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3 years ago

Just 13 a and b please and thank you

Mathematics

1 answer:

ehidna [41]3 years ago

5 0

A)1.625m and b I can't read

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Read 2 more answers

Silicon wafers are scored and then broken into the many small microchips that will he mounted into circuits. Two breaking method

antoniya [11.8K]

Answer:

The estimate is $- 0.0265\le K \le 0.0465$

Method B this is because the faulty breaks are less

Step-by-step explanation:

The number of microchips broken in method A is $n_1 = 400$

The number of faulty breaks of method A is $X_1 = 32$

The number of microchips broken in method B is $n_2 = 400$

The number of faulty breaks of method A is $X_2 = 32$

The proportion of the faulty breaks to the total breaks in method A is

$p_1 = \frac{32}{400}$

$p_1 = 0.08$

The proportion of the faulty to the total breaks in method B is

$p_2 = \frac{28}{400}$

$p_2 = 0.07$

For this estimation the standard error is

$SE = \sqrt{ \frac{p_1 (1 - p_1)}{n_1} + \frac{p_2 (1- p_2 )}{n_2} } }$

substituting values

$SE = \sqrt{ \frac{0.08 (1 - 0.08)}{400} + \frac{0.07 (1- 0.07 )}{400} } }$

$SE = 0.0186$

The z-values of confidence coefficient of 0.95 from the z-table is

$z_{0.95} = 1.96$

The difference between proportions of improperly broken microchips for the two breaking methods is mathematically represented as

$K = [p_1 - p_2 ] \pm z_{0.95} * SE$

substituting values

$K = [0.08 - 0.07 ] \pm 1.96 *0.0186$

$K = - 0.0265 \ or \ K = 0.0465$

The interval of the difference between proportions of improperly broken microchips for the two breaking methods is

$- 0.0265\le K \le 0.0465$

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