Question

A nutritionist claims that the mean tuna consumption by a person is 3.4 pounds per year. A sample of 80 people shows that the mean tuna consumption by a person is 3.3 pounds per year. Assume the population standard deviation is1.14 pounds. At alphaαequals 0.07​,can you reject the​ claim?

1) identify the null hypothesis and alternative hypothesis __________

2) Identify the standardized test statistic.(round to two decimal places)

z= _____

3) find the P value

p=_____

4) decide whether to reject or fail to reject the null hypothesis. Is their sufficient evidence?

Answer 1

Solution:

a)

The null and alternative hypothesis are

H₀ : 3.4

H_a: 3.4

b)

The test statistic t is

z = ( - )/[/n]

= [3.3 - 3.4]/[1.14/80]

= -0.78

z = -0.78

c)

sign in Ha indicates that the test is TWO TAILED.

z = -0.78

So , using calculator ,

p-value = P(z > +0.78) + P(z < -0.78)

= 0.2177+ 0.2177

p value = 0.4354

4)

fail to reject the null hypothesis.

(because p value is greater than given alpha level of 0.07)

There is no sufficient evidence to reject the claim that the mean tuna consumption by a person is 3.4 pounds per year.