Respuesta :
Answer:
A. Fail to reject the null hypothesis.
Step-by-step explanation:
We are given with the Population mean weight of newborn infants at a community hospital, [tex]\mu[/tex] = 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 = [tex]\frac{\sum X_i}{n}[/tex] where Xi are the each data value
and n = sample size
= [tex]\frac{6.0+ 6.6+ 6.8+ 7.3+ 8.4+ 8.8+ 9.0 }{7}[/tex] = 7.56
Sample Standard Deviation, s = [tex]\sqrt{\frac{\sum (X_i-Xbar)^{2}}{n-1}}[/tex] = 1.18
Let Null hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 6.6 pounds
Alternate hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 6.6 pounds
Since we don't know about population standard deviation so the test statistics we use here will be :
[tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] follows t distribution with (n-1) degree of freedom [[tex]t_n_-_1[/tex]]
Test Statistics = [tex]\frac{7.56-6.6}{\frac{1.18}{\sqrt{7} } }[/tex] follows [tex]t_6[/tex]
= 2.1525
At 5% level of significance [tex]t_6[/tex] 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.
