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4 years ago

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Suppose that the probability a seed will germinate is 80%. What is the probability that 7 of these seeds will germinate when 10

are planted? a. 1.0068 b. 25.17 c. 120 d. 0.201

2 answers:

azamat4 years ago

8 0

Your answer will be letter B

Stells [14]4 years ago

6 0

Answer: d. 0.201

Step-by-step explanation:

Given: The probability a seed will germinate = 80%

i.e. $p= 0.8$

Then probability of a seed will not germinate is given by :-

$q= 1-p$

i.e $q=1-0.8$

i.e. $q=0.2$

By Binomial Distribution, the probability of getting r successes in n trials is given by :-

$P(r)=^nC_r\ p^rq^{n-r}$

Now , the probability that 7 (r=7) of these seeds will germinate when 10 (n=10) are planted is given by :-

$P(7)=^{10}C_7\ (0.8)^7(0.2)^{10-7}\\\\=\dfrac{10!}{(10-7)!7!}(0.8)^7(0.2)^3\\\\=\dfrac{10\times9\times8}{3\times2}(0.8)^7(0.2)^3\\\\=0.201326592\approx0.201$

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a) $z=4.658$

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b) Using the significance level assumed $\alpha=0.05$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is higher than 50%.

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Step-by-step explanation:

I )Part a

1) Data given and notation

n=1507 represent the random sample taken

X represent the Americans who thinks that the Civil War is still relevant to American politics and political life

$\hat p$ estimated proportion of Americans who thinks that the Civil War is still relevant to American politics and political life in the sample

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha=0.05$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

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

p= population proportion of Americans who thinks that the Civil War is still relevant to American politics and political life

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =1507*0.5=753.5>10$

$n(1-p_o)=1507*(1-0.5)=753.5>10$

3) Calculate the statistic

The statistic is calculated with the following formula:

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

On this case the value of $p_o=0.5$ is the value that we are testing and n = 1507.

$z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1507}}}=4.658$

The p value for the test would be:

$p_v =P(z>4.65)=1-P(z$

II) Part b

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.

Based on the alternative hypothesis the p value would be given by:

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Using the significance level assumed $\alpha=0.05$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is higher than 50%.

III) Part c

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$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.56 - 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.527$

$0.56 + 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.593$

And the 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

And this result agrees with the result of part b, since the interval not contains the value of 0.5 we can conclude that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life it's higher than 0.5 at 90% of confidence.

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Shkiper50 [21]

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