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In July of 2013, Australians were asked if they thought unemployment would increase, and 47% thought that it would increase. In November of 2013, they were asked again. At that time 570 out of 1300 said that they thought unemployment would increase. At the 8% level, is there enough evidence to show that the proportion of Australians in November 2013 who believe unemployment would increase is lower than the proportion who felt it would increase in July 2013?P: PARAMETER     What is the correct parameter symbol for this problem?     Correct     What is the wording of the parameter in the context of this problem?     CorrectH: HYPOTHESES     Fill in the correct null and alternative hypotheses:     𝐻0: Correct Correct .47Correct     𝐻𝐴: Correct Correct .47CorrectA: ASSUMPTIONS     Since Correct information was collected from each object, what conditions do we need to check?     Check all that apply.     𝑛≥30 or normal population.𝑛(𝑝̂)≥10𝑁≥20𝑛σσ is known.𝑛(1-𝑝̂)≥10𝑛𝑝≥10σσ is unknown.𝑛(1-𝑝)≥10Partially correct     Check those assumptions:          1. 𝑛𝑝 = 611Correct which is Correct 10Correct          2. 𝑛(1-𝑝) = 689Correct which is Correct 10Correct          3. 𝑁 = 1000000Correct which is Correct $30$30​Incorrect              If no N is given in the problem, use 1000000N: NAME THE PROCEDURE     The conditions are met to use a Correct .T: TEST STATISTIC     The symbol and value of the random variable on this problem are as follows:     Leave this answer as a fraction.     Correct = 5701300​Correct      The formula set up of the test statistic is as follows.:       (Leave any values that were given as fractions as fractions) 𝑧=𝑝̂-𝑝𝑝(1-𝑝)𝑛=( 5701300​Correct -.47Correct )/((.47Correct ⋅(1-.47Correct )) /1300Correct )       Final answer for the test statistic from technology.     Round to 2 decimal places:     z = −2.08IncorrectO: OBTAIN THE P-VALUE     Report to 4 decimal places.     It is possible when rounded that a p-value is 0.0000     P-value = .0188IncorrectM: MAKE A DECISION     Since the p-value Correct .08Correct , we Correct .S: STATE A CONCLUSION    There Correct significant evidence to conclude Correct Incorrect

Question

In July of 2013, Australians were asked if they thought unemployment would increase, and 47% thought that it would increase. In November of 2013, they were asked again. At that time 570 out of 1300 said that they thought unemployment would increase. At the 8% level, is there enough evidence to show that the proportion of Australians in November 2013 who believe unemployment would increase is lower than the proportion who felt it would increase in July 2013?P: PARAMETER     What is the correct parameter symbol for this problem?     Correct     What is the wording of the parameter in the context of this problem?     CorrectH: HYPOTHESES     Fill in the correct null and alternative hypotheses:     𝐻0: Correct Correct .47Correct     𝐻𝐴: Correct Correct .47CorrectA: ASSUMPTIONS     Since Correct information was collected from each object, what conditions do we need to check?     Check all that apply.     𝑛≥30 or normal population.𝑛(𝑝̂)≥10𝑁≥20𝑛σσ is known.𝑛(1-𝑝̂)≥10𝑛𝑝≥10σσ is unknown.𝑛(1-𝑝)≥10Partially correct     Check those assumptions:          1. 𝑛𝑝 = 611Correct which is Correct 10Correct          2. 𝑛(1-𝑝) = 689Correct which is Correct 10Correct          3. 𝑁 = 1000000Correct which is Correct 303030​Incorrect              If no N is given in the problem, use 1000000N: NAME THE PROCEDURE     The conditions are met to use a Correct .T: TEST STATISTIC     The symbol and value of the random variable on this problem are as follows:     Leave this answer as a fraction.     Correct = 5701300​Correct      The formula set up of the test statistic is as follows.:       (Leave any values that were given as fractions as fractions) 𝑧=𝑝̂-𝑝𝑝(1-𝑝)𝑛=( 5701300​Correct -.47Correct )/((.47Correct ⋅(1-.47Correct )) /1300Correct )       Final answer for the test statistic from technology.     Round to 2 decimal places:     z = −2.08IncorrectO: OBTAIN THE P-VALUE     Report to 4 decimal places.     It is possible when rounded that a p-value is 0.0000     P-value = .0188IncorrectM: MAKE A DECISION     Since the p-value Correct .08Correct , we Correct .S: STATE A CONCLUSION    There Correct significant evidence to conclude Correct Incorrect

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Solution

The parameter for this problem is the proportion of Australians who believe unemployment will increase.

The null hypothesis (H0) is that the proportion (P) is equal to 0.47, which is the proportion from July 2013. The alternative hypothesis (HA) is that the proportion (P) is less than 0.47.

The assumptions we need to check are that the sample size (n) is greater than or equal to 30, nP is greater than or equal to 10, and n(1-P) is greater than or equal to 10.

Checking these assumptions, we find that nP = 611, which is greater than 10, and n(1-P) = 689, which is also greater than 10. The sample size (n) is 1300, which is greater than 30.

The conditions are met to use a one-sample z-test for a proportion.

The test statistic is calculated as follows: z = (P̂ - P) / sqrt [ P(1 - P) / n ] = (570/1300 - 0.47) / sqrt [ 0.47(1 - 0.47) / 1300 ] = -2.08.

The p-value is 0.0188.

Since the p-value is less than the significance level of 0.08, we reject the null hypothesis.

Therefore, there is significant evidence to conclude that the proportion of Australians in November 2013 who believe unemployment would increase is lower than the proportion who felt it would increase in July 2013.

This problem has been solved

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