Question

lom notsin ns o PROBLEM# 5: A pharmaceutical loader has been set to insert exactly 5 miligrams into each capsule of a particular drug. Periodically a sample of 100 capsules is taken and the contents of each are measured. The standard deviation has been established at .2 milligram per capsule. Corrective action will be taken whenever the computed mean is significantly great or small. Determinea decision rule when a 1% chance is permitted for taking unnecessary a. action. b. Apply the above to find the probability of failing to take needed correction when the true mean is (1) 5.05 milligrams? (2) 4.97 milligrams?

Answer 1

Answer:

Given,

n = 100

xbar = 5

SD = 0.2

a)

Here for the 1% significance level , for two tailed z critical value is 2.58

z = (xbar-mean)/(sd/sqrt(n))

= (5.05-5)/(0.2/10)

= 2.5

Here Zcal < Zcrit , so we reject the null hypothesis Ho.

b)

i)

Null hypothesis Ho : $\mu$ = 5.05

Alternative hypothesis Ha : $\mu$ $\neq$ 5.05

consider,

z = (xbar - mean) / (sd / sqrt(n))

= (5.05 - 5) / (0.2 / 10)

= 2.5

P value = 0.012419.[two tailed]

The P-Value is 0.012419. [since from z table]

So the result is not significant at p < 0.01.

ii)

Now consider,

z = (xbar-mean)/(sd/sqrt(n))

= (4.97-5)/(0.2/10)

= -1.5

p value = 0.1336144 [since from z table]

The P-Value is 0.133614.[two tailed]

Here the result is significant due to that p > 0.01.