Help 11 minutes!!!!!!

Company A makes a large shipment to Company B. Company B can reject the shipment if they can conclude that the proportion of def

boyakko [2]

Answer:

$z=\frac{0.1475-0.1}{\sqrt{\frac{0.1(1-0.1)}{400}}}=3.17$

The p value for this case would be given by:

$p_v =P(z>3.17)=0.00076$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 0.1 and then Company B can reject the shipment

Step-by-step explanation:

Information provided

n=400 represent the random sample taken

X=59 represent number of defectives from the company B

$\hat p=\frac{59}{400}=0.1475$ estimated proportion of defectives from the company B

$p_o=0.1$ is the value to verify

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true proportion of defectives is higher than 0.1 then the system of hypothesis are.:

Null hypothesis: $p \leq 0.1$

Alternative hypothesis: $p > 0.1$

The statistic would be given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.1475-0.1}{\sqrt{\frac{0.1(1-0.1)}{400}}}=3.17$

The p value for this case would be given by:

$p_v =P(z>3.17)=0.00076$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 0.1 and then Company B can reject the shipment

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Answer:

there is not enough info to answer this

Step-by-step explanation:

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