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rosijanka [135]
2 years ago
13

You interview a random sample of 50 adults. the results of the survey show that 46​% of the adults said they were more likely to

buy a product when there are free samples. at alphaequals0.05​, can you reject the claim that at least 51​% of the adults are more likely to buy a product when there are free​ samples?
Business
1 answer:
photoshop1234 [79]2 years ago
7 0

Answer:

z=\frac{0.46 -0.51}{\sqrt{\frac{0.51(1-0.51)}{50}}}=-0.707  

p_v =P(z  

So the p value obtained was a very high 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 of adults who said they were more likely to buy a product when there are free samples is higher or equal than 0.51

Explanation:

Data given and notation

n=50 represent the random sample taken

\hat p==0.46 estimated proportion of adults who said they were more likely to buy a product when there are free samples

p_o=0.51 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is at least 0.51 or no.:  

Null hypothesis:p \geq 0.51  

Alternative hypothesis:p < 0.51  

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{0.46 -0.51}{\sqrt{\frac{0.51(1-0.51)}{50}}}=-0.707  

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 left tailed the p value would be:  

p_v =P(z  

So the p value obtained was a very high 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 of adults who said they were more likely to buy a product when there are free samples is higher or equal than 0.51

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The correct answer is; False, as of 2020.

Further Explanation:

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