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In July of 2013, 47% of Australians thought unemployment would increase.  Assume in a recent poll, 140 out of 300 said that they thought unemployment would increase.  Estimate the proportion of all Australians today who believe unemployment would increase using a 98% confidence interval.P: Parameter    What is the correct parameter symbol for this problem?        What is the wording of the parameter in the context of this problem?    A: AssumptionsSince information was collected from each object, what conditions do we need to check?  Check all that apply.   𝑛≥30 or normal population.𝑁≥20𝑛𝑛(1-𝑝̂)≥10𝑛𝑝≥10σσ is known.𝑛(1-𝑝)≥10σσ is unknown.𝑛(𝑝̂)≥10    Check those assumptions:    1. 𝑛𝑝^= which is     2. 𝑛(1-𝑝^)= which is     3. 𝑁= which is          If no N is given in the problem, use 1000000N: Name the procedure     The conditions are met to use a .I: Interval and point estimate    The symbol and value of the point estimate on this problem are as follows:    = Leave answer as a fraction.    The interval estimate for is ( , )    Round endpoints to 3 decimal places.C: Conclusion We are % confident that the is between % and %

Question

In July of 2013, 47% of Australians thought unemployment would increase.  Assume in a recent poll, 140 out of 300 said that they thought unemployment would increase.  Estimate the proportion of all Australians today who believe unemployment would increase using a 98% confidence interval.P: Parameter    What is the correct parameter symbol for this problem?        What is the wording of the parameter in the context of this problem?    A: AssumptionsSince information was collected from each object, what conditions do we need to check?  Check all that apply.   𝑛≥30 or normal population.𝑁≥20𝑛𝑛(1-𝑝̂)≥10𝑛𝑝≥10σσ is known.𝑛(1-𝑝)≥10σσ is unknown.𝑛(𝑝̂)≥10    Check those assumptions:    1. 𝑛𝑝^= which is     2. 𝑛(1-𝑝^)= which is     3. 𝑁= which is          If no N is given in the problem, use 1000000N: Name the procedure     The conditions are met to use a .I: Interval and point estimate    The symbol and value of the point estimate on this problem are as follows:    = Leave answer as a fraction.    The interval estimate for is ( , )    Round endpoints to 3 decimal places.C: Conclusion We are % confident that the is between % and %

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Solution

P: Parameter

  • The correct parameter symbol for this problem is p p .
  • The wording of the parameter in the context of this problem is: the proportion of all Australians today who believe unemployment would increase.

A: Assumptions Since information was collected from each object, what conditions do we need to check? Check all that apply.

  • n≥30 n \geq 30 or normal population.
  • n(1−p^)≥10 n(1-\hat{p}) \geq 10
  • np^≥10 n\hat{p} \geq 10

Check those assumptions:

  1. np^=140 n\hat{p} = 140 which is 300×140300=140 300 \times \frac{140}{300} = 140
  2. n(1−p^)=160 n(1-\hat{p}) = 160 which is 300×(1−140300)=160 300 \times (1 - \frac{140}{300}) = 160
  3. N=1000000 N = 1000000 which is given as a default if no N is provided.

N: Name the procedure The conditions are met to use a one-proportion z-interval.

I: Interval and point estimate The symbol and value of the point estimate on this problem are as follows: p^=140300=1430=715 \hat{p} = \frac{140}{300} = \frac{14}{30} = \frac{7}{15}

The interval estimate for p p is: p^±z∗p^(1−p^)n \hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Where z∗ z^* for a 98% confidence interval is approximately 2.33.

Calculations: p^=715 \hat{p} = \frac{7}{15} p^(1−p^)n=715×815300=56225300=5667500=0.0008296≈0.0288 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{\frac{7}{15} \times \frac{8}{15}}{300}} = \sqrt{\frac{\frac{56}{225}}{300}} = \sqrt{\frac{56}{67500}} = \sqrt{0.0008296} \approx 0.0288

p^±2.33×0.0288 \hat{p} \pm 2.33 \times 0.0288 715±2.33×0.0288 \frac{7}{15} \pm 2.33 \times 0.0288 715±0.0671 \frac{7}{15} \pm 0.0671

Converting 715\frac{7}{15} to a decimal: 715≈0.467 \frac{7}{15} \approx 0.467

Interval: 0.467−0.0671=0.400 0.467 - 0.0671 = 0.400 0.467+0.0671=0.534 0.467 + 0.0671 = 0.534

The interval estimate for p p is (0.400, 0.534).

C: Conclusion We are 98% confident that the proportion of all Australians today who believe unemployment would increase is between 40.0% and 53.4%.

This problem has been solved

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