machine produces components, which have a standard deviation of 1.6 cm in length. A random sample of 64 parts is selected from the output and this sample has a mean length of 90 cm. The customer will reject the part if it is either less than 88cm or more than 92cm
A. Find the point estimate for the mean length of all components produced. B. Find out the 95% confidence interval for mean length of all components produced.
A. The point estimate for the mean length of all components produced is the sample mean, which is given as 90 cm.
B. The 95% confidence interval can be calculated using the formula:
Confidence Interval = Sample Mean ± Margin of Error
The margin of error can be calculated using the formula:
Margin of Error = Critical Value * Standard Error
First, we need to calculate the critical value. Since the sample size is 64, and we want a 95% confidence interval, we can look up the critical value in the Z-table for a confidence level of 95%. The critical value for a 95% confidence interval is approximately 1.96.
Next, we need to calculate the standard error. The standard error can be calculated using the formula:
Standard Error = Standard Deviation / √Sample Size
Standard Error = 1.6 / √64
Standard Error = 1.6 / 8
Standard Error = 0.2
Now, we can calculate the margin of error:
Margin of Error = 1.96 * 0.2
Margin of Error ≈ 0.392
Finally, we can calculate the confidence interval:
Confidence Interval = 90 ± 0.392
Confidence Interval ≈ (89.608, 90.392)
Therefore, the 95% confidence interval for the mean length of all components produced is approximately (89.608 cm, 90.392 cm).
A. The point estimate for the mean length of all components produced is the sample mean, which is 90 cm.
B. To find the 95% confidence interval for the mean length of all components produced, we can use the formula:
Confidence interval = sample mean ± (critical value) * (standard deviation / √sample size)
First, we need to find the critical value. Since the sample size is 64, the degrees of freedom is (64 - 1) = 63. Looking up the critical value for a 95% confidence level and 63 degrees of freedom in a t-distribution table or using a calculator, we find it to be approximately 1.997.
Next, we can calculate the confidence interval:
Confidence interval = 90 ± (1.997) * (1.6 / √64)
= 90 ± (1.997) * (1.6 / 8)
= 90 ± (1.997) * 0.2
= 90 ± 0.399
= (89.601, 90.399)
Therefore, the 95% confidence interval for the mean length of all components produced is (89.601 cm, 90.399 cm).