Question

A quality inspector is worrying about the defective produced in the production process. He wants to test that the average defectives produced are 145 or less. He collects a sample of 25 production runs and found that a sample mean of 150 defectives were produced with a std deviation of 20. With this information he wishes to perform a hypothesis test at 5% significance level.

The decision is to:

	Reject Null; test statistic < critical value (or t test statistic < t alpha)
	Do not reject null; test statistic < critical value (or t test statistic < t alpha)
	Reject Null; test statistic < -critical value (t test statistic < -t-alpha)
	Do not reject null; test statistic < -critical value (test statistic > -t-alpha)

Answer 1

Answer:

n= 25,  = 145

= 150 , s = 20

= 0.05

null and alternative hypothesis is

Ho:   145

H1:   > 145

formula for test statistics is

t = 1.25

test statistics : t = 1.25

Calculate t critical value for right tailed test with = 0.05

and df = n -1 = 25 -1 = 24

using t table we get critical values as

Critical value =  1.711

decision rule is

Reject Ho if ( test statistics ) > ( Critical value)

here, ( test statistics = 1.25 ) < ( Critical value = 1.711 )

Hence,

Null hypothesis is NOT rejected.

Do not reject null; test statistic < critical value (or t test statistic < t alpha)