Question

In: Statistics and Probability

Birth weights. Suppose an investigator takes a random sample of n = 50 birth weights from...

Birth weights. Suppose an investigator takes a random sample of n = 50 birth weights from
several teaching hospitals located in an inner-city neighborhood. In her random sample, the sample
mean x is 3,150 grams and the standard deviation is 250 grams.
(a) Calculate a 95% confidence interval for the population mean birth weight in these hospitals.
(b) The typical weight of a baby at birth for the US population is 3,250 grams. The investigator suspects
that the birth weights of babies in these teaching hospitals is different than 3,250 grams,
but she is not sure if it is smaller (from malnutrition) or larger (because of obesity prevalence in
mothers giving birth at these hospitals). Carry out the hypothesis test that she would conduct.

Solutions

Expert Solution

Solution:-

a) 95% confidence interval for the population mean birth weight in these hospitals is C.I = (3078.9358,3221.0642).

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 3250
Alternative hypothesis: u 3250

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 35.35534

DF = n - 1

D.F = 49
t = (x - u) / SE

t = - 2.83

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesised population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 49 degrees of freedom is less than -2.83 or greater than 2.83.

P-value = P(t < - 2.83) + P(t > 2.83)

Use the calculator to determine the p-values.

P-value = 0.0034 + 0.0034

Thus, the P-value = 0.0068

Interpret results. Since the P-value (0.0068) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the birth weights of babies in these teaching hospitals is different than 3,250 grams.


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