Question

The new director of a local YMCA has been told by his predecessors that the average member has belonged for 8.7 years. Examining a random sample of 15 membership files, he finds the mean length of membership to be 10.2 years, with a standard deviation of 2.5 years. Assuming the population is approximately normally distributed, and using the 0.05 level, does this result suggest that the actual mean length of membership may be some value other than 8.7 years?

Please show every step explaining how to find the P value.

Answer 1

The new director of a local YMCA has been told by his predecessors that the average member has belonged for 8.7 years.

Here we want to test does sample result suggest that the actual mean length of membership may be some value other than 8.7 years?

Here we need to use one sample t test.

From this the null hypothesis (H₀) and the alternative hypothesis (H_a ) are as follows:

The formula of t test statistic is as follows:

where = sample mean =10.2

s = sample standard deviation = 2.5

n = sample size = 1

So the test statistic value = 2.32

degrees of freedom = n - 1 = 15 - 1 = 14

p-value =2* P( t > 2.32)

using excel:

p-value = "=TDIST(2.32,14,2)" = 0.0360

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.0360 < 0.05 so we used first rule.

That is we reject null hypothesis

Conclusion: At 5% level of significance there are sufficient evidence to conclude that the actual mean length of membership may be some value other than 8.7 years.