A nutritionist claims that the mean tuna consumption by a person is 3.8 pounds p

Question

A nutritionist claims that the mean tuna consumption by a person is 3.8 pounds per year. A sample of 50 people shows that the mean tuna consumption by a person is 3.3 pounds per year with a standard deviation of 1.15 pounds. At alpha = 0.03, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H_0:mu lessthanorequalto 3.8 H_a: mu > 3.8 B. H_0: mu lessthanorequalto 3.3 H_a: mu > 3.3 C. H_0: mu > 3.3 H_a: mu lessthanorequalto 3.3 D. H_0: mu notequalto 3.3 H_a: mu = 3.3 E. H_0: mu = 3.8 H_a: mu notequalto 3.8 F. H_0: mu > 3.8 H_a: mu lessthanorequalto 3.8 (b) Identify the standardized test statistic. z = - 3.07 (Round to two decimal places as needed.) (c) Find the P-value. (Round to three decimal places as needed.)

Explanation / Answer

here std error of mean =std deviation/(n)1/2 =0.1626

hence test stat z=(X-mean)/std error =(3.3-3.8)/0.1626=-3.07

c) p value =0.002

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