Question

In: Statistics and Probability

A Reader’s Digest survey on the drinking habits of Americans estimated the percentage of adults across...

A Reader’s Digest survey on the drinking habits of Americans estimated the percentage of adults across the country who drink beer, wine, or hard liquor at least occasionally. Of the 1516 adults interviewed, 985 said that they drank. Compute the following a. b. c. d. At the e. f. g. h. i. j. Find x, n, and n-x. Find ?̂ and (1−?̂) Find and interpret the 95% confidence interval around your estimate of p. How many samples (n) would you need to have a 99% confidence interval with a margin of error of 2% 2 with no preliminary data on p? 1% significance level, do the majority of adults (more than 50%) drink at least occasionally? What is the null hypothesis? Describe it both verbally and symbolically. What is the alternative hypothesis? Describe it both verbally and symbolically. Is the hypothesis left, right or two-tailed? Compute the z-statistic. Find the p-value or critical value as appropriate. At the 1% significance level, do the data provide sufficient evidence to conclude that the majority of adults (more than 50) drink at least occasionally? Explain your answer.

Solutions

Expert Solution

x = 985

n = 1516

n - x = 1516 - 985 = 531

= 985/1516 = 0.6497

At 95% confidence level, the critical value is z * = 1.96

Thr 95% confidence interval is

+/- z* * sqrt((1 - )/n)

= 0.6497 +/- 1.96 * sqrt(0.6497(1 - 0.6497)/1516)

= 0.6497 +/-  1.96 * sqrt(0.6497(1 - 0.6497)/1516)

= 0.6497 +/- 0.0240

= 0.6257, 0.6737

c) At 99% confidence level, the critical value is z* = 2.575

Margin of error = 0.02

Or, z* * sqrt(p(1 - p)/n) = 0.02

Or, 2.575 * sqrt(0.5 * 0.5/n) = 0.02

Or, n = (2.575 * sqrt(0.5 * 0.5)/0.02)^2

Or, n = 4145

d) H0: p < 0.5

H1: p > 0.5

This is a right tailed test.

The test statistic is z = ( - p)/sqrt(p(1 - p)/n)

= (0.6497 - 0.5)/sqrt(0.5 * 0.5/1516)

= 11.66

P-value = P(Z > 11.66)

= 1 - P(Z < 11.66)

= 1 - 1 = 0

Since the p-value is less than the significance level (0 < 0.01), so we should reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that the majority of adults (more than 50) drink at least occasionally.


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