Question

A group of n = 25 students was selected at random for studies related to the amount of time they spent for exam preparations. Educators assume that individual records are normally distributed with entirely unknown parameters (μ, σ). Sample summaries (in minutes per week) were obtained as

Sample Mean =X =90 and Sample SD =s=20 For hypothesis testing, set significance level = α = 0.01̄

1. To test whether μ < 100, what critical value (or values) are you going to use?

2. Formulate rejection rule that explains whether you are rejecting the null hypothesis or not

3. Evaluate the test statistic

4. State your decision in the form:
   Yes, we have enough evidence to reject the null hypothesis or No, we do not have evidence to reject the null hypothesis

Answer 1

Null hypothesis : Ho : =100

Alternative hypothesis : H1: < 100

Left tailed test;

Sample size : n =25

Degrees of freedom = n-1 =25-1=24

Significance level =0.01

For left tailed test,

For left tailed test :

Critical value is : -t = -t0.01

From standard t-tables , t0.01 for 24 degrees of freedom = 2.492

Critical value are going to use  = -2.492

Rejection Rule

For left tailed test : Rejection Rule : If value of the test statistic is less then critical value ; reject the null hypothesis i.e

if value of the test statistic < -2.492; Reject the null hypothesis ; Otherwise fail to reject the null hypothesis.

Test Statistic :

Hypothesized mean : = 100

Sample mean : =90

Sample standard deviation : s=20

As Calculated Value of t is less than Critical Value i.e. ( -2.5<-2.4922 ); Reject Null Hypothesis

State your decision in the form:
Yes, we have enough evidence to reject the null hypothesis .

Note : If the critical value is round to one decimal, then critical value is equal to the test-statistic, in that case you may fail to reject the null hypothesis.