Question

The earth's temperature (which affects seed germination, crop survival in bad weather, and many other aspects of agricultural production) can be measured using either ground-based sensors or infrared-sensing devices mounted in aircraft or space satellites. Ground-based sensoring is tedious, requiring many replications to obtain an accurate estimate of ground temperature. On the other hand, airplane or satellite sensoring of infrared waves appears to introduce a bias in the temperature readings. To determine the bias, readings were obtained at five different locations using both ground- and air-based temperature sensors. The readings (in degrees Celsius) are listed here. Location Ground Air 1 46.8 47.2 2 45.3 48.0 3 36.5 37.8 4 31.1 32.9 5 24.5 26.0

(a) Do the data present sufficient evidence to indicate a bias in the air-based temperature readings? Explain. (Use α = 0.05. Use μground − μair = μd. Round your answers to three decimal places.)

test statistic=

t > =

t < =

Estimate the difference in mean temperatures between ground- and air-based sensors using a 95% confidence interval. (Round your answers to three decimal places.)

How many paired observations are required to estimate the difference between mean temperatures for ground- versus air-based sensors correct to within 0.2°C, with probability approximately equal to 0.95? (Round your answer up to the nearest whole number.)

Answer 1

Part a)

To Test :-

H0 :-  μd = 0

H1 :-  μd ≠ 0

Test Statistic :-
S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √( 2.772 / 4) = 0.8325
d̅ = Σdi/n = -7.7 / 5 = -1.54
t = d̅ / ( S(d) / √(n) )
t = -1.54 / ( 0.8325 / √(5) )
t = -4.1366

Test Criteria :-
Reject null hypothesis if | t | > t(α/2)
Critical value t(α/2) = t( 0.05 /2 ) = 2.77645
| t | > t(α/2) = 4.13656 > 2.77645
Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 4.1366 ) = 0.01442
Reject null hypothesis if P value < α level of significance
P - value = 0.01442 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that there is bias in the air-based temperature readings.

Part b)

Confidence Interval :-
d̅ ± t(α/2, n-1) Sd / √(n)
t(α/2) = t(0.05 /2) = 2.77645
-1.54 ± t(0.05/2) * 0.8325/√(5)
Lower Limit = -1.54 - t(0.05/2) 0.8325/√(5)
Lower Limit = -2.573
Upper Limit = -1.54 + t(0.05/2}) 0.8325/√(5)
Upper Limit = -0.507
95% Confidence interval is ( -2.573 , -0.507 )

Sample size can be calculated by below formula
n = (( Z(α/2) * S(d) * d̅ ) / e )2
n = (( Z(0.05/2) * 0.8325 * (-1.54) ) / 0.2 )2
Critical value Z(α/2) = Z(0.05/2) = 1.96
n = 158
Required sample size at 95% confident is 158.