nutritionist claims that the mean tuna consumption by a person is 3.6 pounds per year. A sample of 70 people shows that the mean tuna consumption by a person is 3.4 pounds per year. Assume the population standard deviation is 1.19 pounds. At alphaαequals=0.05​, can you reject the​ claim? ​(a) Identify the null hypothesis and alternative hypothesis.     A. Upper H 0H0​: muμless than or equals≤3.6 Upper H Subscript aHa​: muμgreater than>3.6   B. Upper H 0H0​: muμgreater than>3.6 Upper H Subscript aHa​: muμless than or equals≤3.6   C. Upper H 0H0​: muμnot equals≠3.4 Upper H Subscript aHa​: muμequals=3.4   D. Upper H 0H0​: muμgreater than>3.4 Upper H Subscript aHa​: muμless than or equals≤3.4   E. Upper H 0H0​: muμequals=3.6 Upper H Subscript aHa​: muμnot equals≠3.6   F. Upper H 0H0​: muμless than or equals≤3.4 Upper H Subscript aHa​: muμgreater than>3.4 ​(b) Identify the standardized test statistic.   zequals=nothing ​(Round to two decimal places as​ needed.) ​(c) Find the​ P-value.   nothing ​(Round to three decimal places as​ needed.) ​(d) Decide whether to reject or fail to reject the null hypothesis.     A. Fail to reject Upper H 0H0. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.6 pounds.   B. Reject Upper H 0H0. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.6 pounds.   C. RejectReject Upper H 0H0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.6 pounds.   D. Fail to rejectFail to reject Upper H 0H0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.6 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.