A community health association is interested in estimating the average number of

Question

A community health association is interested in estimating the average number of maternity days women stay in the local hospital. A random sample is taken of 36 women who had babies in the hospital during the past year. The following number of maternity days each woman was in the hospital is rounded to the nearest day. Use these data and a population standard deviation of 1.17 to construct a 97% confidence (yes, | you will have to figure out the z value since this is not a common confidence percentage) interval to estimate the average maternity stay in the hospital for all women who have babies in this hospital. Some fast-food chains offer a lower-priced combination meal in an effort. To attract budget-conscious customers. One chain test-marketed a burger, fries and a drink combination for $1.71. The weekly sales volume for these meals was impressive. Suppose the chain wants to estimate the average amount its customers spent on a meal at their restaurant while this combination offer was in effect. An analyst gathers the following data from sample meal totals Use these data to construct a 90% confidence interval to estimate the population mean value. Assume the amounts spent are normally distributed.

Explanation / Answer

Here mean=3.3056 and sd=1.17 (already given)

Now from z table we get P(-2.171<z<2.171)=0.97

so z=2.171

Hence Margin of error=z*sd/sqrt(n)=2.171*1.17/sqrt(36)=0.4233

So CI=mean+/-E=3.3056+/-0.4233=(2.8823,3.7289)

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