Confidence interval using proportions:
CI90 = p + or - (1.645)(√pq/n)
...where + or - 1.645 represents 90% confidence using a z-table.
Note: p = 110/250 (convert to a decimal); q = 1 - p; and n = 250 (sample size).
I'll let you take it from here.
35%of adult Americans are regular voters. A random sample of 250 adults in a medium-size college town were surveyed, and it was found that 110 were regular voters. Estimate the true proportion of regular voters with 90% confidence and comment on your results...
Can you show you work please???
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