Question

3. Dr. Fribance teaches MCSI classes at CCU. One of her classes in particular involves collecting data on the boat. It is well known that increasing temperatures allow for increased conductivity in water. It is important to study this in different settings and types of water (ex. Hayashi, 2004). Using the student collected data consisting of 55 observations, what kind of conclusions can we draw on the effects of temperature on water conductivity? Mean Std. Dev. X (Temperature in Celsius) 21.87 0.3249 Y (Conductivity in s/ 49.48 0.4188 SXY = 7.24 and Σ = 0.26 (a) Obtain a point estimate for the variance of the residuals, σ2. (b) Obtain the LSR line (c) We wish to estimate the average conductivity when the water is 18 °C. What type of interval would be most appropriate (confidence or prediction) Explain. (d) Construct and interpret the indicated interval with 95% confidence. e) Suppose we wished estimate the average conductivity when the water is 22°C Without constructing the interval, explain why it would be wider or narrower than the inteva in d (f) Suppose we wished to estimate the conductivity in a new water source that is 18°C. Would the interval be wider or narrower than the interval in (d. Explain without constructing the interval

Answer 1

Solution

Given

x = temperature (in celsius), y = conductivity (in s/m) ……………………………………………… (1)

n = 55 …………………………………………………………………………………….....……………. (2)

xbar = 21.87, ybar = 48.48 . ……………………………………………………………………………. (3)

S_xy = 7.24 ………………………………………………………………………………....……………… (4)

Now, given

s_x = 0.3249, S_xx = n.s_x² = 5.8415 ………………………………………………………..……………. (5)

s_y = 0.4188, S_yy = n.s_y² = 9.6466 ……………………………………………………..………………. (6)

Part (a)

Point estimate of residual variance, σ²_cap = {S_yy – (S_xy²/S_xx)}/(n – 2)

= 0.0127 [vide (6), (4) and (5)] Answer 1

Part (b)

LSR line is: y_cap = β_0cap + β_1capx, ……………………………………………………….…………….(7)

where β_1cap = Sxy/Sxx and β_0cap = ybar – β_1cap.xbar..………………………………………….…..(8)

So, vide (4), (5), and (3),

LSR line is: y_cap = 23.3742 + 1.2394x Answer 2 …………………………………………………. (9)

Part (c)

Since the average conductivity is required,

the appropriate interval is the confidence interval for y_cap   Answer 3

Part (d)

100(1 - α)% Confidence Interval (CI) for y_cap at x = x₀ is: y_cap at x = x₀ ± t_{n – 2,α/2}xs√[(1/n) + {(x₀ – xbar)²/Sxx}]

So, 95% CI for y_cap at x = 18 is:

[44.3202, 45.0468] Answer 4

Part (e)

The width of the CI is determined by t_{n – 2,α/2}xs√[(1/n) + {(x₀ – xbar)²/Sxx}] and hence all other quantities remaining the same, as x₀ increases/decreases, the width will increase/decrease.

Thus, the width at x = 22 will be wider than the width at x = 18. Answer 5

Part (f)

If the new water source is similar to the water sources used in the study to determine the above regression line, the CI width would not change since x₀ continues to be the same. Answer 6

DONE