In 1990, the mean height of women 20 years of age or older was 63.7 inches based on data obtained from the cdc. suppose that a random sample of 45 women who are 20 years of age or older in 2015 results in a mean height of 63.9 inches with a standard deviation of 0.5 inch. does your sample provide sufficient evidence that women today are taller than in 1990? perform the appropriate test at the 0.05 level of significance

We reject H₀ , we have enough evidence, women today are taller than in 1990

Step-by-step explanation:

The problem is one tail test. We assume height of women follows normal distribution with σ = 0,5 in.

Then    null hypothesis :      H₀    ⇒   μ₀  =  63.7

    Alternative hipothesis:    Hₐ   ⇒    μ   >  63.7

Significance level   0,05       ⇒  z(c)  = 1.645

Statistic z :

z (c)  = ( μ - μ₀ ) / (0,5/√45)    ⇒  z(c)  = (63.9 - 63.7 ) / (0,5 / √45)

z(c) =( 0,2*  √45 ) / 0,5

z(c) =  2,68

Evaluation :  z(s)  > z(c)      2.68 > 1.645

z(c) is in the rejection area  we reject H₀  

i believe your answer is 99.7% of 7-year-old children are between 46.6 inches and 49.5 inches tall.

answer: 270

step-by-step explanation:

you have to times 200 by 1.35 to get your answer