Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 7. the hypotheses h0: μ = 75 and ha: μ < 75 are to be tested using a random sample of n = 25 observations. (a) how many standard deviations (of x) below the null value is x = 72.3? (round your answer to two decimal places.) standard deviations (b) if x = 72.3, what is the conclusion using α = 0.01? (round your answers to two decimal places.) test statistic z = critical value z = what can you conclude? reject the null hypothesis. there is sufficient evidence to conclude that the mean drying time is less than 75. do not reject the null hypothesis. there is sufficient evidence to conclude that the mean drying time is less than 75. do not reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 75. reject the null hypothesis. there is not sufficient evidence to conclude that the mean drying time is less than 75. (c) what is α for the test procedure that rejects h0 when z ≤ −2.8? (round your answer to four decimal places.) α = (d) for the test procedure of part (c), what is β(70)? (round your answer to four decimal places.) β(70) = (e) if the test procedure of part (c) is used, what n is necessary to ensure that β(70) = 0.01? (round your answer up to the next whole number.) n = specimens (f) if a level 0.01 test is used with n = 100, what is the probability of a type i error when μ = 76? (round your answer to four decimal places.)

A paint drying situation, has the standard deviation 1.93. The detailed solution is given below.

Given :

Standard Deviation,

Mean,

Sample, n = 25

Solution :

A) We know that,

    ---- (1)

Now put the values of in equation (1) we get,

z = -1.93

Therefore, 72.3 is 1.93 standard deviations below the mean.

B) From the above part we know that z = - 1.93.Therefore p-value is,

P(-1.93) = 0.03

Since = 0.01 and p-value = 0.03, this means that the p-value is greater than the fail to reject the null hypothesis.

C) The p-value for

is 0.00256

Therefore, the for the test procedure that rejects the null hypothesis when is 0.0026.

D) For alternate hypothesis,

For = 0.0026

E) Now we want n when

n = 51.4

Therefore we need n = 52.

F)

Since p-value is larger that .01 therefore we fail to reject = 76.

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answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab  in the y coordinate 

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

  x coordinate 

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

  y coordinate 

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

  x  coordinate 

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

  y  coordinate 

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

  x  coordinate 

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

  y  coordinate 

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

  x  coordinate 

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

  y  coordinate 

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

  x  coordinate 

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

  y  coordinate 

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

  x  coordinate 

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

  y  coordinate 

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

  x  coordinate 

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

  y  coordinate 

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

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