Question

The average birthweight for babies in the United States is 3,325 grams with σ = 525. A researcher wishes to test whether or not a sample of 250 women in a rural community have a significantly different average birthweight than the national average at the α = 0.05 level. The mean birthweight for the sample of women in the rural community is 3400 grams.

a) what is the degree of freedom in this study?

b)what is the test statistic?

c) is the finding statistically significant?

Answer 1

Answer:

The test statistic is z = $\frac{x-u}{σ/√n}$

the z- value is 2.25 > 1.96 at 5% level of significance

we will not accepted null hypothesis

the mean (x⁻) and μ do not differ significantly

Step-by-step explanation:

Step 1:-

Given n = 250 and

population mean(μ ) = 3325 and

standard deviation of population (σ )= 525

mean of the sample ( x ) = 3400

i) Null hypothesis (H0) : (μ ) = 3325

ii) alternative hypothesis(H1) :(μ )≠ 3325

iii) The level of significance ∝ =0.05 = 1.96

Step 2 :-

The test statistic z = $\frac{x-u}{σ/√n}$

now substitute all these values , we get

   $z = \frac{3400-3325}{525/√250}$  

on simplification

  z = 2.25

The tabulated value z = 1.96 at 5% level of significance (from z- table)

The calculated value  z = 2.25 > tabulated value 1.96 at 0.05 level of significance

There fore the null hypothesis is rejected at 5% level of significance.

Therefore is not statistically significant

conclusion:-

The null hypothesis is rejected at 5% level of significance.

Therefore that is not statistically significant