Question

How would you know to choose a. in part a? As well for part b, how??

A pizza shop claims its average home delivery time is 33 minutes. A sample of 35 deliveries had a sample average of 35.9 minutes. Assume the population standard deviation for the shop's deliveries is 6.7 minutes. Complete parts a and b below. a. Is there support for the shop's claim using the criteria that the sample average of 35.9 minutes falls within the symmetrical interval that includes 95% of the sample means if the true population mean is 33 minutes? Select the correct choice below and fill in the answer box within your choice. (Type an integer or decimal rounded to four decimal places as needed.) CA. The probability P(x235.9) = 0.0052. This indicates that the sample average of 35.9 minutes falls outside the 95% symmetrical interval, contradicting the shop's claim. B. The probability PX235.9) - This indicates that the sample average of 35.9 minutes falls within the 95% symmetrical interval, supporting the shop's claim. b. Identify the symmetrical interval that includes 95% of the sample means if the true population mean is 33 minutes. 30.78 sxs 35.22 (Type integers or decimals rounded to two decimal places as needed.)

Answer 1

Solutions Grent 33 41333 6 = 6.7 n=35 X=35.g PCX235-3) = F-PX 359) =1-P(X-4 , 35:9-33 Sun 6.7/35 I-P(Z <2.5606) = -0.9947 P X235.9) - O-0052 The probability that plx>35.g)=0.0052 This indicates that the sample average of 35.9 minutes falls outside the symmetrical interval that include 95% of the sample mean if the toue.

6 Consbuct the symmetrical internal that includes 95% of the sample mean if the towe population mean is 33 minutes c. I = (Mt = (33 + 1:96(67)) V 35 = ( 33 2.2197) =( 33- 2.2197 33722197) = ( 30.78, 35.22) confidence interval is 30-78 < X< 35.22