Question

A study published in 1993 found that babies born at different times of the year may develop the ability to crawl at different ages! The author of the study suggested that these differences may be related to the temperature at the time the infant is 6 months old. (Benson and Janette, Infant Behavior and Development [1993]. The study found that 32 babies born in January crawled at an average age of 29.84 weeks, with a standard deviation of 7.08 weeks. Among 21 July babies, crawling ages averaged 33.64 weeks, with a standard deviation of 6.91 weeks. Is this difference significant?

Answer 1

`t=\frac{(29.84-33.64)-0}{\sqrt{(7.08^2)/(32)+(6.91^2)/(21)}}}=-1.939`

`p_v =2*P(t_(51)<-1.939)=0.0580`

Comparing the p value with the significance assumed
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different at 1% of significance.

Explanation:

Data given

`\bar X_(1)=29.84` represent the mean for sample January

`\bar X_(2)=33.64` represent the mean for sample July

`s_(1)=7.08` represent the sample standard deviation for 1

`s_(2)=6.91` represent the sample standard deviation for 2

`n_(1)=32` sample size for the group 2

`n_(2)=21` sample size for the group 2

`\alpha=0.01` Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:

Null hypothesis:
`\mu_(1)-\mu_(2)=0`

Alternative hypothesis:
`\mu_(1) - \mu_(2)\\eq 0`

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

`t=\frac{(\bar X_(1)-\bar X_(2))-\Delta}{\sqrt{(\sigma^2_(1))/(n_(1))+(\sigma^2_(2))/(n_(2))}}` (1)

And the degrees of freedom are given by
`df=n_1 +n_2 -2=32+21-2=51`

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

`t=\frac{(29.84-33.64)-0}{\sqrt{(7.08^2)/(32)+(6.91^2)/(21)}}}=-1.939`

P value

Since is a bilateral test the p value would be:

`p_v =2*P(t_(51)<-1.939)=0.0580`

Comparing the p value with the significance assumed
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different at 1% of significance.