Question

I have figured out the first part of this question in that there is no significant difference because 1.78<2.306. But this second part is confusing to me. Can you explain the formula I need to use? Thanks

1. .Menstrual cycle lengths (days) in an SRS of nine women are as follows: {31, 28, 26, 24, 29, 33, 25, 26, 28}. Use this data to test whether mean menstrual cycle length differs significantly from a lunar month using a one sample t-test. (A lunar month is 29.5 days.) Assume that population values vary according to a Normal distribution. Use a two-sided alternative. Show all hypothesis-testing steps.

Answer 1

The hypothesis test is performed in following steps,

Step 1:

The null and alternate hypotheses are,

Null hypothesis: Ho: There is no difference between sample and population mean,

Alternate hypothesis: Ha: There is significant difference between sample and population mean

Where, is sample mean and is the population mean which is lunar month (29.5 days)

This is a two-tailed test.

Step 2:

Level of significance,

The critical value for a two-tailed test statistic is obtained from t distribution table for significance level and degree of freedom, df = n -1 = 9 - 1 = 8.

Step 3:

The one sample t test is used to compare the sample mean to hypothetical population mean (Since the population standard deviation is not known and the sample size, n<30, t statistic is used). The t statistic is computed as follow,,

From the data points,

Mean	27.7778
SD	2.9059
n	9

Step 4:

Since, t-statistic is less than t critical value,  it can be concluded that the null hypothesis is not rejected.

Hence  there is not enough evidence to claim that the population mean is different than 29.5, at 5% significance level.