Question

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What is the decision for a statistical significant change in average weights at birth at the 5% level of significance? rev: 04_14_2015_QC_CS-13524 Reject the null hypothesis and conclude the mean is lower than 6.6 lb. Fail to reject the null hypothesis. Cannot calculate because the population standard deviation is unknown. Reject the null hypothesis and conclude the mean is higher than 6.6 lb.

Answer 1

Explanation:

n = 7

Mean = (9.0 + 7.3 + 6.0 + 8.8 + 6.8 + 8.4 + 6.6)/7 = 7.6

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (9.0 - 7.6)^2 + (7.3 - 7.6)^2 + (6.0 - 7.6)^2 + (8.8 - 7.6)^2 + (6.8 - 7.6)^2 + (8.4 - 7.6)^2 + (6.6 - 7.6)^2 = 8.33

Standard deviation = √(8.33/7

s = 1.1

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 6.6

For the alternative hypothesis,

H1: µ ≠ 6.6

This is a two tailed test.

Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.

Since n = 7

Degrees of freedom, df = n - 1 = 7 - 1 = 6

t = (x - µ)/(s/√n)

Where

x = sample mean = 7.6

µ = population mean = 6.6

s = samples standard deviation = 1.1

t = (7.6 - 6.6)/(1.1/√7) = 2.41

We would determine the p value using the t test calculator. It becomes

p = 0.053

Since alpha, 0.05 < than the p value, 0.053, then we would fail to reject the null hypothesis.