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2 years ago

Consider random samples of size 50 from a population with proportion 0.35.

Mathematics

1 answer:

lisov135 [29]2 years ago

8 0

Answer:

The mean is $\mu_{\^ p } = p = 0.35$

The standard error is $SE = 0.06745$

Step-by-step explanation:

From the question we are told that

The sample size is n = 50

The population proportion is $p = 0.35$

Generally given that the sample size is large enough , then the mean of the distribution is mathematically represented as

$\mu_{\^ p } = p = 0.35$

Generally the standard error of the distribution is mathematically represented as

$SE = \sqrt{\frac{p (1- p )}{n} }$

=> $SE = \sqrt{\frac{0.35 (1- 0.35 )}{50} }$

=> $SE = 0.06745$

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A buyer went to the market to buy strawberries. He purchased 120 randomly selected strawberries from a vendor who claimed that n

mafiozo [28]

Answer:

$z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10$

$p_v =P(z>2.10)=0.018$

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

Step-by-step explanation:

Data given and notation

n=120 represent the random sample taken

X=40 represent the number of strawberries damaged

$\hat p=\frac{40}{120}=0.333$ estimated proportion of strawberries damaged

$p_o=0.25$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that no more than 25% of his total harvest of strawberries was damaged.:

Null hypothesis: $p\leq 0.25$

Alternative hypothesis: $p > 0.25$

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

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

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

Calculate the statistic

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

$z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10$

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 next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2.10)=0.018$

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

6 0

3 years ago

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

Gemiola [76]

Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

$P(Atleast\ 2 \ girls) = 0.5$

iii.

$No\ girl = \{BBB\}$

$P(No\ girl) = 0.125$

Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

7 0

3 years ago

Help plz im sorry if im asking to many questions

Zinaida [17]

Answer:

The 3rd one 50 - 12p = 26

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

To qualify for the championship a runner must complete the race in less than 55 minutes ....... Use "t" to represent the time in

Svet_ta [14]

The inequality is t < 55

Solution:

Given that, To qualify for the championship a runner must complete the race in less than 55 minutes

Let "t" represent the time in minutes of a runner who qualifies for the championship

Here it is given that the value of t is less than 55 minutes

Therefore, "t" must be less than 55, so that the runner qualifies the championship

This is represented by inequality:

$t$

The above inequality means, that time taken to complete the race must be less than 55 for a runner to qualify

Hence the required inequality is t < 55

3 0

3 years ago

Help me on this please

steposvetlana [31]

I got you hun!

27= 3x is the equation you will need

27 is the total the coach needs making it the total. Since each pack comes with 3 you will mutiply the amount of pack, or x, by 3.

X=9

we know this because if you divide 3 by 27 to get it on the other side we get 9. This means the coach needs to buy 9 packs of baseballs for the team.

27=3(9)

Hope this helps!

5 0

3 years ago