Question

An air conditioning company servicing a certain machine room guarantees that the temperature in the room stays below 20o C. Due to malfunctioning of the equipment operating in the machine room, it is suspected that the average temperature actually exceeds 20o for more or less extended periods of time. Seven temperature measurements are collected throughout the day and the following temperatures are observed, 20.8, 20.2, 20.9, 21.5, 22.2, 21.2, 19.8. Assuming that the temperature is at least approximately normally distributed,

a) can you say at a 5% significance level that the room temperature is actually above 20o C?

b) what is the (approximate) p-value for this test? What does it tell you about the conclusion you have just made regarding the room temperature?

c) What is the probability that this test fails to reject Ho even though the true mean temperature is 21o C?

Answer 1

= 20.943

s = 0.8

n = 7

b) H₀: < 20

    H₁: > 20

The test statistic t = ()/(s/)

                             = (20.943 - 20)/(0.8/)

                             = 3.12

At 5% significance level, the critical value is t^* = 1.943

Since the test statistic value is greater than the critical value (3.12 > 1.943), so we should reject the null hypothesis.

So at 5% significance level, there is sufficient evidence to support the claim that the room temprature is actually above 20c.

P-value = P(T > 3.12)

             = 1 - P(T < 3.12)

            = 1 - 0.9897

            = 0.0103

Since the P-value is less than the significance level(0.0103 < 0.05), so we should reject H₀.

So at 5% significance level, there is sufficient evidence to support the claim that the room temprature is actually above 20c.

c) t_crit = 1.943

or, ()/(s/) = 1.943

or, ( - 20)/(0.8/) = 1.943

or, = 1.943 * (0.8/) + 20

or, = 20.6

= P( < 20.6)

    = P(()/(s/) < (20.6 - 21)/(0.8/))

    = P(T < -1.32)

     = 0.1175