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

A. Fail to reject the null hypothesis.

Explanation:

We are given with the Population mean weight of newborn infants at a community hospital,
`\mu` = 6.6 pounds

Now the sample data of seven infants with their weights at birth is given as:

6.0, 6.6, 6.8, 7.3, 8.4, 8.8, 9.0

Sample Mean, Xbar =
`(\sum X_i)/(n)` where Xi are the each data value

and n = sample size

=
`(6.0+ 6.6+ 6.8+ 7.3+ 8.4+ 8.8+ 9.0 )/(7)` = 7.56

Sample Standard Deviation, s =
`\sqrt{(\sum (X_i-Xbar)^(2))/(n-1)}` = 1.18

Let Null hypothesis,
`H_0` :
`\mu` = 6.6 pounds

Alternate hypothesis,
`H_1` :
`\mu` > 6.6 pounds

Since we don't know about population standard deviation so the test statistics we use here will be :

`(Xbar-\mu)/((s)/(√(n) ) )` follows t distribution with (n-1) degree of freedom [
`t_n_-_1`]

Test Statistics =
`(7.56-6.6)/((1.18)/(√(7) ) )` follows
`t_6`

= 2.1525

At 5% level of significance
`t_6` has a value of 1.943 but our test statistics is higher than this so we have sufficient evidence to accept null hypothesis and conclude that mean is 6.6 pounds.

Therefore, option A is correct that we Fail to reject the null hypothesis.