Question

Hypothesis Testing9.Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases, so it is important that the information on packages is accurate. A random sample of n = 12 frozen dinners of a certain type was selected and the calorie content of each one was determined. The stated mean caloric content is 240. Conduct a hypothesis test to determine if the mean caloric content is different than 240. Use a 1% level of significance.Calorie Content255, 244, 239, 242, 265, 245, 248, 225, 226, 251, 233, 259. a)Using technology (calculatoror Excel), find the sample mean and sample standard deviation.  (Round two decimal places).b)Complete the following areas.  Label the critical value and test statistic on the curve and shade the rejection region(s)

.H0: __________________

H1: __________________

Type of Test: ________________

Test Statistic: _________________(2 decimal places)

Critical Value(s): ________________(2 decimal places)

P-Value: __________________(4 decimal places)

Reject or Fail to Reject: ______________________

Conclusion (express in a sentence in the context of the problem)_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Answer 1

6 S.NO (x) (X-X BAR)^2 7 1 255 113.7777778 00 2 244 0.111111111 9 3 239 28.44444444 242 5.444444444 10 4 11 5 12 6 265 427.1111111 245 0.444444444 248 13.44444444 13 7 14 8 15 9 16 10 17 11 18 12 19 SUM 20 MEAN 21 ST. DEV 225 373.7777778 226 336.1111111 251 44.44444444 233 128.4444444 259 215.1111111 2932 1686.666667 244.333 12.3828 STDEV(B7:B18)

Given. no 12 d=1% = 0.01 ll=240 244.333 as 244.33 a) Sample mean standard deviation 8 - 12.3829 SF 12.38 b) Hoo l = 240 1 & 240 Type of test : one sample Hroll t test Test Statistiot 244.333 - 240 all lun 1.212 12.31 28 8/812 t = 1.21 Critical value to df-n-1= 12-1=1 tak, de - to.005,1 ++3.106 (from + fabb) te=3.11 P[t, >121] using face = FDIST.97 (923), 11) 0.9998 P-vake - p = 0.agra P=0.2516 7-).21 < te=3:11 fail to Reject to P = 0.2516 >=0.01 fail to rejec Ho There insufficient evidence to support the stated mean caloric content is different the Claim thom 240