers was selected from a large high school. The data are displayed in the table. conclusion should be made? There is convincing evidence that the proportions of these students, parents, and teachers who Student Parent Teacher volunteer at least once a month differs. Yes 24 12 O There is not convincing evidence that the Volunteer? No 21 13 proportion of these students, parents, and teachers who volunteer at least once a month differs. The researcher would like to test these hypotheses: O There is convincing evidence that the distribution of Họ: There is no difference in the distribution of responses for the…A nutrition store in the mall is selling “Memory Booster” which is a concoction of herbs and minerals that is intended to improve memory performance. To test the effectiveness of the herbal mix, a researcher obtains a sample of n = 16 people and has each person take the suggested dosage every day for 4 weeks. At the end of the four week period, each individual takes a standardized memory test. The scores from the sample produced a mean ??� = 24 with a sample variance s2 = 64. In the general population, the standardized test is known to have a mean μ =20. Do the sample data support the conclusion that the Memory Booster has a significant effect? Test with α = .05. is t value of 2 and a rejection zone t value of 2.131 correct?The z-statistic for the difference between the tvlo sample proportions is 5.266 with a P- value of 0.00000014. What conclusion could you make about the difference between the two population proportions if the level of significance is set to 0.05? O The difference is not statistically significant because the sample proportion difference of 0.05 is equal to the significance level of 0.05 The difference is statistically significant because the P-value of 0.00000014 is less than the significance level of 0.05. O The difference is statistically significant because the z-statistic of 5.266 is greater than the significance level of 0.05. O The difference is not statistically significant because the significance level of 0.05 is greater than the P-value of 0.00000014.
- To evaluate the effect of a treatment, a sample of n= 8 is obtained from a population with a mean of µ = 40, and the %| treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M= 35. a. If the sample variance is s = 32, are the %| data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a = .05? b. If the sample variance is s² = 72, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a = .05? c. Comparing your answer for parts a and b, how does the variability of the scores in the sample influence the outcome of a hypothesis test?A researcher is interested in studying the average number of traffic violations received by male vs. female drivers. The researcher has a sample of 30 male drivers and 32 female drivers and conducts a two-sample t-test (two-tailed, alpha = .05). The researcher finds the following: 1. Male drivers have an average of 4 traffic violations every year. 2. Female drivers have an average of 2 traffic violations every year. 3. The standard error of the mean difference between male and female drivers (i.e., the se) is .50. What are the degrees of freedom for this test? Enter your answer as a whole number with no decimal places (i.e., 10, not 10.01, not 10.0, not 10.1).To evaluate the effect of a treatment, a sample is obtained from a population with a mean of =40, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 44.5 with a variance of s² = 36. If the sample consists of n = 16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a = .05? O For these data, t= 1.50. This value is less than the critical value of 2.131, so we reject the null hypothesis and conclude that there is a significant treatment effect. O For these data, t = 3.00. This value is greater than the critical value of 2.131, so we reject the null hypothesis and conclude that there is a significant treatment effect. O For these data, t = 4.50. This value is greater than the critical value of 2.131, so we reject the null hypothesis and conclude that there is a significant treatment effect. O For these data, t = 1.50. This value is less than the…