Question

2. A random sample of 25 watermelons from New Seasons were weighed, generating a sample mean of xã = 21 pounds with a standard deviation of s = 3 pounds. a) Construct a 98% confidence interval for the population mean weight of watermelons at New Seasons, b) Interpret: c) New Seasons claims that their watermelons have an average weight of 22 pounds. Does your sample from above provide sufficient evidence that the population mean of watermelons is actually less than 22 pounds? Use a = 0.05

Answer 1

n=25 S2 tooliny ulla sided confidence internal = ă ţ toodd solution Given data, ū=21 s=3 c=98% = 0.98 (Confidence) Q- 1-C = 0.02 na 30 use t- dosti budian dt sny 25.1 = 24 (degree of freedang tadt $ 2.492 98%. Two sided confidence internal for men an I tex, dd = 21+ 2,492 * 3 W 25 - 21+ 1.4952 ( 19,5048, 22.4952) 198% confidence interud (19.sos, 22.495) 5 Interpred & we are 98%. Confident that data will fall ander ander confidence interval. Hoi M=22 Ha: 4 <22 Step x=0.05 ta, df = 1.711 (one sided ) O step I Confidence = l-a=0.95-95%

95% one sided confidence interval 21+ 1,711* 3 Jas =2171.027 = (19.993, 22.097) 95% one sided confidence interval (19.973, 22.027) because confidence internal contain 22 valle so we do nos have suffielent evidence to support the claim that popula- tion mean of watermelon in toy than 22 pounds.