Construct a 98% confidence interval to determine whether there is any difference between the mean drug concentration in tablets produced at the two sites. Now analyze the data using a statistical test of hypothesis that the two means are different. Do the inferences drawn from the test of hypothesis and confidence interval agree

Step-by-step explanation:

So, we are given two sites;

Site one: 91.28 , 92.83, 89.35, 91.90 82.85, 94.83, 89.83, 89.00, 84.62, 86.96, 88.32, 91.17, 83.86, 89.74, 92.24, 92.59, 84.21, 89.36, 90.96, 92.85, 89.39, 89.82, 89.91, 92.16, 88.67.

=> The mean = add up all the numbers/ the total numbers (which is 25). = 89.55.

=> The standard deviation = 3.005

SITE TWO: 89.35 86.51 89.04 91.82 93.02 88.32 88.76 89.26 90.36 87.16 91.74 86.12 92.10 83.33 87.61 88.20 92.78 86.35 93.84 91.20 93.44 86.77 83.77 93.19 81.79.

=> The mean= add up all the numbers/ the total numbers (which is 25). = 89.033.

=> Standard deviation = 3.271.

NB: the assumption here is that the variance for both sites are the same.

STEP ONE: DETERMINE THE DEGREE OF FREEDOM AND THE CRITICAL VALUE.

The degree of freedom =( mean of site 1 ) + (mean of site 2) - 2 = (25 + 25) − 2 = 48.

Since the degree of freedom is 48, Recall that the value for alpha given = 0.02. Then, the critical value = 2.407.

STEP TWO: DETERMINE THE STANDARD DEVIATION AND THE STANDARD ERROR FOR BOTH.

Therefore, the standard deviation = √[( 25 - 1) × 3.005^2 + (25 -1) × 3.271^2) ÷ (25 +25 - 2) ] = 3.14.

Also, the standard error = standard deviation × ( √(1/ mean 1 + 1/mean 2) ..

Standard error = 3.14 × [ ✓( 1/25 + 1/25)] = 0.89.

STEP THREE: DETERMINE THE CONFIDENCE INTERVAL.

confidence interval = (89.548 - 89.033 - 2.41 × 0.89, 89.548 - 89.033 + 2.41 × 0.89) = (- 1.62, 2.65).

STEP FOUR: THE HYPOTHESES.

=> THE TWO TAILED TEST.

Jo:μ1 = μ2.

Ja: μ1 ≠ μ2.

Thus, the rejected region is {t : ∣ t ∣ > 2.7}

As critical value = 2.7 for the degree of freedom which is 48 and α=0.01.

If we do the t - test for this,.we will have;

(89.548 - 89.003)÷ ( standard deviation × √ ( 1/25 + 1/25). = 0.58.

So,from our t - value, is less or equals to the critical value, therefore, the null hypothesis is not rejected.

Also, from the p-value table which gives p=0.565. The p- value is greater than or equal to the alpha value(0.01). therefore, the null hypothesis is not rejected.

We are not going to reject the null hypothesis because the information gathered is not enough.

its not b

step-by-step explanation: