Question

A study reported that the average age of menarche of girls entering the first year class of a small American private college in the year 2008 was 11.89. Suppose you want to test whether the age of menarche of the girls in your class is different. After randomly selecting 30 girls and recording their menarche age, we find that this sample of girls were at the average age of 12.98 years old, with a 95% confidence interval from 11.59 year-old to 13.31 year-old. Based on this interval, should you reject or fail to reject the hypothesis that the menarche age of girls in you class is 11.89 on average

Answer 1

Answer:

Null hypothesis: $\mu = 11.89$

Alternative hypothesis: $\mu \neq 11.89$

And we want to test this with a confidence interval:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

After conduct the interval of 95% for the true mean we got: (11.59, 13.31)

And for this case the confidence interval contains the value of 11.89, so then we have enough evidence to FAIL to reject the null hypothesis at 5% of significance that the true mean is 11.89.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=12.98$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n=30 represent the sample size  

Solution to the problem

For this case we want to test the following hypothesis:

Null hypothesis: $\mu = 11.89$

Alternative hypothesis: $\mu \neq 11.89$

And we want to test this with a confidence interval:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

After conduct the interval of 95% for the true mean we got: (11.59, 13.31)

And for this case the confidence interval contains the value of 11.89, so then we have enough evidence to FAIL to reject the null hypothesis at 5% of significance that the true mean is 11.89.