Management Statistics: Analysis Of Data And Run The Data Through SPSS (Coursework Sample)

Management statistics Name Institution Date Question 1 Bottle no. Ounces x2 1 14.23 202.4929 2 14.32 205.0624 3 14.98 224.4004 4 15.00 225.0000 5 15.11 228.3121 6 15.21 231.3441 7 15.42 237.7764 8 15.47 239.3209 9 15.65 244.9225 10 15.74 247.7476 11 15.77 248.6929 12 15.80 249.6400 13 15.82 250.2724 14 15.87 251.8569 15 15.98 255.3604 16 16.00 256.0000 17 16.02 256.6404 18 16.05 257.6025 19 16.21 262.7641 20 16.21 262.7641 21 16.23 263.4129 22 16.25 264.0625 23 16.31 266.0161 24 16.32 266.3424 25 16.34 266.9956 26 16.46 270.9316 27 16.47 271.2609 28 16.51 272.5801 29 16.91 285.9481 30 16.96 287.6416 n 457.62 7553.165 * Mean 457.6230 = 15.85 * Median The middle variables are 15.98 &16.00 therefore, 15.98+162 = 15.99 * Standard deviation 7553.165-(457.622/30) 30-1 = 4.44362 Question 2 95% confidence level 4.4436230 = 0.81129 Question 3 1 H0: µ ≥16 2 Ha: μ < 16 3 α = 0.01 4 Sample statistics Standard error - sn =4.4436230 = 0.811290 Degree of freedom (DF) – n-1 30-1= 29 T calculated - (x - μ) / SE (15.85-16)0.81129 = -0.18489 Given the alternative hypothesis (μ < 16), we want to know whether the observed sample mean is small enough to cause us to reject the null hypothesis. The observed sample mean produced a t statistic test statistic of -0.18489. We use the t Distribution Calculator to find P (t < -0.18489) = 0.253. This means we would expect to find a sample mean of 15.85 or smaller in 25.3% of our samples, if the true ounces were 16. Thus, the P-value in this analysis is 0.19. Interpret results. Since the P-value (0.253) is greater than the significance level (0.01), we cannot reject the null hypothesis. Conclusion: There is evidence from these data and results to suggest that a bottle contains less than 16 ounces. Question 4 The causes for less ounces in the soda can be due to inefficiency with the machines, due to old age of the machines it can lead ...