Question

Suppose we have two thermometers. One thermometer is very precise but is delicate and heavy (X). We have another thermometer that is much cheaper and lighter, but of unknown precision (Y). We would like to know if we can (reliably) bring the lighter thermometer with us into the field. So, we set up an experiment where we expose both thermometers to 31 different temperatures and measure the temperature with each. We get the following observations:
x = 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120
y = 0.02, 3.99, 7.91, 12.03, 16.09, 20.00, 23.98, 28.09, 31.94, 36.03, 40.00, 44.05, 47.95, 52.00, 55.87, 59.90, 63.91, 67.95, 72.11, 76.02, 80.01, 84.10, 88.06, 91.74, 96.02, 99.95, 103.87, 108.01, 111.99, 116.04, 120.03
We want to test if these thermometers seem to be measuring the same temperatures. Let's use the threshold.
(a) Write down the appropriate hypothesis tests for β1.
H0: β1(≠ / = / > / <) 1 and Ha: β1( ≤ / = / > / ≠ / ≥ / <) 1.
(b) The test statistic is____.
(c) The p-value is____.
(d) Therefore, we can conclude that:_____.
1) The data provides evidence at the 0.1 significance level that these thermometers are not consistent.
2) The data provides no evidence at the 0.1 significance level that these thermometers are not consistent.
3) The probability that the null hypothesis is true is equal to the p-value.
4) The probability that we have made a mistake is equal to 0.1.

Answer 1

Answer:

a)H0: β1= 1 and Ha: β1 ≠1

b) The test statistic is____.t= b- β/ sb

c) The p-value is____.0.999144

(d) Therefore, we can conclude that:_____. there is not enough evidence to reject the null hypothesis.    

2) The data provides no evidence at the 0.1 significance level that these thermometers are not consistent.

Step-by-step explanation:

The null and alternate hypotheses are

H0: β1= 1 and Ha: β1 ≠1

The significance level is  alpha= 0.1 but ∝/2 =0.1/2= 0.05 for two tailed test.

The critical region at t ∝/2 (29)=  t ≤ - 1.699 , t ≥ 1.699

The test statistic is

t= b- β/ sb

which has t distribution with υ= 31-2= 29 degrees of freedom.

Calculations

Ŷ = a +bX

b = SPxy /SS x= Σ(xi-x`)(yi-y`)/Σ(xi-x`)²

b = 39671.6/39680= 0.9998

a = y` - bx`

x`= 60

y`= 59.989

a = 59.989 -0.9998*60 = 0.001734

Syx²= Σ( yi -y`)²/ n-2= 39663.3857/ 29=1367.68965

Syx= 36.9822

Sb= Syx/ √∑  (x-x`)²

Sb= 36.9823/ √39680

Sb= 36.9823/ 199.984

Sb= 0.18493

x-x`                 y-y`                    (x-x`)²     (x-x`)(y-y`)

-60                 -59.969            3600              3598.1419

-56                  -55.999           3136          3135.9458

-52                 -52.079             2704          2708.1097

-48                -47.959              2304           2302.0335

-44                 -43.899              1936          1931.5574

-40                -39.989               1600          1599.5613

-36              -36.009                1296            1296.3252

-32               -31.899              1024              1020.769

-28               -28.049             784                785.3729

-24                -23.959             576               575.0168

-20               -19.989             400                  399.7806

-16                 -15.939            256                  255.0245

-12               -12.039             144                    144.4684

-8                  -7.989             64                       63.9123  

-4                 -4.119                16                      16.4761

0                 -0.08903           0                         0

4                  3.921              16                      15.6839

8                7.961                64                       63.6877

12                12.121             144                     145.4516

16                16.031             256                   256.4955

20               20.021           400                  400.4194

24               24.111              576                   578.6632

28                28.071           784                     785.9871

32                 31.751            1024                    1016.031

36               36.031           1296                    1297.1148

40                39.961          1600                    1598.4387

44                43.881           1936                     1930.7626

48               48.021           2304                     2305.0065

52              52.001             2704                     2704.0503

56             56.051              3136                       3138.8542

60            60.041                3600                     3602.4581

∑0          0                 39680 (SSx)     39671.6 (SPxy)

Putting the values

The test statistic is

t= b- β/ sb

t= 0.9998-1/ 0.18493      ( β=1 given)

t=-0.0010815

Since the calculated t=-0.0010815  does not lie in the critical region  t ∝/2 (29)=  t ≤ - 1.699 , t ≥ 1.699 we conclude that these thermometers seem to be measure temperatures.

The p-value is  ≈ 0.999144

there is not enough evidence to reject the null hypothesis.