4. Use calc. Scientific method please answer each with number responses.. ty...

Question

4. Use calc. Scientific method please answer each with number responses.. ty... please round per question

The piston diameter of a certain hand pump is 0.7 inch. The manager determines that the diameters are normally distributed, with a mean of 0.7 inch and a standard deviation of 0.004 inch. After recalibrating the production machine, the manager randomly selects 25 pistons and determines that the standard deviation is 0.0035 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the alpha equals =0.10 level of significance?

What are the correct hypotheses for this test?

The null hypothesis is H0: (1)_______(2)_______ (3)_______

The alternative hypothesis is H1: (4)_______ (5)________ (6)________

Calculate the value of the test statistic. 2 =________ (Round to three decimal places as needed.)

Use technology to determine the P-value for the test statistic. The P-value is ________ (Round to three decimal places as needed.)

What is the correct conclusion at the alpha equals =0.10 level of significance?

Since the P-value is (7) _________than the level of significance, (8) ___________the null hypothesis.

There (9) ________sufficient evidence to conclude that the standard deviation has decreased at the 0.10 level of significance.

Answer accordingly

(1)u//p

(2)>//=/<

(3)0.0035./0.004.

(4)/p/

(5)>/=/</

(6)0.0035./0.004.

(7)greater/less

(8)reject/do not reject

(9)is/is not

Explanation / Answer

Given that,
population standard deviation ()=0.7
sample standard deviation (s) =0.0035
sample size (n) = 25
we calculate,
population variance (^2) =0.49
sample variance (s^2)=0.00001225
null, Ho: =0.7
alternate, H1 : <0.7
level of significance, = 0.05
from standard normal table,left tailed ^2 /2 =36.415
since our test is left-tailed
reject Ho, if ^2 o < -36.415
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(25 - 1 ) * 0.00001225 / 0.49 = 24*0.00001225/0.49 = 0
| ^2 cal | =0
critical value
the value of |^2 | at los 0.05 with d.f (n-1)=24 is 36.415
we got | ^2| =0 & | ^2 | =36.415
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =1

ANSWERS
---------------
null, Ho: =0.7
alternate, H1 : <0.7
test statistic: 0
critical value: -36.415
p-value:1
decision: do not reject Ho

we don't hav evidence that it is decreased

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