The mean weight of newborn infants at a community hospital 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 significant increase in the average birthrate at a 5% level of significance?
A)Fail to reject the null hypothesis and conclude the mean is 6.6 lb.
B)Reject the null hypothesis and conclude the mean is lower than 6.6 lb.
C)Reject the null hypothesis and conclude the mean is greater than 6.6 lb.
D)Cannot calculate because population standard deviation is unknown.

Question

Quizplus · Accepted Answer

We first set up the null and alternative hypotheses:
Null hypothesis: The average birthweight is 6.6 lb.
Alternative hypothesis: The average birthweight is greater than 6.6 lb.
We then calculate the sample mean, which is:
(9.0+7.3+6.0+8.8+6.8+8.4+6.6)/7 = 7.6 lb
Next, we calculate the t-statistic:
t = (7.6-6.6)/(s/√n)
where s is the sample standard deviation and n is the sample size. Since the population standard deviation is unknown, we estimate it using the sample standard deviation:
s = √{(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)²/6} = 1.145 lb
Plugging this into the t-statistic formula, we get:
t = (7.6-6.6)/(1.145/√7) = 2.19
Using a t-distribution table with 6 degrees of freedom and a 5% level of significance, we find that the critical value is 1.943. Since the calculated t-value (2.19) is greater than the critical value (1.943), we reject the null hypothesis and conclude that the mean birthweight is greater than 6.6 lb. Therefore, the answer is (C).

The Mean Weight of Newborn Infants at a Community Hospital