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weqwewe [10]
3 years ago
13

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

ective items in the shipment is larger than 0.1. In a sample of 400 items from the shipment, Company B finds that 59 are defective. Conduct the appropriate hypothesis test for Company B using a 0.05 level of significance.
Mathematics
1 answer:
boyakko [2]3 years ago
8 0

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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p_v =P(t_{(5)}>0.499) =0.319

The p value is higher than any significance level given for example \alpha=0.05,0.1, so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations we can use it.  

Let put some notation  

x=Auscultatory method , y = Oscillatory method

x: 74 86 84 79 70 78 79 70

y: 70 85 90 110 71 80 69 74

The system of hypothesis for this case are:

Null hypothesis: \mu_x- \mu_y \leq 0

Alternative hypothesis: \mu_x -\mu_y >0

The first step is define the difference d_i=x_i-y_i, that is given so we have:

d: 6.6, 4.2, -5.5, -3.1, 9.3, -3.9

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}=1.267

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =6.215

The fourth step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{1.267 -0}{\frac{6.215}{\sqrt{6}}}=0.499

The next step is calculate the degrees of freedom given by:

df=n-1=6-1=5

Now we can calculate the p value, since we have a right tailed test the p value is given by:

p_v =P(t_{(5)}>0.499) =0.319

The p value is higher than any significance level given for example \alpha=0.05,0.1, so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.

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