MATHEMATICS COLLEGE

A random sample of 150recent donations at a certainblood bank reveals that
45 were type A blood.Does this suggest that the
actual percentage of type A
donations is less than40%, the percentage of the
population having type A
blood? Carry out a testof the appropriate
hypotheses using a significance
level of0.01.

Answers

Answer 1

Answer:

Answer:

Null hypothesis: $p\geq 0.4$

Alternative hypothesis: $p < 0.4$

$z=\frac{0.3 -0.4}{\sqrt{(0.4(1-0.4))/(150)}}=-2.5$

$p_v =2*P(Z<-2.5)=0.0124$

If we compare the p value obtained and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that the true proportion is not significantly lower than 0.4 or 40% at 1% of significance.

Step-by-step explanation:

1) Data given and notation

n=150 represent the random sample taken

X=45 represent the people with type A blood

$\hat p=(45)/(150)=0.3$ estimated proportion of people with type A blood

$p_o=0.4$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of people type A blood is less than 0.4:

Null hypothesis: $p\geq 0.4$

Alternative hypothesis: $p < 0.4$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.3 -0.4}{\sqrt{(0.4(1-0.4))/(150)}}=-2.5$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(Z<-2.5)=0.0124$

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Answers

Answer:

ffbhjdhjdfahdsfsdfakfawlefgwaheghfgawlygfailwegfalawegfuwlgweilfygweiyaglifwafawegwegerg

Step-by-step explanation:

Pfvr alguem me ajuda !!!

Answers

Answer:

...i cant see the pic

Step-by-step explanation:

Paul owns a mobile wood-fired pizza oven operation. A couple of his clients complained about his dough at a recent catering, so he changed his dough to a newer product. Using the old dough, there were 6 complaints out of 385 pizzas. With the new dough, there were 16 complaints out of 340 pizzas. Let p 1 be the proportion of customer complaints with the old dough and p 2 be the proportion of customer complaints with the new dough. State the competing hypotheses to determine if the proportion of customer complaints using the old dough is less than the proportion of customer complaints using the new dough g

Answers

Answer:

$z=\frac{0.0156-0.0471}{\sqrt{0.0303(1-0.0303)((1)/(385)+(1)/(340))}}=-2.469$

Now we can calculate the p value with this probability:

$p_v =P(Z<-2.469)= 0.0068$

Since the p value is a very low value and using any significance level 5% or 10% we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of customer complaints using the old dough is less than the proportion of customer complaints using the new dough.

Step-by-step explanation:

Information provided

$X_(1)=6$ represent the complaints with the old dough

$X_(2)=16$ represent the complaints with the new dough

$n_(1)=385$ sample 1 selected

$n_(2)=340$ sample 2 selected

$p_(1)=(6)/(385)=0.0156$ represent the proportion of complaints with the old dough

$p_(2)=(16)/(340)=0.0471$ represent the proportion of complaints with the new dough

$\hat p$ represent the pooled estimate of p

z would represent the statistic

$p_v$ represent the value

Hypothesis to test

We want to verify if the proportion of customer complaints using the old dough is less than the proportion of customer complaints using the new dough, the system of hypothesis would be:

Null hypothesis: $p_(1) \geq p_(2)$