Module #6 - Sampling & Confidence Interval Estimation

This week's topic covered confidence interval estimations.

### Question # 1 : A publishing company has just published a new textbook. Before the company decides the price at which to sell this textbook, it wants to know the average price of all such textbooks in the market. The research department at the company took a sample of 25 comparable textbooks and collected information on their prices. This information produced a mean price of \$145 for this sample. It is known that the standard deviation of the prices of all such textbooks is \$35 and the population of such prices is normal.

1. #### What is the point of estimate of the mean price all such textbooks?

```  The point estimate is equivalent to the sample mean.
x̄ = 145
```
1. #### Construct a 90% confidence interval for the mean price of all such textbooks

```> z <- qnorm(0.9)
> z
 1.281552
> marginerror <- z * (35 / sqrt(25))
> marginerror
 8.970861
> upperbound <- 145 + marginerror
> lowerbound <- 145 - marginerror
> cat("The 90% confidence interval is:
(", lowerbound, ", ", upperbound, ")")
The 90% confidence interval is: ( 136.0291 ,  153.9709 )
```

### Question # 2 : According to Mobes Services Inc. an individual checking his/her account at major U.S banks via cellphones cost the banks between \$350 and \$450. A recent random sample of 600 such checking accounts produced a mean annual cost of \$500 to major U.S banks. Assume that the standard deviation of annual costs to major US banks of all such checking account is \$40. Make a 99% confidence interval for the current mean annual cost to major banks all such checking account.

```> z <- qnorm(0.99)
> z
 2.326348
> marginerror <- z * (40 / sqrt(600))
> marginerror
 3.79891
> upperbound <- 500 + marginerror
> lowerbound <- 500 - marginerror
> cat("The 99% confidence interval is:
(", lowerbound, ", ", upperbound, ")")
The 99% confidence interval is: ( 496.2011 ,  503.7989 )
```