A nutritionist claims that the mean tuna consumption by a person is 3.9 pounds per year. A sample of 80 people shows that the mean tuna consumption by a person is 3.6 pounds per year. Assume the population standard deviation is 1.05 pounds. At a 0.08, can you reject the claim? ..... (a) Identify the null hypothesis and alternative hypothesis. O A. Ho:H>3.6 O B. Ho: Hs3.9 O C. Ho: u> 3.9 Ha:us3.6 Haiu>3.9 Haius3.9 OD. Ho:H=3.9 Ha: u#3.9 O E. Ho: HS3.6 Ha:u> 3.6 OF. Ho: u#3.6 Hai=3.6 (b) Identify the standardized test statistic. O (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. Fail to reject Ho. There is not sufficient evidence to reject the claim that mean O B. Reject Ho. There is not sufficient evidence to reject the claim that mean tuna tuna consumption is equal to 3.9 pounds. consumption is equal to 3.9 pounds. O C. Reject Ho. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.9 pounds. O D. Fail to reject Ho. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.9 pounds.

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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.