The statistic software output for this problem is:
(a) Degrees of freedom = 5
(b) Critical value = -3.365
(c) Test statistics = -1.3608
option B) is correct
(1 point) A sample of 6 measurments, randomly selected from a normally distributed population, resulted in...
(1 point) A sample of 6 measurments, randomly selected from a normally distributed population, resulted in a sample mean, t = 7.7 and sample standard deviation s = 1.2. Using a = 0.05, test the null hypothesis that the mean of the population is 7.2 against the alternative hypothesis that the mean of the population, j < 7.2 by giving the following: (a) the degree of freedom (b) the critical t value (c) the test statistic The final conclustion is...
A sample of 5 measurements, randomly selected from a normally distributed population, resulted in a sample mean, x¯¯¯=6.9 and sample standard deviation s=1.28. Using α=0.05, test the null hypothesis that the mean of the population is 7.7 against the alternative hypothesis that the mean of the population, μ<7.7 by giving the following: find: (a) the degree of freedom: (b) the critical t value: (c) the test statistic:
A sample of five measurements, randomly selected from a normally distributed population, resulted in the following summary statistics: x overbar equals4.4, sequals 1.3. Complete parts a through c. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, mu less than6 . Use alpha equals0.05 . If alpha equals0.05 , find the rejection region for the test. Choose the correct answer below. A. tgreater than 2.776 B. tgreater than 2.132 C. tless than minus2.776or...
The sample of six measurements shown below was randomly selected from a normally distributed population. Complete parts a throughc 1,3, 1, 5, 1,2 a. Test the null hypothesis that the mean of the population is 3 against the alternative hypothesis, 3. Use a 0.10. If a = 0.10, find the rejection region for the test. Choose the correct answer below O A. t 2.015 or t> 2.015 O C. t-2015 O E. t1476 O B. t-1.476 O D. t 1.476...
A random sample of size n= 15 obtained from a population that is normally distributed results in a sample mean of 45.8 and sample standard deviation 12.2. An independent sample of size n = 20 obtained from a population that is normally distributed results in a sample mean of 51.9 and sample standard deviation 14.6. Does this constitute sufficient evidence to conclude that the population means differ at the a = 0.05 level of significance? Click here to view the...
19 25 The sample of six measurements shown below was randomly selected from a normally distributed population. Complete parts a through c. 1,2,3,3,4,1 a. Test the null hypothesis that the mean of the population is 3 against the alternative hypothesis. p < 3. Use a = 0.05 Ifq=0.05, find the rejection region for the test. Choose the correct answer below. % 1994 1994 OA. <-2015 or t> 2015 Oct-2571 O E. > 2571 OB < -2015 OD < -2571 ort...
(1 point) A certain manufactured product is supposed to contain 23% potassium by weight. A sample of 10 specimens of this product had an average of 23.24 and and a standard deviation of 8 = 0.24. If the mean percentage is found to differ from 23, the manufacturing process will be recalibrated. Using a = 0.01, test an appropriate null and alternate hypotheses. (a) the degree of freedom (b) the critical t value (c) the test statistic The final conclusion...
If we wish to carry out inference procedures on the mean of a normally distributed population where sigma is known, we a) Click for List - should use the distribution. b) Click for List Both the t distribution and the standard normal distribution have a median of O. c) Click for List For the t distribution, the mean and variance are always equal. d) Click for List The t distribution has more area in the tails than the standard normal...
come from populations (1 point) Test t mean. Assume that the samples are independent simple random samples. Use a significance level of a 0.01 Sample 1: n1 15, 1-28.4, 81-6.07 Sample 2: n2 10, 2 22, 82 8.92 (a) The degree of freedom is (b) The standardized test statistic is (c) The final conclusion is O A. We can reject the null hypothesis that (14-Ha) 0 and accept that (M1-μ2) 0 B. There is not sufficient evidence to reject the...
Consider the following summary statistics, calculated from two independent random samples taken from normally distributed populations. Sample 1 F1 = 23.65 = 2.50 p1 = 18 Sample 2 F2 = 25.62 = 3.28 p2 = 20 Test the null hypothesis Ho: P1 = r2 against the alternative hypothesis HA : H1 CH2 a) Calculate the test statistic for the Welch Approximate procedure. Round your response to at least 3 decimal places. Number b) The Welch-Satterthwaite approximation to the degrees of...