Karen wants to advertise how many chocolate chips are in...

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 56 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 17.6 and a standard deviation of 3.7. What is the 98% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place). ____< μ < ____

Transcript

00:01 Our question says karen wants to advertise how many chocolate chips are in the eye in each big big chip cookie at a bakery.

00:08 She randomly select a sample of 56 cookies and find that the mean number of chocolates per cookie.

00:13 The mean number of chocolate chip per cookie in the sample has a mean of 17 .6 and a standard division of 3 .7.

00:20 So what is the 98 % confidence in favor for the number of chocolate chips per cookie for the big chip cookies? so we are supposed to enter our answers to 1 .1.

00:30 Decimal place so let's go into our worksheet to extract out all the necessary details so we have our end to be cost to we have that to be 58 we have that to be 56 rather we have our sample mean xb to be cost to 17 .6 and we have our sample standard division to be cost to 3 .7 so to construct a 98 % confidence interval for population mean we have the formula that says mule is equals to x bar plus or minus the critical value time x divided by the square root of n now we have all of our details ready aside the critical value so the critical value is dependent on the type of distribution that defines a data set now we have to go back to what we call the central limit theorem which states that when the sample size of a sample distribution is greater than or equals to 30 we can definitely see that the data set is normally distributed at a sample size of 56 this value is greater than 30 so we can definitely say that our data sets is a normally distributed data and our critical value which is cv is going to be a z score so that means that we have to get our critical value based on a zsco our confidence level is 98 % so that means our alpha level is going to be 2 % which is 0 .02 so that being said it's time for us to get the critical value and to get the critical value i would be using a calculator so here is my calculator we have a standard normal we have it to be a two -tail test and this is 0 .02 and that is 2 .3 to 6 3 so our critical value is 2 .3 to 663 so let's substitute our parameters into the formula so we have that meal is across to 17 .6 plus or minus 2 .3260 times 3 .7 divided by square root of 56 so we have meal to be close to 17 .6 plus or minus 2 .3 to 36 times.

02:35 So 3 .7, the variable is square root of 56.

02:40 We have this to be 0 .4944...

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