Question

A hospital director believes that 75% of the test tubes contain errors. A sample of 390 tubes found 273 errors. Is there sufficient evidence at the 0.10 level to refute the hospital director's claim? State the null and alternative hypotheses for the above scenario.

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H0 : p =0.75

Ha : p 0.75

= x / n = 273/390=0.7

1 - P0 = 1-0.75=0.25

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

=0.7-0.75 / [(0.75*0.25) / 390]

= -2.28

P(z <-2.28 ) = 0.0113

P-value = 2*0.0133=0.0266

= 0.10

P-value <

Reject the null hypothesis .

There is sufficient evidence to suggest that