Please answer B. You are interested in finding a 98% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 14 randomly selected college students. **Round answers to 3 decimal places where possible.**Data:\[ 8, \, 2, \, 2, \, 4, \, 3, \, 7, \, 6, \, 8, \, 12, \, 9, \, 1, \, 4, \, 3, \, 5 \]---**a.** To compute the confidence interval use a \[ t \] distribution.---**b.** With 98% confidence the population mean number of days of class that college students miss is between \_\_\_\_ and \_\_\_\_ days.---**c.** If many groups of 14 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About \[ 98 \] percent of these confidence intervals will contain the true population mean number of missed class days and about \[ 2 \] percent will not contain the true population mean number of missed class days.

From the provided information, Sample size (n) = 14 The mean and standard deviation of this sample data can be obtained as:Confidence level = 98% Since the population standard deviation is unknown therefore, the t distribution will be used. Level of significance (α) = 1 – 0.98 = 0.02 The degree of freedom = n – 1 = 14 – 1 = 13 The critical value at 13 degree of freedom with 0.02 level of significance from the t value table is 2.650. The required 98% confidence interval can be obtained as: Thus, the required 98% confidence interval is 3.039 to 7.533. With 98% confidence the population mean number of days of class that college students miss is between 3.039 and 7.533.