Consider the following results for independent random samples taken from two populations. Sample 1 Sample 2 n 1 = 10 n 2 = 30 x 1 = 22.5 x 2 = 20.3 s 1 = 2.1 s 2 = 4.9

Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.05 significance level to test for a difference between the measurements from the two arms. What can be concluded? Right arm Left arm 150 142 120 131 167 160 179 156 In this example, Hd is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis test? O A. Ho: Hd = 0 H₁: Hd 0 Since the P-value is than the significance level, the null hypothesis. There sufficient evidence to support the claim of a difference in measurements between the two arms.

Longevities of Archbishops and Monarchs 16 Archbishops 16 15 15 16 17 1 18 18 10 13 13 11 13 11 11 17 18 10 18 12 19 17 Monarchs 14 18 14 13 19 16 21 14 16 17 15 16

Consider the following ANOVA table below. Some parts are blank intentionally. The number of observations is 249. Source Model (Regression) Residual (Error) Total The MSE is: SS 1,332 DF MS 921 Round your answer to two decimal places.

The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. Table 10.4 shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance. Engine      Sample Mean Number of RPM         Population Standard Deviation  1                      1,500                                                       50  2                      1,600                                                       60

An experiment was conducted to determine whether giving candy to dining parties resulted in greater tips. The mean tip percentages and standard deviations are given in the accompanying table along with the sample sizes. Assume that the two samples are independent simple random samples selected from normally distributed​ populations, and do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b).   ...   Question content area top right Part 1     μ n x s   No candy μ1 36 18.61 1.39 Two candies μ2 36 21.26 2.34   * find the t stat * find the p value * State the conclusion * Construct a confidence interval suitable for testing the claim that the two samples are from populations with the same mean.

The average house has 13 paintings on its walls.  Is the mean different for houses owned by teachers? The data show the results of a survey of 12 teachers who were asked how many paintings they have in their houses. Assume that the distribution of the population is normal. 12, 14, 14, 14, 15, 14, 14, 14, 12, 13, 15, 14 What can be concluded at the  αα = 0.01 level of significance? For this study, we should use     The null and alternative hypotheses would be:       H0:H0:                   H1:H1:               The test statistic     =  (please show your answer to 3 decimal places.) The p-value =  (Please show your answer to 4 decimal places.) The p-value is     αα Based on this, we should      the null hypothesis. Thus, the final conclusion is that ... The data suggest the population mean is not significantly different from 13 at αα = 0.01, so there is sufficient evidence to conclude that the population mean number of paintings that are in teachers' houses is equal to 13. The data…

A one-sample z-test compares a sample with 50 participants with mean 83 to a population with mean 80 and standard deviation 10. What is the effect size d?     0.1     0.3     0.5     0.7

To determine whether there is a significant difference in the weights of boys and girls beginning kindergarten, random samples of 50 boys and 50 girls aged 5 years produced the following information. Mean Standard Deviation Sample Size Boys 43.5 lb 6.8 50 Girls 38.7 lb 3.5 50 (a) What is the point estimate of the difference in the average weights for 5-year old boys and girls? (Use ?Boys − ?Girls.) lb What is the 95% margin of error for the point estimate you found? (Round your answer to three decimal places.) lb (b) What is the 98% confidence interval estimate of ?Boys − ?Girls? (Round your answers to two decimal places.) lb to lb

Calculate the standard score (z-score) of the given sample mean. Round your answer to two decimal places. mu=57" and "sigma=14;n=38; bar(x)=62