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.

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.

========================

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.

1) Checking conditions: In order to construct a confidence interval, we need to check certain conditions. One of the assumptions is that the sample should be randomly selected. In this case, it is mentioned that the sample is a random sample of 78 teenagers, so this condition seems to be met. Additionally, the sample size is sufficiently large (n = 78), which allows us to assume that the sampling distribution of the proportion is approximately normal.

2) Confidence interval formula: The formula to calculate a confidence interval for a population proportion (p) is given by:
CI = p-hat ± z * sqrt((p-hat * (1 - p-hat)) / n)

3) Value of critical z-value: The critical z-value is obtained from the standard normal distribution and is based on the desired confidence level. In this case, we want a 98% confidence interval, so the critical z-value can be found using a z-table or calculator. For a 98% confidence level, the critical z-value is approximately 2.33.

4) Confidence interval: To construct the confidence interval, plug in the given values into the formula:
CI = 0.41 ± 2.33 * sqrt((0.41 * (1 - 0.41)) / 78)

5) Interpretation of the 98% confidence interval: The 98% confidence interval is a range of values within which we are 98% confident that the true proportion of teenagers who text while driving lies. In this case, the confidence interval can be computed to be (0.313, 0.507). Therefore, we can say that we are 98% confident that the true percentage of teenagers that text while driving falls between 31.3% and 50.7%.

6) To determine the sample size needed to estimate the percentage of teenagers who text while driving within 5% with 98% confidence, we can rearrange the confidence interval formula:
n = (z^2 * p-hat * (1 - p-hat)) / E^2

where:
n = required sample size
z = critical z-value
p-hat = estimated proportion (0.41 in this case)
E = desired margin of error (5% = 0.05)

Plugging in the values, we can calculate:
n = (2.33^2 * 0.41 * (1 - 0.41)) / (0.05^2)

Calculating this expression, we find that the required sample size is approximately 347 teenagers. Therefore, we would need to sample about 347 teenagers to estimate the percentage that text while driving to within 5% with 98% confidence.