In a survey, 23 people were asked how much they spent on...

In a survey, 23 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $49 and standard deviation of $12. Construct a confidence interval at a 98% confidence level. Give your answers to one decimal place. + You measure 26 textbooks' weights, and find they have a mean weight of 79 ounces. Assume the population standard deviation is 14.8 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places <?<

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00:01 Hello student, given that we measure a 46 textbook weight and find mean weight that is x bar is equal to 70 ounces and population st sigma is equal to 5 .6.

00:20 Now based on this we have to construct a 90 % confidence interval.

00:27 So formula for calculating confidence interval for population mean textbook weight.

00:36 So your confidence interval is equal to x bar plus minus z 1 minus alpha by 2 into sigma upon square root of n.

00:49 Now here we substitute all the values in the given formula.

00:53 So your x bar is equal to 70 plus minus z 0 .95 into sigma is equal to 5 .6 divided by square root of 46.

01:08 We have 70 plus minus z 0 .95 is equal to 1 .6449 into 0 .8257.

01:21 Now solving it will gives the value 70 plus minus 1 .3581.

01:33 Therefore 90 % confidence interval for population mean textbook weight.

01:42 Therefore confidence interval is equal to 68 .64 comma 71 .35.

01:53 Now in the next bit we have given n is equal to 16.

01:59 So in the next bit n is equal to 16 x bar is equal to 41 and s is equal to 8.

02:11 Then here we have to find out what is the margin of error at a 90 % confidence level.

02:19 So here formula for calculating margin of error is equals to t 1 minus alpha by 2 comma degrees of freedom into s divided by root n.

02:35 Now here t 1 minus alpha by 2 is equals to t 0 .95...

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