Question

A road inspector checks the with a 45 random stripes to see if the machine has slipped out of adjustment the mean diameter for this sample is X bar 3.88" with a standard deviation of 0.5" Does this indicate that the machine S slipped out of adjustment and the average with of stripes is no longer 4" use a 5% level of significance conducted AT test to examine whether to me with have stripes is different from 4" calculate the P value

Answer 1

H₀: = 4

H₁: 4

The test statistic z = ()/(s/)

                           = (3.88 - 4)/(0.5/)

                          = -1.61

At 5% significance level, the critical values are +/- z_0.025 = +/- 1.96

Since the test statistic value is not less than the negative critical value(-1.61 > -1.96), so we should not reject the null hypothesis.

So there is not sufficient evidence to conclude that the average with of stripes is no longer 4.

P-value = 2 * P(Z < -1.61)

             = 2 * 0.0537

            = 0.1074