Question

Question 2: 10 points An Article in Food Testing and Analysis, "Improving Reproducibility of Refractometry Measurements of Fruit Juices" measured the sugar concentration (Brix) in clear apple juice. All readings were taken at 20°C: 10.07 8.37 10.79 10.74 10.38 10.24 11.30 11.75 9.98 10.36 9.56 11.45 10.33 9.55 10.15 11.70 9.16 10.04 10.46 9.96 1. Test the hypothesis Ho: p=10.5 versus He: x10.5, using a=0.01. Find the P-value 2. Compute the power of the test if the true mean is 10.4 3. What sample size would be required to detect a true mean sugar concentration of 10.45 if we want the power of the test to be at least 0.9 4. Explain how the question in part (1) could be answered by constructing a two-sided confidence interval on the mean sugar concentration 5. Is there an evidence to support the assumption that the sugar concentration is normally distributed? Sketch the required plot and comment on it.

Answer 1

To test

a)

H0 : $\mu$ = 10.5 vs    H1 : $\mu$ $\neq$ 10.5

All given readings are as follow

     10.07 ,8.37,10.79,10.74,10.38,10.24,11.30,11.75 , 9.98,10.36,9.56,11.45,10.33,9.55,10.15,11.70,

     9.16,10.04,10.46,9.96

Here n = 20 ( 20 readings )

Now we will use t-test to test weather $\mu$ = 10.5 or $\mu$ $\neq$ 10.5

Test Statistics :

                          T.S = $\frac{\bar{x}-\mu }{s/\sqrt{n}}$

where $\bar{x}$ = mean of 20 readings =          ( mean of sample )

Σ τi/η

and   s = $\sqrt{\frac{\sum (xi-\bar{x})^{2}}{n-1}}$       ( standard deviation of sample )

Rejection criteria -

If Calculated |T.S| > $t_{\alpha /2,n-1}$ for $\alpha$ = 0.01 and n-1 = 19 , or if calculated P-value is less than 0.01 we reject null hypothesis at 1% of level of significane .

Calculation -

$\bar{x}$    =
Σ τi/η
    = 206.34 / 20 = 10.317

s = $\sqrt{\frac{\sum (xi-\bar{x})^{2}}{n-1}}$    

Now ${\frac{\sum (xi-\bar{x})^{2}}{n-1}}$ = 13.34862 / ( 20-1 ) = 0.7025589

Thus , s = $\sqrt{\frac{\sum (xi-\bar{x})^{2}}{n-1}}$      = = 0.8381879

0.7025589

hence   $\bar{x}$ = 10.317   and   s = 0.8381879

Thus

     T.S = $\frac{\bar{x}-\mu }{s/\sqrt{n}}$ = $\frac{10.317-10.5 }{0.8381879/\sqrt{20}}$ = -0.97639

Calculated Test statistics is T.S = -0.97639

t-table value is $t_{\alpha /2,n-1}$ at $\alpha$ = 0.01 and n-1 = 19 degree of freedom

                         $t_{\alpha /2,n-1}$ = 2.86

Thus | -0.97639 | < 2.86

|T.S| < $t_{\alpha /2,n-1}$

We do not reect null hypothesis at 5% of level of significance .

Calculating P-value for two sided test .

P-value = P ( T.S > $t_{\alpha /2,n-1}$ ) + P ( -T.S < $t_{\alpha /2,n-1}$ )

              = 0.17057 + 0.17057 =  0.34114

hence P-value = 0.34114 > 0.01 hence we do not reject null hypothesis at 1% of given level of significance .

Conclusion - At $\alpha$ = 0.01 we do not have evidence to reject null hypothesis . hence $\mu$ = 10.5 might be true .

b)

To Compute power of test .

It is given by

Power = 1 - $\beta$

where $\beta$ = P(Type II error ) = P(Accept H0 | H1)

Given true mean is 10.4

Thus under H1 $\mu$ = 10.4

Thus diference is d = 10.5 - 10.4 = 0.1

Thus d =0.1

Now standard Error S.E is

S.E = $\hat{\sigma}} /\sqrt{n}$ = $s /\sqrt{n}$ = 0.8381879 / $\sqrt{20}$ = 0.04411515

Thus   S.E = 0.1874245

t-table value with n-1= 19 degree of freedom and 1% of l.o.s is

$t_{\alpha /2,n-1}$ = 2.86

Hence Error is

Error = $t_{\alpha /2,n-1}$ * S.E

Error = 2.86 *   0.1874245

         = 0.5360341

hence left critical region is

Left = 10.5 - S.E = 10.5 - 0.5360341 = 9.963966

And right critical region is

Right = 10.5 + S.E = 10.5 - 0.5360341 = 11.03603

Next we find the t-scores for the left and right values assuming that the true mean is 10.4:

        TS_left = (10.4 - left)/S.E = (10.4 -9.963966 )/0.1874245 = 2.326452

And TS_right = = (right -10.4)/S.E = ( 11.03603-10.4 )/0.1874245 = 3.393548

Thus type II error is given by

$\beta$ = P(Type II error )

    = P( TS_left < t-value < TS_right )

where t-vale = $t_{\alpha /2,n-1}$ at n-1 = 19 df

$\beta$ = P (t-value < TS_right) - P (t-value < TS_left)

     = P (t-value < 3.393548) - P (t-value < 2.326452)

     = 0.9984757 - 0.9843948

$\beta$ =    0.01408089

Hence Power of test is

Power = 1 - $\beta$ = 1 - 0.01408089

            = 0.9859191

c)

To find sample size

Given power of test is atleast 0.9

Thus $\beta$ = 1 - Power = 1 - 0.9 = 0.1

hence $\beta$ = 0.1

Let given level of significance $\alpha$ = 0.01

now $t_{\alpha /2,n-1}$ = 2.86   ,    and   $t_{\beta,n-1}$ = 1.32      at n-1 = 19 degree of freedom

Under null hypothesis   H0 : $\mu$ = 10.5

Given true mean 10.45

Thus d= 10.5 - 10.45 = 0.05

Formual

n = 2*s * ($t_{\alpha /2,n-1}$ + $t_{\beta,n-1}$ ) / d2

     = 2 * 0.8381879 * (2.86 + 1.32) / 0.052

     = 2802.9

d)

to calculate Confidence interval

Now margin of error M.E it given by

M.E = $t_{\alpha /2,n-1}$ * S.E

We have calculated S.E = $\hat{\sigma}} /\sqrt{n}$ = $s /\sqrt{n}$ = 0.8381879 / $\sqrt{20}$ = 0.04411515

Thus   S.E = 0.1874245

and t-table value with n-1= 19 degree of freedom and 1% of l.o.s is

$t_{\alpha /2,n-1}$ = 2.86

Thus   M.E = $t_{\alpha /2,n-1}$ * S.E

M.E = 2.86 *   0.1874245

       = 0.5360341

Thus 99% Confidence interval is given by

C.I = ( $\bar{x}$ - M.E , $\bar{x}$ + M.E )

       = ( 10.317 - 0.5360341 , 10.317+ 0.5360341 )

C.I = ( 9.780966 , 10.85303 )

So these is 100(1-$\alpha$)% or 99% confidence interval

Interpretation

from these confidence interval we can say that for any value outside these interval we will reject null hypothesis at 1 % of level of ignificance .

So question in part 1 can be explaines as any value(testing given mean ) outside these confidence inteval will result into rejection of null hypothesis .