The owner of a coffee shop has found that the average amount...

The owner of a coffee shop has found that the average amount spent by 25 customers at the shop is $6.80. Based on a large number of similar studies, it is known that the variance for the amount spent by all customers is $1.96. What is the 90% confidence interval for the population mean amount spent by customers? 6.15 < µ < 7.44 (a) 6.34 < µ < 7.26 6.67 < µ < 6.93 6.71 < µ < 6.89

Transcript

00:01 Hi, i'm david and i'm here to help you and change your question.

00:03 Now let me bring up your question here.

00:05 In the question here, we are going to discuss about the confidence interval for the mean.

00:11 Let me remind you that the formula for the confidence interval for the mew, it will be the x bar plus and minus z and 4 over 2, time when the sigma, the value square root n.

00:25 Now in this question we're given the sample size n equal to the 25, the x -par equal to the 6 .80 and the standard division i believe this will be the 1 .96 and from here the question asks me to find the 90 % confidence interval.

00:44 Look at the formula we have given the x -par the 6 -5 the n the only quantity we need to find will be the z and far over 2 and the 90 % confidence interval and 5 equal to the 10 % deviant by 2 equal to the 5 % which is equal to the subon 05.

01:01 If we look up the z table, this value equals 1 .645.

01:07 Now with this information, we are ready to write down the 90 % confidence interval for the mu.

01:15 X bar will be the 6 .a plus and minus the critical value, tamund the sigma overscor root the n.

01:25 Now if we compute the quantity on the right, we will have 1 .645 times 1 .96 divided by 5 equal to the 0 .644.

01:38 So it will use the minus 6 .a minus the answer equal to the 6 .15516.

01:47 It will use the plus.

01:48 We have the 6 .a plus 0 .644484.

01:53 Equal to the 7 .44484...