Answer:
The correct option is A
Step-by-step explanation:
From the question we are told that
The population is [tex]\mu = 6.6[/tex]
The level of significance is [tex]\alpha = 5\% = 0.05[/tex]
The sample data is 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds
The Null hypothesis is [tex]H_o : \mu = 6.6[/tex]
The Alternative hypothesis is [tex]H_a : \mu > 6.6[/tex]
The critical value of the level of significance obtained from the normal distribution table is
[tex]Z_{\alpha } = Z_{0.05 } = 1.645[/tex]
Generally the sample mean is mathematically evaluated as
[tex]\=x = \frac{\sum x_i }{n}[/tex]
substituting values
[tex]\=x = \frac{9.0 + 7.3 + 6.0+ 8.8+ 6.8+ 8.4+6.6 }{7}[/tex]
[tex]\=x = 7.5571[/tex]
The standard deviation is mathematically evaluated as
[tex]\sigma = \sqrt{\frac{\sum [ x - \= x ]}{n} }[/tex]
substituting values
[tex]\sigma = \sqrt{\frac{ [ 9.0-7.5571]^2 + [7.3 -7.5571]^2 + [6.0-7.5571]^2 + [8.8- 7.5571]^2 + [6.8- 7.5571]^2 + [8.4 - 7.5571]^2+ [6.6- 7.5571]^2 }{7} }[/tex][tex]\sigma = 1.1774[/tex]
Generally the test statistic is mathematically evaluated as
[tex]t = \frac{\= x - \mu } { \frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
[tex]t = \frac{7.5571 - 6.6 } { \frac{1.1774 }{\sqrt{7} } }[/tex]
[tex]t = 1.4274[/tex]
Looking at the value of t and [tex]Z_{\alpha }[/tex] we see that [tex]t < Z_{\alpha }[/tex] hence we fail to reject the null hypothesis
What this implies is that there is no sufficient evidence to state that the sample data show as significant increase in the average birth rate
The conclusion is that the mean is [tex]\mu = 6.6 \ lb[/tex]
