Shot peening is used to compress the surface area of metal parts to make them more resistant to fractures. It is done by bombarding the surface with small particles hurled at high velocity. Each time a particle hits, it puts a small dent in the surface and compresses the area directly beneath the surface. The bombardment continues until eventually the entire surface is compressed. Tests are conducted on a particular part to see if shot peening reduces the proportion of parts that fractice when put into use. These data result:

Not shot peened: n1 = 35 and number fractured = 7.

Shot peened: n2 = 40 and number fracture = 3.

Set up the appropriate null and alternative hypotheses. based on these data, do you think that shot peening reduces the probability that a part will fracture when put into use? Explain based on the P value of the test.

Question

Mathematics

Kayden Key

18 August, 02:03

Shot peening is used to compress the surface area of metal parts to make them more resistant to fractures. It is done by bombarding the surface with small particles hurled at high velocity. Each time a particle hits, it puts a small dent in the surface and compresses the area directly beneath the surface. The bombardment continues until eventually the entire surface is compressed. Tests are conducted on a particular part to see if shot peening reduces the proportion of parts that fractice when put into use. These data result:

Not shot peened: n1 = 35 and number fractured = 7.

Shot peened: n2 = 40 and number fracture = 3.

Set up the appropriate null and alternative hypotheses. based on these data, do you think that shot peening reduces the probability that a part will fracture when put into use? Explain based on the P value of the test.

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Answers (1)

Know the Answer?

Reed Vasquez · Answer 1

Step-by-step explanation:

This is a test of 2 population proportions. Let 1 and 2 be the subscript for metal parts not shot peened and metal parts shot peened metal parts. The population proportions of fractures on metal parts not shot peened and fractures on metal parts shot peened would be p1 and p2

p1 - p2 = difference in the proportion of fractures on not shot peened and shot peened metal parts.

The null hypothesis is

H0 : p1 = p2

p1 - p2 = 0

The alternative hypothesis is

H1 : p1 > p2

p1 - p2 > 0

it is a right tailed test

Sample proportion = x/n

Where

x represents number of success (number of fractures)

n represents number of samples

For metal parts not shot peened,

x1 = 7

n1 = 35

p1 = 7/35 = 0.2

For metal parts shot peened,

x2 = 3

n2 = 40

p2 = 3/40 = 0.075

The pooled proportion, pc is

pc = (x1 + x2) / (n1 + n2)

pc = (7 + 3) / (35 + 40) = 0.13

1 - pc = 1 - 0.13 = 0.87

z = (p1 - p2) / √pc (1 - pc) (1/n1 + 1/n2)

z = (0.2 - 0.075) / √ (0.13) (0.87) (1/35 + 1/40) = - 0.031/0.07783911979

z = 1.61

Since it is a right tailed test, we would find the p value for the area to the right of the z score. From the normal distribution table,

p value = 1 - 0.9463 = 0.0537

Let us assume a significance level of 0.05

Since 0.05 < 0.0537, we would accept the null hypothesis

Therefore, we can conclude that shot peening does not reduce the probability that a part will fracture when put into use.