Question

Example 19.8. The frequency of chirping of a cricket is thought to be related to temperature. This suggests the possibility that temperature can be estimated from the chirp frequency. Let the following data on the number chirps per second, r by the striped ground cricket and the temperature, y in Fahrenheit is shown below: 20 1620 18 17 16 15 17 15 16 8972 93 8481 75 7082 69 83 Find the normal regression line that approximates the regression of tempera- ture on the number chirps per second by the striped ground cricket. Further test the hypothesis Ho : 8 4 versus Ha B#4 at the significance level 0.1.

Answer 1

Minitab - Untitled - [Session] File Edit Data Calc Stat Graph Editor Tools Window Help Assistant || 日 曼|晶噛范ㄧ勺 O @ E |幻喦倫 t Regression Analysis: y versus x Model Summary S R-sq R-sq (adj) R-sq (pred) 75.92% 3.41809 84.15% 82. 17% Coefficients Coef SE Coef T-Value P-Value VIF 10.7 Term Constant 10.7 0.624 0.346 0.000 1.00 1.00 6.52 4.067 Regression Equation y = 10.7 + 4.067 x To test H0: beta = 4 MTB > let k1 (4.067-4)/0·624 MTB > print kl Data Display K1 MTB CDF K1; value of test statistic 0.107372 SUBC T 8 Cumulative Distribution Function Student's t distribution with 8 DF 0.541431 0.107372 MTB > let k2 = 2*(1-0.541431) MTB > print k2 Data Display K20.917138 P-value Current Worksheet: Worksheet 1 O Type here to search

The estimated regression equation is:

10.7 + 4.0672. y

To test
H0 : β = 4
against $H_a:\beta \neq 4$

The test statistic can be written as

which under H₀ follows a t distribution with n-2 df.

β-4 SE(B)

We reject H₀ at 0.10 level of significance if P-value < 0.10

Now,

3 = 4.067, SE( 3) = 0.624. n = 10

The value of the test statistic

tots = 0.107372

and P-value =

P(t-2 >tobsPlts > 0.107372) 2 Pts > 0.107372) 2(1 0.541431)0.917138

Since p-value > 0.10, so we fail to reject H₀ at 0.10 level of significance and we can conclude that the slope coefficient $\beta$ is not significantly different from 4.