Question

Please solve it the easy way.

Several drugs are used to treat diabetes. A sales specialist for a leading pharmaceutical company needs an estimate of the number of new prescriptions that were written during a particular month for his company’s new diabetes drug. The numbers of new prescriptions in a random sample of 25 sales districts are as follows:
210 240 190 275 290 185 223 190 185 192
265 312 284 261 243 168 240 170 187 190
215 240 210 235 290
1. Find a 90% confidence interval for the average number of new prescriptions written for this new drug among all the sales districts. State the assumptions.
2. Calculate the widths for 95% and 98% confidence intervals.

Answer 1

part 1

assumptions:

1.sample values must have been selected through simple random sampling method.

2. The sample values must be independent of each other.

3. data should be drawn from a normally distributed population.

4.sample size must be less than 10% of the population size

For 90 % CI, since population standard deviation is unknown and sample size is less than 30, hence we will use t distribution.

The sample mean (x̄) = 227.6

The sample standard deviation(s) = 41.864

Number of observations (n) = 25

Alpha = 1-0.90=0.1,alpha/2 =0.05 , degree of freedom = df = n-1 = 24

t at 0.1(two tailed) = 1.711

Standard error = SE = s /sqrt (n) = 41.864 / sqrt (25) = 8.3728

Margin of error = t * standard error = 1.711*8.3728 = 14.326

So, lower bound = sample mean - margin of error = 227.6 - 14.326 = 213.274 = 213(round to nearest integer)

And upper bound = sample mean + margin of error = 227.6 + 14.326 = 241.926 = 242(round to nearest integer)

So, 90% CI for the average number of new prescriptions written for this new drug among all the sales districts is (213 , 242)

part 2

interval width = 2* t * standard error

for 95% CI, alpha = 0.05 , critical t value at 0.05(two tailed) = 2.064 for df = 24

so, interval width for 95% CI for mean = 2*2.064*8.3728 = 34.563 = 35(closest integer value)

similarly, for 98% CI, alpha = 0.02 , critical t value at 0.02 (two tailed) = 2.49216 for df = 24

so, interval width for 98% CI for mean = 2*2.49216*8.3728 = 41.733 = 42(closest integer value)