Find the critical value z that α 2 corresponds to the confidence level 87%. Z = (Round to two decimal α 2 places as needed.) K Find the critical value za/2 that corresponds to the confidence level 87%. Za/2= (Round to two decimal places as needed.)

Find the 95% z-interval or t-interval for the indicated parameter. (a) u x= 152, s = 31, n= 35 (b) p p= 0.4, n =71 (a) The 95% confidence interval for µ is | to (Round to two decimal places as needed.)

Find the lower bound of the confidence interval, rounding the result to four decimal places. Р(1 — р) p - E = P – Z«j2\ %3D 0.57(1 – 0.57) = 0.57 1.960 1,007 Find the upper bound of the confidence interval, rounding the result to four decimal places. P(1 — р) p + E : p + z 0.57(1 – 0.57) = 0.57 + 1.960 1,007 Therefore, a 95% confidence interval for the population proportion of adults who lack confidence they will be able to afford health insurance in the future is from a lower bound of X to an upper bound of

The owner of a currency exchange bureau on an airport wishes to estimate the mean value of cash necessary for the exchange of PLN and EUR. Experience tells the owner that the demand for euro is distributed normally, with a standard deviation equal to 4 (thousand PLN). Having observed the demand for 10 days, the owner obtained a mean of 24.4 ( thousand PLN). Calculate the realization of the confidence interval for the expected demand with a confidence level of 0.90.

Zip Code, Age and Number of Dogs by Person in Petland Person Zip Code Number Age of Dogs Ann 402 20 0 Bill 805 35 55 2 Carmelo 805 40 5 Dan 402 50 1 Eric 402 20 20 Ferdinand 402 35 55 3 3 Gert 402 20 2 Homer 402 20 0 In-choi 805 100 3 Jimmy 805 60 1

Given that X̄=20, σ=9 and n=81, construct the following confidence intervals: the 90% confidence interval for μ is:  a (18.355, 21.645)   b (18.040, 21.960) c (17.840, 36.450) d (18.750, 23.450)

Consider the estimated equation from Example 4.3, which can be used to study the effects of skipping class on college GPA: colGPA = 1.39.412 hsGPA+.015 ACT- .083 skipped (.33) (.094) (.011) (.026) n = 141, R² = .234. (ii) (i) Using the standard normal approximation, find the 95% confidence interval for BhsGPA- Can you reject the hypothesis Ho: ßhsGPA .4 against the two-sided alternative at the 5% level? Can you reject the hypothesis Ho: PhsGPA = 1 against the two-sided alternative at the 5% level? - (iii)

Tom wants to estimate the percentage of students who live more than 3 miles from the school.He wants to create a 95% confidence interval which has a margin of error of at most 3% .How many students must be polled to create the confidence interval?

Consider the estimated equation from Example 4.3, which can be used to study the effects of skipping class on college GPA: 1.39 + .412 hsGPA + .015 ACT – 083 skipped (.33) (.094) n = 141, R = .234. colGPA (.026) (.011) Using the standard normal approximation, find the 95% confidence interval for BraGPA: |(ii) Can you reject the hypothesis H,: (iii) Can you reject the hypothesis Hg: BisGPA = 1 against the two-sided alternative at the 5% level? (i) 4 against the two-sided alternative at the 5% level?

A manufacturer of cell phone batteries wants to estimate the useful life of it'sthe battery (in thousand of hours.) The estimate is the be within 0.10 (100) hours.Assume a 95 percent level of confidence and that the standard deviation of theuseful life of the battery is 0.90 (900 hours). Determine the required sample size.