Question

1 - Menstrual cycle length. Problem 3 calculated the mean length of menstrual cycles in an SRS of 9 women. The data revealed days with standard deviation s = 2.906 days.

a - Calculate a 95% confidence interval for the mean menstrual cycle length.

b - Based on the confidence interval you just calculated, is the mean menstrual cycle length significantly different from 28.5 days at α = 0.05 (two sided)? Is it significantly different from μ = 30 days at the same α-level? Explain your reasoning. (Section 10.4 in your text considered the relationship between confidence intervals and significance tests. The same rules apply here.)

Answer 1

Solution :

Given that,

= 28.5

s = 2.906

n = 9

a ) Degrees of freedom = df = n - 1 = 9 - 1 = 8

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,8 =2.306

Margin of error = E = t/2,df * (s /n)

= 2.306* (0.9 / 25)

= 2.234

Margin of error = 2.234

The 95% confidence interval estimate of the population mean is,

- E < < + E

28.5 - 2.234 < < 28.5 + 2.234

26.266 < < 30.734

(26.266, 30.734)

b ) This is the two tailed test .

The null and alternative hypothesis is ,

H0 :    = 30

Ha :     30

Test statistic = t

= ( - ) / S / n

= (28.5 -30) / 2.906 / 9

= −1.549

Test statistic = t =−1.549  

P-value =0.1601

= 0.05  

P-value ≥

0.1601 ≥ 0.05

Reject the null hypothesis .

There is sufficient evidence to suggest that