Question

A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.2 pounds per year. Assume the population standard deviation is 1.03 pounds. At c =0.03, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. O A. Ho: 53.5 Ha: > 3.5 OD. Hou#3.2 Hau = 3.2 O B. Ho: > 3.5 Ha $3.5 O E. Ho: 53.2 Ha > 3.2 0 C. Ho: H= 3.5 Hau#3.5 OF. Hou>3.2 Ha: 53.2 (b) Identify the standardized test statistic. Z= (Round to two decimal places as needed.) (c) Find the P-value. (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. O A. Reject Ho. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5 pounds. OC. Fail to reject H. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5 pounds. OB. Reject Ho. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5 pounds. O D. Fail to reject Ho. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5 pounds.

Answer 1

Solution Given data 1.03 L=0.03 n =90 X=3.2 (a) Now to state the null and alternative hypothesis: Null hypothesis Ho: M = 3.5 Alternative hypothesis Hai M + 3.5 This is "two- taited" test Therefore the "option-ch is the correct answer (6) Now to find the standardized test statistic We know that The formula of test Statistic, zis given by T-M sun 3.2-3.5 Z 1.031190 -0.3 D.1086 Z=-2.76 Test statistic 1 Z=-276

2 R-value (c) Now to find the P-value 2 xP(Z >2.16) 2x[1-P(Z<2.76) ex [1-0.9971] P-value = 2* 0.0029 0.0058 P-value = 0.006 (d) Decision P-value 0.006 < 0.03 Since the P-value is less than significance level. So we reject the hall hypothesis (4) Conclusion : Reject Ho. There is sufficient evidence to reject the to tuna consumption is equal mean 3.5 pounds. Therefore the "option -A" is the correct answer. claim that