You are interested in determining the effects of a recent coupon promotion on customer purchases at a local grocery store. You sample of 50 customer receipts one day after the coupon is distributed and find the customers' mean spending to be $97.50 with a standard deviation of $4.50. A 90% confidence interval for the population mean would require the use of which critical value: a. t = 1.6766 b. t = 1.2991 c. z = 1.2800 d. z = 1.6450

(a) t = 1.6766

Step-by-step explanation:

n = 50

degree of freedom = n - 1 = 50 - 1 = 49

confidence level = 90%

The critical value corresponding to 90% confidence level and 49 degrees of freedom is obtained by interpolating between 48 and 50 degrees of freedom (df)

Let the critical value at 49 df be y

dof. critical value

48. 1.677

49. y

50. 1.676

(49 - 48)/(50 - 48) = (y - 1.677)/(1.676 - 1.677)

0.5 × -1×10^-3 = y - 1.677

y - 1.677 = -5×10^-4

y = -5×10^-4 + 1.677 = 1.6765

The closest option is (a)

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation:

cheater

step-by-step explanation: