Question

Please provide excel functions to answer!! Thanks!

An experiment was conducted to investigate the machine performance
of deep hole drilling when chip congestion exists. The length, in mm, of 50 drill chips had
a sample mean of 71.2 mm. Assume the population standard deviation is 15.2 mm.
Conduct a test to determine whether the true mean drill chip length is less than 75 mm at α=.01.

Sample size

xbar

sigma

null hypothesis

alternative hypothesis

standard error

test statistic (z)

p value - lower tail

p value - upper tail

p value two tail

significance level (alpha)

appropriate p value

conclusion (write in terms with alternative hypothesis)

Answer 1

Here we have given that,

n=Sample size i.e. the number of drill chips = 50

= sample mean =71.2 mm

= Population standard deviation= 15.2 mm

Here, the population standard deviation is known we are using the one-sample z-test to test the hypothesis.

Claim: To check whether the mean drill chip length is less than 75 mm.

The null and alternative hypothesis is as follows

Versus

where = Population mean of drill chip length

This is left one-tailed test.

Now, we can find the test statistic

z-statistics =

                  =

                  = Here, we get the standard error ==2.150

                   = -1.77

The Test statistic is -1.77

Now we can find the P-value

This is left one-tailed test

P-value = P(Z < z-statistics)

               =P( Z < -1.77)

               =0.0384 Using standard normal z table see the value corresponding to the z=-1.77

we get the P-value is 0.0384

= level of significance=0.01

Decision:

P-value (0.0384) greater than (>) 0.01 ()

Conclusion:

we fail to reject the Ho (Null Hypothesis).

There is not sufficient evidence to conclude that the mean drill chip length is less than 75 mm.