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.
If Alpha = 0.05,
1. What is the critical t-value?
2. What is the decision for a statistically significant change in average weights at birth at the 5% level of significance?

Answer 1

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.

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:

`H_0: \mu=6.6\\\\H_a:\mu\\eq 6.6`

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:

`s_M=(s)/(√(n))=(1.18)/(√(7))=0.446`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(7.56-6.6)/(0.446)=(0.96)/(0.446)=2.152`

The degrees of freedom for this sample size are:

`df=n-1=7-1=6`

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:

`M=(1)/(n)\sum_(i=1)^n\,x_i\\\\\\M=(1)/(7)(9+7.3+6+. . .+6.6)\\\\\\M=(52.9)/(7)\\\\\\M=7.56\\\\\\s=\sqrt{(1)/(n-1)\sum_(i=1)^n\,(x_i-M)^2}\\\\\\s=\sqrt{(1)/(6)((9-7.56)^2+(7.3-7.56)^2+(6-7.56)^2+. . . +(6.6-7.56)^2)}\\\\\\s=\sqrt{(8.32)/(6)}\\\\\\s=√(1.39)=1.18\\\\\\`