Question

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9. (20 pts) A fishing line manufacturer claims that his new X-25 series has an average tensile strength (H) of 25 psi following a normal distribution. To verify his claim, a sample of 25 specimens was tested and yielded the following results: 32, 19, 27, 24, 28, 29, 38, 30, 25, 33, 18, 25, 37,28, 20, 32, 23,21, 25, 19, 23,27, 32, 26,30 this fishing line not meeting the specifications? Based on the 90% confidence interval for μ is the manufacturer's claim valid? How large a sample is needed a. Is the sample data following a normal distribution? If the specification is 25 + 5 psi, what's the percentage of b. if we wish to be 90% confident that our sample mean will be within 1 psi of the true mean?

Answer 1

(a)

Following is the box plot of the data:

Following is the normal quantile plot of the data:

The box plot is symmetric and normal quantile plot shows approximate straight line pattern so we can assume that data is normally distributed.

The interval corresponding to 25 +/- 5 psi is (20, 30). Out of 25 data values, 16 lies in this interval and 25-16 = 9 does not lie in this interval. So the percentage of finishing time not meeting the specifications is

(9/25) *100% = 36%

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(b)

The 90% confidence interval.

Here we have

Since there 16 data values in the sample so degree of freedom is df=25-1=24 and critical value of t for 90% confidence interval, using excel function "=TINV(0.10,24)" is 1.711.

The required confidence interval is

Therefore, a 90% confidence interval for the mean is (24.99, 28.69).

Since 25 lies in the above interval so it seems that the manufacturer's claim is valid.

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Assuming sample standard deviation is a good estimate of population standard deviation we have

$sigma=5.41$

Critical value of z for 90% confidence interval , , is . The margin of error is E = 1 so required sample size will be

So required sample size for 90% confidence interval is 80.