Statistics Question

Question

Assume the average weight of a full-term newborn infant (i.e. 39-40 weeks gestation) in the US is 3400g. Suppose we measure the birth weight of 1000 (full-term) infants born to alcoholic mothers. We find that the sample mean birth weight is 3200g, and the sample standard deviation is 500g.

a. Calculate a 95% confidence interval for the population mean birth weight of infants born to alcoholic mothers.

1.96*500sqrt(1000)= 30, 3200+/- 98= (3102,3298) <---- Is this right?

b. Suppose we want to test the hypothesis

H0: μ = 3400g

H1: μ ≠ 3400g

where μ is the population mean birth weight for offspring of alcoholic mothers. Using the confidence interval you calculated in (a), would you reject or fail to reject H0 at α = 0.05? (Recall the close relationship between confidence intervals and hypothesis testing. Suppose you want to test a hypothesis H0: μ = μo versus H1: μ ≠ μo using α = .05. You can test this hypothesis using a confidence interval with this rule: If μo is OUTSIDE the 95% CI then REJECT the null hypothesis; if μo is INSIDE the 95% CI then DO NOT REJECT the null hypothesis.)

The null would be rejected, correct?

c. Compute the following test statistic for the two-sided hypothesis test in part (b):

t = x-μ_o/s/sqrtn

 

(X has a line over it)

 

where μo is the assumed value of μ when H0 is true and for large n, t ~ N(0,1).

Is the p-value based on this test statistic less than 0.05?

I'm not sure how to go about doing this one...

Answer 1

Hello,

 

The standard error = SD/sqrt(n). In your case, this is 500/sqrt(1000) = 15.81. Then we compute the margin of error by multiplying by 1.96 (this value is because you want the 95% confidence interval; the values can be found in the normal distribution table - technically you would want a t-table, but since the n is so large, it won't make much of a difference). This gives us: 1.96*15.81 = 30.99. To get the confidence interval, you add and subtract this value from the sample mean:

3200 + 30.99 = 3230.99

3200 - 30.99 = 3169.01

 

So the confidence interval is 3169.01 to 3230.99. You can think of this as meaning that the study found that the mean weight was 3200g but, because of sampling error, it is reasonable to expect that the true (population) birth weight is anywhere between 3169.01 and 3230.99. Notice how this interval does not include 3400. This means that it is not reasonable to expect that the true birth weight is 3400g (at least, not at the 95% level of confidence). In this case, the null would be rejected.

 

If we plug your numbers into the formula for the one-sample t-test we get (3200 - 3400)/(500/sqrt(1000)). Simplifying that formula, we get: -200/(500/31.62) = 200/(15.81) = -12.65. so t = -12.65. Lastly, we compare this value against the t-critical value for 999 degrees of freedom (for the one-sample t-test, the degrees of freedom is n - 1 or, in your case, 1000 - 1 = 999). The t-critical value can be found in the t-table. Your value of -12.65 is larger than the critical value so we say that 3200 is significantly different from 3400 (you have rejected the null hypothesis).