Question

1.) An online retailer wants to estimate the true proportion of orders made, in a given month, by new customers. In a simple random sample 40 orders, it was found that 12 of them are from new customers Construct a 93% confidence interval for the true proportion. a. Use the calculator to determine the b. State the Conclusion. confidence interval 2.) Suppose boiling temperature of a certain liquid, in degrees Celsius, are normally distributed. A student gathers 42 different readings, with a mean temperature of 101.25 and a standard deviation of 1.23. Determine the 98% confidence interval for the true standard deviation boiling temperatures of this particular liquid a. Draw a density curve and label it with b. Use the formula to construct the CI. the critical values. MacBook Air

Answer 1

93% CI for p is given as: ľ + Z P (1-P) For 932CI. Z.81205 Given, n = 40 p = 12/40 = 0.3 So we have 02- 0.3+ 1. 81205X0.3x0.7 yo = co. 169 0.431 ) . We are 93% confident that the true proportion of orders by new customers in a given month I lies in the above interval 42 From the Central limit the oren. X ~ N( 101.5 123) - i e.N(101.25, 0.190) distribution u=101.25 uto=101.25 +0.19=101.44 utao= 101.63 ut38= 101.82 u-6= 101. 06, u-202100.87 Lu-3 62 100.68 so the density curve is as follows: n Z 100.68 -3 10087 101.06 101 is -2 -1 . 101.44 101.63 t +2 101.82 +3

Here 98% since ob unknown, C.I is given as ! t t X - xs 0.obn-i in = 101.25 +2.422 x 1.23 142 = ( 100.79 Aus We are 98% confident that the temperature of the liquid lies above interval true mean in the