Question

A health administration recommends that individuals consume 1000 mg of calcium daily. After an advertising campaign aimed at male​ teenagers, a dairy association states that male teenagers on average consume more than the recommended daily amount of calcium. To support this​ statement, the association obtained a random sample of 40 male teenagers and found that the mean amount of calcium consumed was 1015 ​mg, with a standard deviation of 100 mg. Is there significant evidence to support the statement of the association at the alphaαequals=0.05 level of​ significance? what does t subscript0 equal?

Answer 1

Solution :

Given that ,

= 1000  

= 1015

= 100

n = 40

The null and alternative hypothesis is ,

H₀ :   = 1000

H_a : > 1000

This is the right tailed test .

Test statistic = z

= ( - ) / / n

= ( 1015 - 1000) / 100 / 40

= 0.949

The test statistic = 0.949

P - value = P(Z > 0.949 ) = 1 - P (Z < 0.949 )

= 1 - 0.8287

= 0.1713

P-value = 0.1713

= 0.05  

0.1713 > 0.05

P-value >

fail to reject the null hypothesis .

There is not sufficient evidence to test the claim .

No