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Anna71 [15]
3 years ago
8

A political candidate has asked you to conduct a poll to determine what percentage of people support him. If the candidate only

wants a 5% margin of error at a 99.5% confidence level, what size of sample is needed?
Mathematics
1 answer:
lara31 [8.8K]3 years ago
7 0

Answer:

900

Step-by-step explanation:

Margin of error = critical value × standard error

The margin of error is given to be 0.05.

At 99.5% confidence, the critical value is z = 3.

The standard error for a proportion is √(pq/n).  Since p is not given, we will assume p = 0.5.

0.05 = 3 √(0.5 × 0.5 / n)

n = 900

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Answer:

z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179    

p_v =2*P(z>1.179)=2*[1-P(Z    

Step-by-step explanation:

Data given and notation    

\bar X=4000 represent the average score for the sample    

s=12000 represent the sample standard deviation    

n=200 sample size    

\mu_o =39000 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to apply a two tailed test.    

What are H0 and Ha for this study?    

Null hypothesis:  \mu = 39000    

Alternative hypothesis :\mu \neq 39000    

Compute the test statistic  

The sample size is large enough to assume the distribution for the statisitc normal. The statistic for this case is given by:    

z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

We can replace in formula (1) the info given like this:    

z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179    

Give the appropriate conclusion for the test  

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p_v =2*P(z>1.179)=2*[1-P(Z    

Conclusion    

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