Here are a few hints.
Formula for confidence interval:
CI98 = p + or - (2.33)(√pq/n) ...where + or - 2.33 represents 98% confidence using a z-table. (q = 1 - p; n = sample size)
p = 32/78
q = 1 - p
n = 78
Change all fractions to decimals. Plug values into formula and calculate the interval.
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Formula to find sample size:
n = [(z-value)^2 * p * q]/E^2
... where n = sample size, z-value is 2.33; p = 32/78 (when no value is stated in the problem), q = 1 - p, ^2 means squared, * means to multiply, and E = .05.
Change all fractions to decimals. Plug values into the formula and calculate the sample size.
I hope this will help get you started.
In a random sample of 78 teenagers, 32 admit to texting while driving. Construct a 98% confidence interval for the percentage of teenager that text while driving. Start by showing that conditions for constructing a confidence interval are met.
1) Check conditions
2) confidence interval formula
3) value of critical z-value used in confidence interval formula
4) confidence interval
5) interpret the 98% confidence interval
6) How many teenagers would we need to sample to estimate the percentage that text while driving to within 5% with 98% confidence? Use the value of p-hat from above.
1 answer
1 answer