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OverLord2011 [107]
2 years ago
5

A grocery store manager claims that 75% of shoppers purchase bananas as least once a month. Technology was used to simulate choo

sing 150 SRSs of size n = 100 from a population of shoppers where 75% buy bananas. The dotplot shows pˆ = the sample proportion of shoppers who bought bananas in the past month. A random sample of 100 shoppers from the store were selected and 64 bought bananas in the past month. Does this sample provide evidence that the grocery store manager overstated the true proportion? Justify your answer.
Mathematics
1 answer:
Keith_Richards [23]2 years ago
7 0

Answer:

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

 

Step-by-step explanation:

 Given the data in the question;

To test whether population proportion p is overstated;

Null hypothesis H₀ : p = (75%) = 0.75

Alternative hypothesis H₁ : = < (75%) < 0.75

now, sample proportion p" = 64 / 100 = 0.64

from the dot plot below, we will determine the p-value for test { P(p" < 0.64)}

so, the number of times p"<0.64 in 150 simulations is 5

Hence; P(p" < 0.64 ) = 5 / 150 = 0.0333

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

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Read 2 more answers
Test the hypothesis using the P value approach. Be sure the verify the requirements of the test.
Andreas93 [3]

Answer:

p_v =2*P(z  

If we compare the p value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly different from 0.77.  

Step-by-step explanation:

1) Data given and notation

n=500 represent the random sample taken

X=380 represent the number of people with some characteristic

\hat p=\frac{380}{500}=0.76 estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

p_o=0.76 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

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 is 0.7 .:  

Null hypothesis:p=0.77  

Alternative hypothesis:p \neq 0.77  

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.

<em>Check for the assumptions that he sample must satisfy in order to apply the test </em>

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 =500*0.77=385>10

n(1-p_o)=384*(1-0.77)=115>10

3) Calculate the statistic  

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

z=\frac{0.76-0.77}{\sqrt{\frac{0.77(1-0.77)}{500}}}=-0.531  

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=0.05. 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  

If we compare the p value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly different from 0.77.  

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Suppose that you are headed toward a plataeu 70m high. If the angle of elevation to the top of the platau is 35 degree how far a
SIZIF [17.4K]

Answer:

The distance you are from the base of plateau = 99.97\ m\approx 100\ m

Step-by-step explanation:

Given:

Height of plateau = 70 m

Angle of elevation to the top of plateau = 35°

To find the distance you are from the base of plateau.

We will construct a triangle ABC to model the given situation. The triangle would be a right triangle for which we know an angle and its opposite side. We need to find the adjacent side of the triangle.

We will apply trigonometric ratio to find the adjacent side.

\tan\theta=\frac{Opposite\ Side}{Adjacent\ Side}

where \theta represents the angle of reference.

Plugging in the values from the triangle.

\tan 35\° =\frac{BC}{AC}

\tan 35\° =\frac{70}{AC}

Multiplying both sides by AC.

AC\times \tan 35\° =AC\times \frac{70}{AC}

AC\times \tan 35\° =70

Dividing both sides by \tan 35\°

\frac{AC\times \tan 35\°}{\tan 35\°} =\frac{70}{\tan 35\°}

∴ AC=99.97\ m\approx 100\ m

The distance you are from the base of plateau = 99.97\ m\approx 100\ m

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