HERE IS THE TABLE OF DATA YOU NEED TO SOLVE A, B AND C Ball Scratches 1 0 2 1 3

Question

HERE IS THE TABLE OF DATA YOU NEED TO SOLVE A, B AND C

Ball Scratches 1 0 2 1 3 0 4 0 5 0 6 0 7 1 8 0 9 0 10 2 11 0 12 0 13 0 14 0 15 0 16 1 17 0 18 0 19 0 20 0 21 0 22 0 23 0 24 0 25 0 26 1 27 0 28 0 29 0 30 0 31 0 32 0 33 0 34 0 35 0 36 1 37 0 38 0 39 1 40 0 41 0 42 0 43 0 44 0 45 0 46 1 47 0 48 0 Cliff is the quality-assurance engineer for a particular manufacturer of billiard supplies. One of the items that the manufacturer produces is sets of pocket billiard balls. Cliff has been monitoring the finish of the pocket billiard balls. He is concemed that sets of billiard balls have been shipped with an increasing number of scratches. The company's goal is to have no more than an average of one scratch per set of pocket billiard balls. A set contains 16 balls. over the last week, Cliff selected a sample of 48 billiard balls and inspected them to determine the number of scratches. The data collected by Cliff are displayed in the accompanying data table. Complete parts a through d below. BEE Click on the icon to view the data table. a. Detemine the number of scratches in the sample. The number of scratches in the sample is

Explanation / Answer

Solution:-

a) The number of scratches in the sample is 9.

b) The average number of scratches for 48 pockets billard balls if manufacturer has met its goal is 3.

For 16 balls there are maximum of 1 balls

For 48 balls there are 3 scratches

c)The probability of observing atleast 1 scratches in each sample is 0.644.

p = 1 / 16 = 0.0625

n = 16

x = 1

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

P(x > 1) = 0.644

d)

P = 1/16 = 0.0625

p = 9/48 = 0.1875

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.0625

Alternative hypothesis: P > 0.0625

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.0349

z = (p - P) /

z = 3.582

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is more than 3.582. We use the Normal Distribution Calculator to find P(z > 3.582) = 0.0002

Thus, the P-value = 0.0002

Interpret results. Since the P-value (0.0002) is less than the significance level (0.05), we have to reject the null hypothesis.

From the test we do not have sufficient evidence that manufacturer has met his goal.

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