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mars1129 [50]
3 years ago
11

A poll indicates that ​% of a​ country's adults favor a ban on assault weapons. Suppose adults in a certain region are randomly

surveyed and say they favor a ban. Based on this​ sample, test using alpha whether the proportion of the​ region's residents who favor the ban is different from the proportion given in the poll for the country as a whole. Report the​ p-value for this hypothesis test. Identify the null and alternative hypotheses for this test. Let p be the population proportion of the​ region's residents who favor a ban on assault weapons.
Mathematics
1 answer:
hammer [34]3 years ago
6 0

Complete Question

poll indicates that 47% of a country's adults favor a ban on assault weapons. Suppose 330 adults in a certain region are randomly surveyed and 161 say they favor a ban. Based on this sample, test using alpha = 0.025 whether the proportion of the region's residents who favor the ban is different from the proportion given in the poll for the country as a whole. Report the p-value for this hypothesis test.

Identify the null and alternative hypotheses for this test. Let p be the population proportion of the region's residents who favor a ban on assault weapons.

A.

H_o: p \ge  0.47\\\\ H_A\  p < 0.47

B

 H_o: p \le\  0.47\\\\ H_A\  p > 0.47

C

 H_o: p \ne  \  0.47\\\\ H_A\  p =  0.47

D

 H_o: p =  \ 0.47\\\\ H_A\  p \ne  0.47

Perform the test using a normal approximation. Identify the test statistic. (Round to two decimal places as needed.)

Identify the p-value. (Round to three decimal places as needed.)

State the conclusion for this hypothesis test.

Answer:

The correct option is D

The test statistics is  z = 0.6116

The p-value  p-value= 0.5408

The conclusion

There no sufficient evidence to state that the proportion of the region's residents who favor the ban is different from the proportion given in the poll for the country as a whole.

Step-by-step explanation:

From the question we are told that  

  The population proportion is  p = 0.47

  The sample size is  n  =  330

  The number that favor the ban is k= 161

  The level of significance is \alpha = 0.025

The null hypothesis is  H_o : p =0.47

The alternative hypothesis is H_a : p \ne  0.47

Generally sample proportion is mathematically represented as

        \^p = \frac{k}{n}

=>     \^p = \frac{161}{330}

=>     \^p = 0.4879

Generally the test statistics is mathematically represented as

       z = \frac{\^ p - p }{\sqrt{\frac{p(1 - p)}{n}  } }

=>    z = \frac{0.4879 - 0.47 }{\sqrt{\frac{ 0.47 (1 - 0.47)}{ 330}  } }

=>    z = 0.6116

From the z table the area under the normal curve to the right corresponding to  0.6116 is  

       P( Z >  0.6116 ) = 0.2704

given that is a two tailed test the p-value is mathematically represented as

     p-value= 2 * 0.2704

      p-value= 0.5408

From the values obtained we see that  p-value > \alpha  so  

The decision rule  is

Fail to reject the null hypothesis

The conclusion

There no sufficient evidence to state that the proportion of the region's residents who favor the ban is different from the proportion given in the poll for the country as a whole

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