6. A manufacturer of sprinkler systems used for fire pro claims that the true is

Question

6. A manufacturer of sprinkler systems used for fire pro claims that the true is 130°p. A lll tested, yields a sample average activation temperature o stribution of activation times is normal with standard deviation e average system-activation temperature sys 131.08°. T , does the data contradict the manufacturer's claim? 7. The quality engineer of a company that manufactures synthetic rubber wants to examine the hardness of the rubber, measured in degrees Shore. A random sample of 12 pieces of rubber was evaluated with the following results: 62.4 66.0 67.6 63.2 66.3 61.8 61.8 67.5 61.3 65.0 64.6 61.3 Set up a 99% confidence interval for the average hardness of the rubber. a. o. What assumption did you have to make in part (a) to construct the confidence interval?

Explanation / Answer

Sample mean = 64.0667; Sample variance = 5.7515

Estimate of the standard error of the mean = sqrt(5.7515) / sqrt(12) = 0.6923

We use a t-distribution to calculate the confidence interval

Degrees of freedom = 12-1 = 11

Since the confidence interval is 99%, area on each side = 0.01/2=0.005

We need to find t0.995,11 = 3.106 (using a standard t-table)

The confidence interval is given as:

Lower Limit: 64.0667 - 3.106 x 0.6923 = 61.9165

Upper Limit: 64.0667 + 3.106 x 0.6923 = 66.2168

The 99% confidence interval for the population mean is [61.9165,66.2168]

b. The assumption is that the hardness of rubber population is normally distributed.

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