5. use z* from the normal distribution using the large sample method. 6. A group

Question

5.

use z* from the normal distribution using the large sample method.

6.

A group of students in a seminar are interested in investigating what proportion of freshman students had visited the campus prior to enrolling. They interview a group of n = 25 freshman and find that 14 had made a visit to the campus. Since the sample size is small the confidence interval for p should: use t* with degrees of freedom df = 24. use z* with the plus four method. Use all of the above.

use z* from the normal distribution using the large sample method.

6.

A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. They randomly selected 850 residents in the community and contacted them by telephone. Of the 850 residents surveyed, 410 supported the property tax levy. Let p represent the proportion of residents in the community that support the property tax levy. A 90% confidence interval for p is: 0.4463 to 0.5185. 0.4489 to 0.5159. 0.4487 to 0.5161. 0.4542 to 0.5105

Explanation / Answer

5) Here first option is correct. As sample size is small use t* with degrees of freedom df = 24. Here degree of freedom = sample size - 1 = 25 - 1 = 24

6) Here the sample size = n = 850

Residents who supported the property tax = 410

So ratio = p = 410/850 = 0.4823529

To calculate confidence interval, we need to calculate margin of error.

ME = critical value * standard error

SE = sqrt [ p(1 - p) / n ]

= sqrt[ 0.4823529 * (1 - 0.4823529) / 850 ] = sqrt[ 0.0002937513]

= 0.01713917

Confidence level = 90%

So alpha = 1 - 0.90 = 0.10

Critical probability = 1 - 0.10/2 = 1 - 0.05 = 0.95

From the z table, we can find the critical value corresponding to probability of 0.95 = 1.645

So ME = critical value * standard error = 1.645 * 0.01713917= 0.02819393

The range of the confidence interval is defined by the sample statistic + margin of error.

Here sample statistic = 0.4823529

So 90% confidence interval is 0.4823529  + 0.02819393

= 0.454159 to 0.5105468

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