The diameter of a ball bearing is to have an average of 2.0 Cm. A random sample of 10 bearings is selected every hour to check the diameters. The sample selected at 9 a.m. gave the following measurements: 2.02, 2.04, 2.00, 1.98, 1.98, 2.04, 2.04, 2.00, 2.02, 1.96. Calculate the t - value and draw your conclusions, if any.
Ave e Sum / 10.
To calculate the t-value and draw conclusions, we first need to determine a few things. The given problem is asking about the average diameter of the ball bearings, so we are dealing with a one-sample t-test.
Let's break down the steps to solve this problem:
Step 1: Calculate the sample mean
To calculate the sample mean, sum up all the measurements and divide by the sample size:
(2.02 + 2.04 + 2.00 + 1.98 + 1.98 + 2.04 + 2.04 + 2.00 + 2.02 + 1.96) / 10 = 19.98 / 10 = 1.998 cm
Step 2: Calculate the sample standard deviation
To calculate the sample standard deviation, we need to find the variance first. Subtract the sample mean from each measurement, square the differences, sum them up, and divide by (n-1):
[(2.02-1.998)^2 + (2.04-1.998)^2 + (2.00-1.998)^2 + (1.98-1.998)^2 + (1.98-1.998)^2 + (2.04-1.998)^2 + (2.04-1.998)^2 + (2.00-1.998)^2 + (2.02-1.998)^2 + (1.96-1.998)^2] / (10-1)
= (0.000004 + 0.000004 + 0.000004 + 0.000004 + 0.000004 + 0.000004 + 0.000004 + 0.000004 + 0.000004 + 0.000004) / 9
= 0.000036 / 9
≈ 0.000004 cm^2
The square root of the variance gives us the standard deviation:
√0.000004 ≈ 0.002 cm
Step 3: Calculate the t-value
To calculate the t-value, we need the formula:
t = (sample mean - population mean) / (sample standard deviation / √n)
In this case, the population mean is given as 2.0 cm, the sample mean is 1.998 cm, the sample standard deviation is 0.002 cm, and the sample size is 10:
t = (1.998 - 2.0) / (0.002 / √10)
≈ -0.002 / (0.002 / √10)
≈ -1 / (√10)
≈ -0.316
Step 4: Draw conclusions
To interpret the t-value, we need to compare it with the critical t-value at a particular significance level and degrees of freedom. Since the degrees of freedom in this case are 10-1 = 9 and we don't have a specific significance level mentioned, we'll assume a 95% confidence level.
Referring to the t-distribution table, at a 95% confidence level with 9 degrees of freedom, the critical t-value is approximately 2.262.
Since our calculated t-value (-0.316) is less than the critical t-value (2.262), we fail to reject the null hypothesis. This means that there is no statistically significant difference between the sample mean and the population mean. In other words, the sample data suggests that the average diameter of the ball bearings is not significantly different from the desired average of 2.0 cm.
Note: It's important to keep in mind that this analysis assumes the sample is truly random and representative of the entire population.