In Case Study 19.1, we learned that about 56% of American adults actually voted in the presidential election of 1992, whereas about 61% of a random sample claimed that they had voted. The size of the sample was not specified, but suppose it were based on 1600 American adults, a common size for such studies. (a) Into what interval of values should the sample proportion fall 68%, 95%, and almost all of the time? (Round your answers to three decimal places.) 68% to 95% to Almost all to (b) Is the observed value of 61% reasonable, based on your answer to par

a) Confidence interval for 68% confidence level

= (0.548, 0.572)

Confidence interval for 95% confidence level

= (0.536, 0.584)

Confidence interval for 99.99% confidence level = (0.523, 0.598)

b) The sample proportion of 0.61 is unusual as falls outside all of the range of intervals where the sample mean can found for all 3 confidence levels examined.

c) Standardized score for the reported percentage using a sample size of 400 = 2.02

Since, most of the variables in a normal distribution should fall within 2 standard deviations of the mean, a sample mean that corresponds to standard deviation of 2.02 from the population mean makes it seem very plausible that the people that participated in this sample weren't telling the truth. At least, the mathematics and myself, do not believe that they were telling the truth.

Step-by-step explanation:

The mean of this sample distribution is

Mean = μₓ = np = 0.61 × 1600 = 976

But the sample mean according to the population mean should have been

Sample mean = population mean = nP

= 0.56 × 1600 = 896.

To find the interval of values where the sample proportion should fall 68%, 95%, and almost all of the time, we obtain confidence interval for those confidence levels. Because, that's basically what the definition of confidence interval is; an interval where the true value can be obtained to a certain level.of confidence.

We will be doing the calculations in sample proportions,

We will find the confidence interval for confidence level of 68%, 95% and almost all of the time (99.7%).

Basically the empirical rule of 68-95-99.7 for standard deviations 1, 2 and 3 from the mean.

Confidence interval = (Sample mean) ± (Margin of error)

Sample Mean = population mean = 0.56

Margin of Error = (critical value) × (standard deviation of the distribution of sample means)

Standard deviation of the distribution of sample means = √[p(1-p)/n] = √[(0.56×0.44)/1600] = 0.0124

Critical value for 68% confidence interval

= 0.999 (from the z-tables)

Critical value for 95% confidence interval

= 1.960 (also from the z-tables)

Critical values for the 99.7% confidence interval = 3.000 (also from the z-tables)

Confidence interval for 68% confidence level

= 0.56 ± (0.999 × 0.0124)

= 0.56 ± 0.0124

= (0.5476, 0.5724)

Confidence interval for 95% confidence level

= 0.56 ± (1.960 × 0.0124)

= 0.56 ± 0.0243

= (0.5357, 0.5843)

Confidence interval for 99.7% confidence level

= 0.56 ± (3.000 × 0.0124)

= 0.56 ± 0.0372

= (0.5228, 0.5972)

b) Based on the obtained intervals for the range of intervals that can contain the sample mean for 3 different confidence levels, the sample proportion of 0.61 is unusual as it falls outside of all the range of intervals where the sample mean can found for all 3 confidence levels examined.

c) Now suppose that the sample had been of only 400 people. Compute a standardized score to correspond to the reported percentage of 61%. Comment on whether or not you believe that people in the sample could all have been telling the truth, based on your result.

The new standard deviation of the distribution of sample means for a sample size of 400

√[p(1-p)/n] = √[(0.56×0.44)/400] = 0.0248

The standardized score for any is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.61 - 0.56)/0.0248 = 2.02

Standardized score for the reported percentage using a sample size of 400 = 2.02

Since, most of the variables in a normal distribution should fall within 2 standard deviations of the mean, a sample mean that corresponds to standard deviation of 2.02 from the population mean makes it seem very plausible that the people that participated in this sample weren't telling the truth. At least, the mathematics and myself, do not believe that they were telling the truth.

Hope this Helps!!!

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step-by-step explanation: