MATH 25 CO3 (Fall 2017) Test: EXAM 3 (required) Time Remaining: 01.2359 Submit T

Question

MATH 25 CO3 (Fall 2017) Test: EXAM 3 (required) Time Remaining: 01.2359 Submit Test This Question: 5 pls | 11 of 1409 complete) This Test: 47 pls po A nutritionist claims that the mean tuna consumption by a person is 3 3 pounds per year A sample of 70 people shows that the mean tuna consumption by a person is 32 pounds per year Assume the population standard deviation is 1 02 pounds At -005, can you re-d te dain? (a) Identify the null hypothesis and alternative hypothesis. H H32 D. Ho 3.2 Ha =32 Ha 3.3 OF. Ho 32 Ha #3.2 Ha > 3.3 (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 Click to select your answers) Type here to search

Explanation / Answer

Using minitab we do this Z test '

Stat >>>> basic statistics >>>> 1-sample Z

One-Sample Z

Test of = 3.3 vs 3.3
The assumed standard deviation = 1.02

N   Mean SE Mean      95% CI          Z      P
70 3.200    0.122 (2.961, 3.439) -0.82 0.412

a) so our cliam staement is mean tuna consumption by a pearson is 3.3 pound .

So this cliam goes to null hypothesis with equal (=) sign and exaclty appositive null is altenative hypothesis .

So Ho : = 3.3

vs

H1: 3.3

b) Test satistics

Z= -0.82

c) P-value = 0.412.

D) the correct Answer is B)

Becuase we P-value is 0.412 which is grater than 0.05 so fail to Reject Ho that means we Accept Ho

that's why Answer is B)

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