Answer:
1. Critical value t=±2.447
2. The null hypothesis is failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the birth weight significantly differs from 6.6 lbs.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the birth weight significantly differs from 6.6 lbs.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=6.6\\\\H_a:\mu\neq 6.6[/tex]
The significance level is 0.05.
The sample has a size n=7.
The sample mean is M=7.56.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.18.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1.18}{\sqrt{7}}=0.446[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.56-6.6}{0.446}=\dfrac{0.96}{0.446}=2.152[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=7-1=6[/tex]
For a two-tailed test with 5% level of significance and 6 degrees of freedom, the critical value for t is ±2.447.
As the test statistic t=2.152 is under 2.447 and over -2.447, it falls in the acceptance region, so the effect is not significant. The null hypothesis is failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the birth weight significantly differs from 6.6 lbs.
Sample mean and standard deviation calculations:
[tex]M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{7}(9+7.3+6+. . .+6.6)\\\\\\M=\dfrac{52.9}{7}\\\\\\M=7.56\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{6}((9-7.56)^2+(7.3-7.56)^2+(6-7.56)^2+. . . +(6.6-7.56)^2)}\\\\\\s=\sqrt{\dfrac{8.32}{6}}\\\\\\s=\sqrt{1.39}=1.18\\\\\\[/tex]
