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. When evaluated using the 5% level of significance, which of the following would be the best decision/interpretation of the null hypothesis?

A. Fail to reject the null hypothesis.

B. Reject the null hypothesis and conclude the mean is higher than 6.6 lbs.

C. Reject the null hypothesis and conclude the mean is lower than 6.6 lbs.

D. Cannot calculate because the population standard deviation is unknown.

Answer #1

Solution:

Here, we have to use one sample t test for population mean.

H_{0}: µ = 6.6 versus H_{a}: µ ≠ 6.6

This is a two tailed test.

We are given

Level of significance = α = 0.05

Test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

Xbar = 7.557142857

S = 1.177365906

n = 7

df = n – 1 = 6

t = (7.557142857 – 6.6)/[ 1.177365906/sqrt(7)]

t = 2.1509

P-value = 0.0750

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the weight of newborn infants at a community hospital is that the mean is 6.6 pounds.

**A. Fail to reject the null hypothesis.**

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. If
α = 0.05, what is the critical t-value?
Multiple Choice
−2.365
±1.96
±2.365
±2.447

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