Answer to Question #255551 in Statistics and Probability for Tine

Question

In a hospital’s shipment of 3 500 insulin syringes, 14 were unusable due to defects. (a) at alpha = 0.05, is this sufficient evidence to reject future shipments from this supplier if the hospital’s quality standard requires 99.7 percent of the syringes to be acceptable?

Answer 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$p\geq0.997$

$p<0.997$

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z: z < -1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{3486/3500-0.997}{\sqrt{\dfrac{0.997(1-0.997)}{3500}}}$$

$$=-1.081747$$

Since it is observed that $z = -1.081747 \ge -1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p =P(Z<-1.081747)= 0.139682,$ and since $p = 0.139682 \ge 0.05,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is less than $0.997,$ at the $\alpha = 0.05$ significance level.

It is not sufficient evidence to reject future shipments from this supplier at the $\alpha = 0.05$ significance level.