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guajiro [1.7K]
2 years ago
13

An inspector inspects a shipment of medications to determine the efficacy in terms of the proportion p in the shipment that fail

ed to retain full potency after 60 days of production. Unless there is clear evidence that this proportion is less than 0.05, she will reject the shipment. To reach a decision, she will test the following hypotheses using the large-sample test for a population proportion:
H0 : p = 0.05, Ha : p < 0.05
To do so, she selects an SRS of 200 pills. Suppose that eight of the pills have failed to retain their full potency. The proportion 'p-hat' for the proportion of pills that have failed to retain their potency is:_____.
Mathematics
1 answer:
Dafna1 [17]2 years ago
4 0

Solution :

It is given that :

$H_0:p \geq 0.05$     (Null hypothesis, $H_0 : p = 0.05$ is also correct)

$H_a:p     (Alternate hypothesis, also called $H_1$)

This is a lower tailed test,

Therefore,

Sample proportion, $\hat p = \frac{x}{n}$

                                    $=\frac{8}{200}$

                                    = 0.04

And claimed proportion, P = 0.05

Significance level, $\alpha$ = 0.01

Therefore, calculating the statistics, we get

Standard deviation of $\hat p, \sigma_{\hat p}=\sqrt{\frac{P\times (1-P)}{n}}$

                                             $=\sqrt{\frac{0.05\times (1-0.05)}{200}}$

                                               $\approx 0.0154$

Test statistic,

$z_{observed}=\frac{\hat p -0.05}{\sigma_{\hat p}}$

              $=\frac{0.04 -0.05}{0.0154}$

               $\approx -0.65$

Now since this is a lower tailed test, p-value = $P(Z < z_{observed})=P(Z < -0.65)=0.2578$

Rejection Criteria : Reject $H_0$ if p-value < α .

Conclusion : Since the p-value $\geq \alpha$, we fail to reject the null hypothesis. There is insufficient evidence that p is less than 0.05

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<u>Step-by-step explanation:</u>

<u>Given :- </u>

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\implies Distance =\sqrt{ (x_2-x_1)^2+(y_2-y_1)^2} \\\\\implies Distance =\sqrt{ (18-7)^2+(42-15)^2 }  \\\\\implies Distance =\sqrt{ 11^2 + 27^2 } \\\\\implies Distance =\sqrt{ 121 + 729 } \\\\\implies Distance = \sqrt{ 850} \\\\\implies \boxed{ Distance = 29.15 \ units }

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