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?

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:

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:

Then, we can calculate the t-statistic as:

The degrees of freedom for this sample size are:

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: