Question

The weights of aluminum castings produced by a process are normally distributed. A random sample of 5 castings is selected; the sample mean weight is 2.21 pounds; and the sample standard deviation is 0.12 pound. The 98% confidence interval for the population mean casting weight is _________.

Answer 1

Answer: [2.00891488, 2.411085121

Step-by-step explanation:

Step 1: Subtract 1 from your sample size. 5 – 1 = 4. This gives you degrees of freedom, which you’ll need in step 3.

Step 2: Subtract the confidence level from 1, then divide by two.

(1 – .98) / 2 = 0.02

Step 3: Look up your answers to step 1 and 2 in the t-distribution table. For 9 degrees of freedom (df) and α = 0.02, my result is 3.747

Step 4: Divide your sample standard deviation by the square root of your sample size.

0.12 / √(5) = 0.0536656315

Step 5: Multiply step 3 by step 4.

3.747 × 0.0536656315 = 0.201085121

Step 6: For the lower end of the range, subtract step 5 from the sample mean.

2.21 – 0.201085121 = 2.00891488

Step 7: For the upper end of the range, add step 5 to the sample mean.

2.21 + 0.201085121 = 2.411085121

That's the answer. Cheers.

Answer 2

Answer:

(2.085, 2.335)

Step-by-step explanation:

See calculations in the attached file below.

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